Matrices Exercise 3.1
Question1. In the matrix \(\left[\begin{array}{cccc}
2 & 5 & 19 & -7 \\
35 & -2 & \frac{5}{2} & 12 \\
\sqrt{3} & 1 & -5 & 17
\end{array}\right]\), Write
The order of the matrix (2) the number of elements (3) write the elements \(a_{13}, a_{21}, a_{33}, a_{24}, a_{23}\)
Solution:
1. In the given matrix, the number of rows is 3 and the number of columns is 4. Therefore, the order of the matrix is 3 x 4.
2. Since, the order of the matrix is 3 x 4, so there are 3 x 4 = 12 elements in it
3. Let \(\left[\begin{array}{cccc}
2 & 5 & 19 & -7 \\
35 & -2 & \frac{5}{2} & 12 \\
\sqrt{3} & 1 & -5 & 17
\end{array}\right]=\left[\begin{array}{llll}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34}
\end{array}\right]\)
On comparing the corresponding elements, we get \(a_{13}=19; a_{21}=35; a_{33}=-5, a_{24}=12; a_{23}=\frac{5}{2}\)
Question 2. If a matrix has 24 elements, what are the possible orders it can have? What, if it has 1 3 elements? Sol. We know that if a matrix is of the order m x n, it has mn elements. Hence, a matrix containing 24 elements can have any one of the following orders:
Solution:
1 x 24, 2 x 12, 3 x 8, 4 x 6, 6 x 4, 8 x 3, 12 x 2 or 24 x 1.
Similarly, a matrix containing 13 elements can have an order of 1 x 13 or 13 x 1.
Question 3. If a matrix has 1 8 elements, what are the possible orders it can have? What, if it has 5 elements?
Solution: A matrix containing 1 8 elements can have any one of the following orders:
1 X 18, 18 x 1,2 x 9,9 x 2,3 x 6,6 x 3
Similarly, a matrix containing 5 elements can have an order of 1 x 5 or 5 x 1.
Question 4. Construct 2 \(\times matrix, A=\left[a_{i j}\right]\), whose elements are given by:
- \(\mathrm{a}_{\mathrm{ij}}=\frac{(\mathrm{i}+\mathrm{j})^2}{2}\)
- \(\mathrm{a}_{\mathrm{ij}}=\frac{\mathrm{i}}{\mathrm{j}}\)
- \(\mathrm{a}_{\mathrm{ij}}=\frac{(\mathrm{i}+2 \mathrm{j})^2}{2}\)
Solution:
(1) The order of the given matrix is 2 \(\times\) 2, so A=\(\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]_{2 \times 2}\)
where \(a_{i j}=\frac{(i+j)^2}{2}\).
⇒ \(\mathrm{a}_{11}=\frac{(1+1)^2}{2}=2, \mathrm{a}_{12}=\frac{(1+2)^2}{2}=\frac{9}{2}, \mathrm{a}_{21}=\frac{(2+1)^2}{2}=\frac{9}{2}, \mathrm{a}_{22}=\frac{(2+2)^2}{2}=8\)
Read and Learn More Class 12 Maths Chapter Wise with Solutions
Hence, the required matrix is A=\(\left[\begin{array}{cc}2 & 9 / 2 \\ 9 / 2 & 8\end{array}\right]_{2 \times 2}\)
2. Here, A=\(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]_{2 \times 2}\) where \(a_{i j}=\frac{i}{j}\)
⇒ \(a_{11}=\frac{1}{1}=1, a_{12}=\frac{1}{2}, a_{21}=\frac{2}{1}=2, a_{22}=\frac{2}{2}\)=1
Hence, the required matrix is A=\(\left[\begin{array}{cc}
1 & 1 / 2 \\
2 & 1
\end{array}\right]_{2 \times 2}\)
3. Here A =\(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]_{2 \times 2}\) where \(a_{i j}=\frac{(i+2 j)^2}{2}\)
⇒ \(a_{11}=\frac{(1+2)^2}{2}=\frac{9}{2}, a_{12}=\frac{(1+4)^2}{2}=\frac{25}{2}, a_{21}=\frac{(2+2)^2}{2}=8, a_{22}=\frac{(2+4)^2}{2}=18\)
Hence The Required Matrix A=\(\left[\begin{array}{cc}
9 / 2 & 25 / 2 \\
8 & 18
\end{array}\right]_{2 \times 2}\)
Question 5. Construct a 3 x 4 matrix whose elements are given by:
- \(a_{i j}=\frac{1}{2}|-3 i+j|\)
- \(\mathrm{a}_{\mathrm{ij}}=2 \mathrm{i}-\mathrm{j}\)
Solution:
1. The order of the given matrix is 3 x 4, so the required matrix is
A=\(\left[\begin{array}{llll}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34}
\end{array}\right]_{3 \times 4}\) where \(a_{i j}=\frac{1}{2}|-3 i+j|\)
⇒ \(\mathrm{a}_{11}=\frac{1}{2}|-3+1|=1, \mathrm{a}_{12}=\frac{1}{2}|-3+2|=\frac{1}{2}, \mathrm{a}_{13}=\frac{1}{2}|-3+3|=0, \mathrm{a}_{14}=\frac{1}{2}|-3+4|=\frac{1}{2}\)
⇒ \(\mathrm{a}_{21}=\frac{1}{2}|-6+1|=\frac{5}{2}, \mathrm{a}_{22}=\frac{1}{2}|-6+2|=2, \mathrm{a}_{23}=\frac{1}{2}|-6+3|=\frac{3}{2}, \mathrm{a}_{24}=\frac{1}{2}|-6+4|=1 \)
⇒ \(\mathrm{a}_{31}=\frac{1}{2}|-9+1|=4, \mathrm{a}_{32}=\frac{1}{2}|-9+2|=\frac{7}{2}, \mathrm{a}_{33}=\frac{1}{2}|-9+3|=3 and \mathrm{a}_{34}=\frac{1}{2}|-9+4|=\frac{5}{2}\)
Here the Required Matrix Is A=\(\left[\begin{array}{cccc}
1 & 1 / 2 & 0 & 1 / 2 \\
5 / 2 & 2 & 3 / 2 & 1 \\
4 & 7 / 2 & 3 & 5 / 2
\end{array}\right]_{3 \times 4}\)
2. Here, \(\left[\begin{array}{llll}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34}
\end{array}\right]_{3 \times 4}\) Where, \(a_{i j}\)=2 i-j
⇒ \(a_{11}=2-1=1, a_{12}=2-2=0, a_{13}=2-3=-1, a_{14}=2-4\)=-2,
⇒ \(a_{21}=4-1=3, a_{22}=4-2=2, a_{23}=4-3=1, a_{24}=4-4\)=0,
⇒ \(a_{31}=6-1=5, a_{32}=6-2=4, a_{33}=6-3\)=3 and \(a_{34}=6-4\)=2
Here, The Required Matrix A=\(\left[\begin{array}{cccc}
1 & 0 & -1 & -2 \\
3 & 2 & 1 & 0 \\
5 & 4 & 3 & 2
\end{array}\right]_{3 \times 4}\)
Question 6. Find the values of x, y, and z from the following equations :
1. \(\left[\begin{array}{ll}
4 & 3 \\
\mathrm{x} & 5
\end{array}\right]=\left[\begin{array}{ll}
\mathrm{y} & \mathrm{z} \\
1 & 5
\end{array}\right]\)
2. \(\left[\begin{array}{cc}
x+y & 2 \\
5+z & x y
\end{array}\right]=\left[\begin{array}{ll}
6 & 2 \\
5 & 8
\end{array}\right]\)
3. \(\left[\begin{array}{c}
x+y+z \\
x+z \\
y+z
\end{array}\right]=\left[\begin{array}{l}
9 \\
5 \\
7
\end{array}\right]\)
Solution:
1. Given, \(\left[\begin{array}{ll}
4 & 3 \\
\mathrm{x} & 5
\end{array}\right]=\left[\begin{array}{ll}
\mathrm{y} & \mathrm{z} \\
1 & 5
\end{array}\right]\) By definition of equality of matrix as the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get 4- % 3- z and x = 1 ⇒ x = 1., y = 4 and z = 3
2. Given, \(\left[\begin{array}{cc}
x+y & 2 \\
5+z & x y
\end{array}\right]=\left[\begin{array}{ll}
6 & 2 \\
5 & 8
\end{array}\right]\) By definition of equality of matrix as the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements,
we get x + y- 6 → Equation 1
5 + 2=5 → Equation 2
xy- 8 → Equation 3
From Eq. (2), we get z = 0
From Eq, (1), y = 6- x Equation 4
Substituting this value of y in Eq. (3), we obtain
x(6- x) – 8 ⇒ x‘- 6x – 8 – 0 (x- 2)(x- 4) = 0 ⇒ x = 2 or x = 4
When x = 2, then from Eq, (4), y = 6- 2 = 4 and
when x = 4, then from Eq. (4), y = 6- 4 = 2.
So, either x = 2, y = 4 and z = 0 or x = 4, y = 2 and z = 0
3. \(\left[\begin{array}{c}
x+y+z \\
x+z \\
y+z
\end{array}\right]=\left[\begin{array}{l}
9 \\
5 \\
7
\end{array}\right]\) By definition of equality of matrix as the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements,
we get x + y + z = 9 → Equation 1
x + z = 5 → Equation 2
y + z = 7 →Equation3
Subtracting Eq. (2) from Eq. ( 1 ), we get y = 4
Subtracting Eq. (3) from Eq.(l), we get x = 2
Substituting y = 4 in Eq. (3), we get 4 +z = 7 ⇒ z = 7-4 = 3
Hence; x = 2, y = 4, z = 3
Question 7. Find the values of a, b. c and d from the equation \(\left[\begin{array}{cc}
a-b & 2 a+c \\
2 a-b & 3 c+d
\end{array}\right]=\left[\begin{array}{cc}
-1 & 5 \\
0 & 13
\end{array}\right]\)
Solution:
Given, \(\left[\begin{array}{cc}
a-b & 2 a+c \\
2 a-b & 3 c+d
\end{array}\right]=\left[\begin{array}{cc}
-1 & 5 \\
0 & 13
\end{array}\right]\)
By definition of equality of matrix, as the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get:
a – b = – 1 → Equation 1
2a- b = 0 → Equation 2
2a + c = 5 → Equation 3
and 3c + d= 13
Subtracting Eq. (1) from Eq.(2), we get a =1
Putting a = 1 in Eq. (1) and Eq.(3), we get 1 – b =- 1 and 2 c 5 => b 2 and c = 3
Substituting c = 3 in Eq. (4),
We obtain 3 x 3 + d = 13 => d- 13 – 9 = 4
Hence; a = 1 , b = 2, c = 3 and d = 4.
Question 8. A=\(\left[a_1\right]_{n \rightarrow n} \text { is a square matrix, if ? }\)
- m n
- m > n
- m = n
- None of these
Solution: 3.
Since, in a square matrix, the number of rows is equal to the number of columns; therefore, we must have m = n
Question 9. Which of the given values of x and y make the following pairs of matrices equal?
\(\left[\begin{array}{cc}
3 x+7 & 5 \\
y+1 & 2-3 x
\end{array}\right] \cdot\left[\begin{array}{cc}
0 & y-2 \\
8 & 4
\end{array}\right]\)
- x=\(\frac{-1}{3}\), y=7
- Not possible to find
- y=7, x=\(\frac{-2}{3}\)
- x=\(\frac{-1}{3}, y=\frac{-2}{3}\)
Solution: 2. Not possible to find
y=7, x=\(\frac{-2}{3}\)
x=\(\frac{-1}{3}, y=\frac{-2}{3}\)
According to the question, \(\left[\begin{array}{cc}
3 x+7 & 5 \\
y+1 & 2-3 x
\end{array}\right] \cdot\left[\begin{array}{cc}
0 & y-2 \\
8 & 4
\end{array}\right]\)
By definition of equality of matrices, we have
3x +7 = 0 → Equation 1
5 = y – 2 → Equation 2
y + 1 = 8 → Equation 3
2- 3x = 4 → Equation 4
From Eq. (2), y=7
From Eq. (1), 3 x+7=0 \(\Rightarrow x=\frac{-7}{3}\)
From Eq. (4), 2-3 x=4 \(\Rightarrow\) x=\(\frac{-2}{3}\)
‘x’ can have only one value at a time.
Hence, it is not possible to find the values of x and y for which the given matrices are equal.
Question 10. The number of all possible matrices of order 3 x 3 with each entry or 0 or 1 is
- 27
- 18
- 81
- 512
Solution: 4. 512
We know that a matrix having an order of 3 x 3 contains 9 elements.
Each element can be selected in 2 ways (it can be either 0 or 1).
Hence, All the nine entries can be chosen by 2° ways =512 ways.
Matrices Exercise 3.2
Question 1. Let A= \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\), B= \(\left[\begin{array}{cc}
1 & 3 \\
-2 & 5
\end{array}\right]\), C=\(\left[\begin{array}{cc}
-2 & 5 \\
3 & 4
\end{array}\right]\).Find Each Of The Following
- A + B
- A – B
- 3A – C
- AB
- BA
Solution:
1. A + B = \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]+\left[\begin{array}{cc}
1 & 3 \\
-2 & 5
\end{array}\right]=\left[\begin{array}{cc}
2+1 & 4+3 \\
3+(-2) & 2+5
\end{array}\right]=\left[\begin{array}{ll}
3 & 7 \\
1 & 7
\end{array}\right]\)
2. A – B =\(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]-\left[\begin{array}{cc}
1 & 3 \\
-2 & 5
\end{array}\right]=\left[\begin{array}{cc}
2-1 & 4-3 \\
3-(-2) & 2-5
\end{array}\right]=\left[\begin{array}{cc}
1 & 1 \\
5 & -3
\end{array}\right]\)
3. 3A – C = \(3\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]-\left[\begin{array}{cc}
-2 & 5 \\
3 & 4
\end{array}\right]=\left[\begin{array}{cc}
6 & 12 \\
9 & 6
\end{array}\right]-\left[\begin{array}{cc}
-2 & 5 \\
3 & 4
\end{array}\right]=\left[\begin{array}{ll}
8 & 7 \\
6 & 2
\end{array}\right]\)
4. AB= \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\left[\begin{array}{cc}
1 & 3 \\
-2 & 5
\end{array}\right]=\left[\begin{array}{ll}
2 \times 1+4 \times(-2) & 2 \times 3+4 \times 5 \\
3 \times 1+2 \times(-2) & 3 \times 3+2 \times 5
\end{array}\right]=\left[\begin{array}{cc}
-6 & 26 \\
-1 & 19
\end{array}\right]\)
5. BA =\(\left[\begin{array}{cc}
1 & 3 \\
-2 & 5
\end{array}\right]\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]=\left[\begin{array}{cc}
1 \times 2+3 \times 3 & 1 \times 4+3 \times 2 \\
(-2) \times 2+5 \times 3 & (-2) \times 4+5 \times 2
\end{array}\right]=\left[\begin{array}{cc}
11 & 10 \\
11 & 2
\end{array}\right]\)
Question 2. Compute the following:
(1) \(\left[\begin{array}{cc}a & b \\ -b &a\end{array}\right]+\left[\begin{array}{ll}a & b \\ b & a\end{array}\right]\)
(2) \(\left[\begin{array}{ll}a^2+b^2 & b^2+c^2 \\ a^2+c^2 & a^2+b^2\end{array}\right]+\left[\begin{array}{cc}2 a b & 2 b c \\ -2 a c & -2 a b\end{array}\right]\)
(3) \(\left[\begin{array}{ccc}-1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5\end{array}\right]+\left[\begin{array}{ccc}12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4\end{array}\right]\)
(4) \(\left[\begin{array}{ll}\cos ^2 x & \sin ^2 x \\ \sin ^2 x & \cos ^2 x\end{array}\right]+\left[\begin{array}{ll}\sin ^2 x & \cos ^2 x \\ \cos ^2 x & \sin ^2 x\end{array}\right]\)
Solution:
1. \(\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]+\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]=\left[\begin{array}{cc}
a+a & b+b \\
-b+b & a+a
\end{array}\right]=\left[\begin{array}{cc}
2 a & 2 b \\
0 & 2 a
\end{array}\right]\)
2. \(\left[\begin{array}{ll}
a^2+b^2 & b^2+c^2 \\
a^2+c^2 & a^2+b^2
\end{array}\right]+\left[\begin{array}{cc}
2 a b & 2 b c \\
-2 a c & -2 a b
\end{array}\right]=\left[\begin{array}{ll}
a^2+b^2+2 a b & b^2+c^2+2 b c \\
a^2+c^2-2 a c & a^2+b^2-2 a b
\end{array}\right] \)
= \(\left[\begin{array}{ll}
(a+b)^2 & (b+c)^2 \\
(a-c)^2 & (a-b)^2
\end{array}\right] \quad\left\{\begin{array}{ll}
& (a+b)^2=a^2+2 a b+b^2 \\
\text { and } & (a-b)^2=a^2-2 a b+b^2
\end{array}\right\}\)
3. \(\left[\begin{array}{ccc}
-1 & 4 & -6 \\
8 & 5 & 16 \\
2 & 8 & 5
\end{array}\right]+\left[\begin{array}{ccc}
12 & 7 & 6 \\
8 & 0 & 5 \\
3 & 2 & 4
\end{array}\right]=\left[\begin{array}{ccc}
-1+12 & 4+7 & -6+6 \\
8+8 & 5+0 & 16+5 \\
2+3 & 8+2 & 5+4
\end{array}\right]=\left[\begin{array}{ccc}
11 & 11 & 0 \\
16 & 5 & 21 \\
5 & 10 & 9
\end{array}\right]\)
4. \(\left[\begin{array}{ll}
\cos ^2 x & \sin ^2 x \\
\sin ^2 x & \cos ^2 x
\end{array}\right]+\left[\begin{array}{cc}
\sin ^2 x & \cos ^2 x \\
\cos ^2 x & \sin ^2 x
\end{array}\right]=\left[\begin{array}{ll}
\cos ^2 x+\sin ^2 x & \sin ^2 x+\cos ^2 x \\
\sin ^2 x+\cos ^2 x & \cos ^2 x+\sin ^2 x
\end{array}\right]=\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
Question 3. Compute identical products
1. \(\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]\left[\begin{array}{cc}
a & -b \\
b & a
\end{array}\right]\)
2. \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\left[\begin{array}{lll}
2 & 3 & 4
\end{array}\right]\)
3. \(\left[\begin{array}{cc}
1 & -2 \\
2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right]\)
4. \(\left[\begin{array}{lll}
2 & 3 & 4 \\
3 & 4 & 5 \\
4 & 5 & 6
\end{array}\right]\left[\begin{array}{ccc}
1 & -3 & 5 \\
0 & 2 & 4 \\
3 & 0 & 5
\end{array}\right]\)
5. \(\left[\begin{array}{cc}
2 & 1 \\
3 & 2 \\
-1 & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & 0 & 1 \\
-1 & 2 & 1
\end{array}\right]\)
6. \(\left[\begin{array}{ccc}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\left[\begin{array}{cc}
2 & -3 \\
1 & 0 \\
3 & 1
\end{array}\right]\)
Solution:
1. \(\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]\left[\begin{array}{cc}
a & -b \\
b & a
\end{array}\right]=\left[\begin{array}{cc}
a \times a+b \times b & a \times(-b)+b \times a \\
(-b) \times a+a \times b & (-b) \times(-b)+a \times a
\end{array}\right]=\left[\begin{array}{cc}
a^2+b^2 & 0 \\
0 & b^2+a^2
\end{array}\right]\)
2. \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\left[\begin{array}{lll}
2 & 3 & 4
\end{array}\right]=\left[\begin{array}{lll}
1 \times 2 & 1 \times 3 & 1 \times 4 \\
2 \times 2 & 2 \times 3 & 2 \times 4 \\
3 \times 2 & 3 \times 3 & 3 \times 4
\end{array}\right]=\left[\begin{array}{ccc}
2 & 3 & 4 \\
4 & 6 & 8 \\
6 & 9 & 12
\end{array}\right]\)
3. \(\left[\begin{array}{cc}
1 & -2 \\
2 & 3
\end{array}\right]\left[\begin{array}{ccc}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right]=\left[\begin{array}{ccc}
1 \times 1+(-2) \times 2 & 1 \times 2+(-2) \times 3 & 1 \times 3+(-2) \times 1 \\
2 \times 1+3 \times 2 & 2 \times 2+3 \times 3 & 2 \times 3+3 \times 1
\end{array}\right]\)
=\(\left[\begin{array}{ccc}
-3 & -4 & 1 \\
8 & 13 & 9
\end{array}\right]\)
4. \(\left[\begin{array}{lll}
2 & 3 & 4 \\
3 & 4 & 5 \\
4 & 5 & 6
\end{array}\right]\left[\begin{array}{ccc}
1 & -3 & 5 \\
0 & 2 & 4 \\
3 & 0 & 5
\end{array}\right]=\left[\begin{array}{ccc}
2+0+12 & -6+6+0 & 10+12+20 \\
3+0+15 & -9+8+0 & 15+16+25 \\
4+0+18 & -12+10+0 & 20+20+30
\end{array}\right]\)
=\(\left[\begin{array}{ccc}
14 & 0 & 42 \\
18 & -1 & 56 \\
22 & -2 & 70
\end{array}\right]\)
5. \(\left[\begin{array}{cc}
2 & 1 \\
3 & 2 \\
-1 & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & 0 & 1 \\
-1 & 2 & 1
\end{array}\right]=\left[\begin{array}{ccc}
2-1 & 0+2 & 2+1 \\
3-2 & 0+4 & 3+2 \\
-1-1 & 0+2 & -1+1
\end{array}\right]=\left[\begin{array}{ccc}
1 & 2 & 3 \\
1 & 4 & 5 \\
-2 & 2 & 0
\end{array}\right]\)
6. \(\left[\begin{array}{ccc}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\left[\begin{array}{cc}
2 & -3 \\
1 & 0 \\
3 & 1
\end{array}\right]=\left[\begin{array}{cc}
6-1+9 & -9+0+3 \\
-2+0+6 & 3+0+2
\end{array}\right]=\left[\begin{array}{cc}
14 & -6 \\
4 & 5
\end{array}\right]\)
Question 4. If A =\(\left[\begin{array}{ccc}
1 & 2 & -3 \\
5 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]+\left[\begin{array}{ccc}
3 & -1 & 2 \\
4 & 2 & 5 \\
2 & 0 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
3 & -1 & 2 \\
4 & 2 & 5 \\
2 & 0 & 3
\end{array}\right]\), C =\(\left[\begin{array}{ccc}
4 & 1 & 2 \\
0 & 3 & 2 \\
1 & -2 & 3
\end{array}\right]\), then compute (A + B) and (B – C).
Also, verify that A + (B- C) = (A + B)- C.
Solution:
A+B=\(\left[\begin{array}{ccc}
1 & 2 & -3 \\
5 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]+\left[\begin{array}{ccc}
3 & -1 & 2 \\
4 & 2 & 5 \\
2 & 0 & 3
\end{array}\right]=\left[\begin{array}{ccc}
4 & 1 & -1 \\
9 & 2 & 7 \\
3 & -1 & 4
\end{array}\right]\)
And B-c = \(\left[\begin{array}{ccc}
3 & -1 & 2 \\
4 & 2 & 5 \\
2 & 0 & 3
\end{array}\right]-\left[\begin{array}{ccc}
4 & 1 & 2 \\
0 & 3 & 2 \\
1 & -2 & 3
\end{array}\right]=\left[\begin{array}{ccc}
-1 & -2 & 0 \\
4 & -1 & 3 \\
1 & 2 & 0
\end{array}\right]\)
⇒ \(A+(B-C)=\left[\begin{array}{ccc}
1 & 2 & -3 \\
5 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]+\left[\begin{array}{ccc}
-1 & -2 & 0 \\
4 & -1 & 3 \\
1 & 2 & 0
\end{array}\right]=\left[\begin{array}{ccc}
0 & 0 & -3 \\
9 & -1 & 5 \\
2 & 1 & 1
\end{array}\right]\)
⇒ \((A+B)-C=\left[\begin{array}{ccc}
4 & 1 & -1 \\
9 & 2 & 7 \\
3 & -1 & 4
\end{array}\right]-\left[\begin{array}{ccc}
4 & 1 & 2 \\
0 & 3 & 2 \\
1 & -2 & 3
\end{array}\right]=\left[\begin{array}{ccc}
0 & 0 & -3 \\
9 & -1 & 5 \\
2 & 1 & 1
\end{array}\right]\)
Hence, A + (B- C) = (A + B) ~ C is verified.
Question 5. If A= \(=\left[\begin{array}{ccc}
2 / 3 & 1 & 5 / 3 \\
1 / 3 & 2 / 3 & 4 / 3 \\
7 / 3 & 2 & 2 / 3
\end{array}\right]\) and B=\(\left[\begin{array}{ccc}
2 / 5 & 3 / 5 & 1 \\
1 / 5 & 2 / 5 & 4 / 5 \\
7 / 5 & 6 / 5 & 2 / 5
\end{array}\right]\),Then Compute 3A-5B
Solution:
⇒ \(3 \mathrm{~A}-5 \mathrm{~B}=3\left[\begin{array}{ccc}
2 / 3 & 1 & 5 / 3 \\
1 / 3 & 2 / 3 & 4 / 3 \\
7 / 3 & 2 & 2 / 3
\end{array}\right]-5\left[\begin{array}{ccc}
2 / 5 & 3 / 5 & 1 \\
1 / 5 & 2 / 5 & 4 / 5 \\
7 / 5 & 6 / 5 & 2 / 5
\end{array}\right]\)
= \(\left[\begin{array}{lll}
2 & 3 & 5 \\
1 & 2 & 4 \\
7 & 6 & 2
\end{array}\right]-\left[\begin{array}{lll}
2 & 3 & 5 \\
1 & 2 & 4 \\
7 & 6 & 2
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)=0 (Zero matrix)
Question 6. Simplify \(\cos \theta\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]+\sin \theta\left[\begin{array}{cc}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right]\).
Solution:
⇒ \(\cos \theta\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]+\sin \theta\left[\begin{array}{cc}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\cos ^2 \theta & \cos \theta \sin \theta \\
-\cos \theta \sin \theta & \cos ^2 \theta
\end{array}\right]+\left[\begin{array}{cc}
\sin ^2 \theta & -\sin \theta \cos \theta \\
\sin \theta \cos \theta & \sin ^2 \theta
\end{array}\right] \)
= \(\left[\begin{array}{cc}
\cos ^2 \theta+\sin ^2 \theta & 0 \\
0 & \cos ^2 \theta+\sin ^2 \theta
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) (\(\cos ^2 \theta+\sin ^2 \theta\)=1)
Question 7. Find X and Y if:
X+Y=X+Y=\(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) and X-Y=\(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\)
\(\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right]\) and 3 X+2 Y=\(\left[\begin{array}{cc}
2 & -2 \\
-1 & 5
\end{array}\right]\)
Solution:
Given, X+Y=\(\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]\) → Equation 1
and X-Y=\(\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]\) → Equation 2
Adding equation (1) and (2), we get
1. 2 \(\mathrm{X}=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]+\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]=\left[\begin{array}{cc}
10 & 0 \\
2 & 8
\end{array}\right] \Rightarrow \mathrm{X}=\frac{1}{2}\left[\begin{array}{cc}
10 & 0 \\
2 & 8
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
1 & 4
\end{array}\right]\)
Subtracting equation (2) from equation (1), we get
2 \(\mathrm{Y}=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]-\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]=\left[\begin{array}{ll}
4 & 0 \\
2 & 2
\end{array}\right] \Rightarrow \mathrm{Y}=\frac{1}{2}\left[\begin{array}{ll}
4 & 0 \\
2 & 2
\end{array}\right]=\left[\begin{array}{ll}
2 & 0 \\
1 & 1
\end{array}\right]\)
2. Given, 2 \(\mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]\)
and 3 X+2 Y=\(\left[\begin{array}{cc}2 & -2 \\ -1 & 5\end{array}\right]\)
Multiplying equation (1) by 2, \mathrm{Eq}. (2) by 3 and then subtracting, we get
⇒ \( 2(2 \mathrm{X}+3 \mathrm{Y})-3(3 \mathrm{X}+2 \mathrm{Y})=2\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right]-3\left[\begin{array}{cc}
2 & -2 \\
-1 & 5
\end{array}\right]\)
⇒ \(\Rightarrow 4 \mathrm{X}+6 \mathrm{Y}-9 \mathrm{X}-6 \mathrm{Y}=\left[\begin{array}{cc}
4 & 6 \\
8 & 0
\end{array}\right]-\left[\begin{array}{cc}
6 & -6 \\
-3 & 15
\end{array}\right]\)
⇒ \(\Rightarrow-5 \mathrm{X}=\left[\begin{array}{cc}
4-6 & 6+6 \\
8+3 & 0-15
\end{array}\right]=\left[\begin{array}{cc}
-2 & 12 \\
11 & -15
\end{array}\right]\)
⇒ \(\Rightarrow \mathrm{X}=-\frac{1}{5}\left[\begin{array}{cc}
-2 & 12 \\
11 & -15
\end{array}\right]=\left[\begin{array}{cc}
2 / 5 & -12 / 5 \\
-11 / 5 & 3
\end{array}\right]\)
Then, from Eq. (1),
⇒ \(3 \mathrm{Y}=\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right]-2 \mathrm{X}\)
=\(\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right]-2\left[\begin{array}{cc}
2 / 5 & -12 / 5
-11 / 5 & 3
\end{array}\right]\)
=\(\left[\begin{array}{cc}
2-\frac{4}{5} & 3+\frac{24}{5}
4+\frac{22}{5} & 0-6
\end{array}\right]=\left[\begin{array}{cc}
6 / 5 & 39 / 5
42 / 5 & -6
\end{array}\right]\)
⇒ \(\mathrm{Y}=\frac{1}{3}\left[\begin{array}{cc}
6 / 5 & 39 / 5
42 / 5 & -6
\end{array}\right]=\left[\begin{array}{cc}
2 / 5 & 13 / 5
14 / 5 & -2
\end{array}\right]\)
Question 8. Find X , if Y=\(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\) and 2 X+Y=\(\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]\)
Solution:
⇒ \(2 X=\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]-Y=\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]-\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]=\left[\begin{array}{cc}
1-3 & 0-2 \\
-3-1 & 2-4
\end{array}\right]=\left[\begin{array}{cc}
-2 & -2 \\
-4 & -2
\end{array}\right]\)
X=\(\frac{1}{2}\left[\begin{array}{ll}
-2 & -2 \\
-4 & -2
\end{array}\right]=\left[\begin{array}{cc}
-1 & -1 \\
-2 & -1
\end{array}\right]\)
Question 9. Find x and y ,if 2\(\left[\begin{array}{ll}
1 & 3 \\
0 & x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\).
Solution:
Given, \(2\left[\begin{array}{ll}
1 & 3 \\
0 & \mathrm{x}
\end{array}\right]+\left[\begin{array}{ll}
\mathrm{y} & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right] \Rightarrow\left[\begin{array}{cc}
2+\mathrm{y} & 6+0 \\
0+1 & 2 \mathrm{x}+2
\end{array}\right]=\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
By definition of equality of matrix, as the given matrices are equal, their corresponding elements are also equal. Comparing these corresponding elements, we get
2+Y = S → Equation (1)
and 2x+2=8 → Equation (2)
y=5-2=3 and 2 x=8-2 \(\Rightarrow\) y=3 and x=\(\frac{6}{2}\)=3
Question 10. Solve the equation for x, y, z And t,- if 2\(\left[\begin{array}{ll}
x & z \\
y & t
\end{array}\right]+3\left[\begin{array}{cc}
1 & -1 \\
0 & 2
\end{array}\right]=3\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right]\)
Solution:
By definition of equality of matrix, as the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get
2x + 3 = 9,2y + 0 = 12,2z – 3 = 15 and 2t + 6 = 18
⇒ \(x=\frac{9-3}{2}, y=\frac{12}{2}, z=\frac{15+3}{2} \text { and } t=\frac{18-6}{2}\)
x=3, y=6, 2=9 and t=6.
Question 11. If \(x\left[\begin{array}{l}
2 \\
3
\end{array}\right]+y\left[\begin{array}{c}
-1 \\
1
\end{array}\right]=\left[\begin{array}{c}
10 \\
5
\end{array}\right]\),Then find the values of x And y.
Solution:
Given that \(x\left[\begin{array}{l}
2 \\
3
\end{array}\right]+y\left[\begin{array}{c}
-1 \\
1
\end{array}\right]=\left[\begin{array}{c}
10 \\
5
\end{array}\right] \Rightarrow\left[\begin{array}{c}
2 x-y \\
3 x+y
\end{array}\right]=\left[\begin{array}{c}
10 \\
5
\end{array}\right]\)
By definition of equality of matrix as the given matrices are equal, their corresponding elements are also equal” Comparing the corresponding elements, we get
2x – y = 19 Equation (1)
and 3x + Y = 5 Equation (2)
Adding Eq. (1) and (2), we get; 5x = 15 =* x = 3
Substituting x= 3 in Eq. (1), we get : 2 x 3 -y = 10 + y =6 -10 =-4
Question 12. \(3\left[\begin{array}{cc}
x & y \\
z & w
\end{array}\right]=\left[\begin{array}{cc}
x & 6 \\
-1 & 2 w
\end{array}\right]+\left[\begin{array}{cc}
4 & x+y \\
z+w & 3
\end{array}\right]\),Find the values of x,y,z And w.
Solution:
Given, \(3\left[\begin{array}{ll}
x & y \\
z & w
\end{array}\right]=\left[\begin{array}{cc}
x & 6 \\
-1 & 2 w
\end{array}\right]+\left[\begin{array}{cc}
4 & x+y \\
z+w & 3
\end{array}\right] \Rightarrow\left[\begin{array}{cc}
3 x & 3 y \\
3 z & 3 w
\end{array}\right]=\left[\begin{array}{cc}
x+4 & 6+x+y \\
-1+z+w & 2 w+3
\end{array}\right]\)
By definition of equality of matrix, as the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get
3 x=x+4 \(\Rightarrow 2 x=4 \Rightarrow\) x=2 and 3 y=6+x+y \(\Rightarrow 2 y=6+x \Rightarrow y=\frac{6+x}{2}=\frac{6+2}{2}=\frac{8}{2}\)=4
Now, 3 z=-1+z+w, 2 z=-1+w \(\Rightarrow z=\frac{-1+w}{2}\)
Now, 3 w=2 w+3 \(\Rightarrow\) w=3
Putting the value of w in Eq. (1), we get: z=\(\frac{-1+3}{2}\)=1
Hence, the values of x, y, z, and w are 2,4,1 and 3, respectively.
Question 13. If F(x)= \(\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right]\) , then show that F(x) \(\cdot \)F(y)=F(x+y).
Solution:
⇒ \(F(x) F(y)=\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right]\)
⇒ \(\left[\begin{array}{ccc}
\cos y & -\sin y & 0 \\
\sin y & \cos y & 0 \\
0 & 0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
\cos x \cos y-\sin x \sin y & -\sin y \cos x-\sin x \cos y & 0 \\
\sin x \cos y+\cos x \sin y & -\sin x \sin y+\cos x \cos y & 0 \\
0 & 0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
\cos (x+y) & -\sin (x+y) & 0 \\
\sin (x+y) & \cos (x+y) & 0 \\
0 & 0 & 1
\end{array}\right] \left\{\begin{array}{c}
\cos (A+B)=\cos A \cos B-\sin A \sin B \\
\sin (A+B)=\sin A \cos B+\sin B \cos A
\end{array}\right\}\)
F(x), F(y)=F(x+y)
Question 14. Show That
1. \(\left[\begin{array}{cc}
5 & -1 \\
6 & 7
\end{array}\right]\left[\begin{array}{ll}
2 & 1 \\
3 & 4
\end{array}\right] \neq\left[\begin{array}{ll}
2 & 1 \\
3 & 4
\end{array}\right]\left[\begin{array}{cc}
5 & -1 \\
6 & 7
\end{array}\right]\)
2. \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right] \neq\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]\)
Solution:
1. \({\left[\begin{array}{cc}
5 & -1 \\
6 & 7
\end{array}\right]\left[\begin{array}{ll}
2 & 1 \\
3 & 4
\end{array}\right]=\left[\begin{array}{cc}
10-3 & 5-4 \\
12+21 & 6+28
\end{array}\right]=\left[\begin{array}{cc}
7 & 1 \\
33 & 34
\end{array}\right]}\)
⇒ \({\left[\begin{array}{cc}
2 & 1 \\
3 & 4
\end{array}\right]\left[\begin{array}{cc}
5 & -1 \\
6 & 7
\end{array}\right]=\left[\begin{array}{cc}
10+6 & -2+7 \\
15+24 & -3+28
\end{array}\right]=\left[\begin{array}{cc}
16 & 5 \\
39 & 25
\end{array}\right]}\)
Hence, \(\left[\begin{array}{cc}
5 & -1 \\
6 & 7
\end{array}\right]\left[\begin{array}{ll}
2 & 1 \\
3 & 4
\end{array}\right] \neq\left[\begin{array}{ll}
2 & 1 \\
3 & 4
\end{array}\right]\left[\begin{array}{cc}
5 & -1 \\
6 & 7
\end{array}\right]\)
2. Here, \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right]\)
=\(\left[\begin{array}{ccc}
-1+0+6 & 1-2+9 & 0+2+12 \\
0+0+0 & 0+(-1)+0 & 0+1+0 \\
-1+0+0 & 1-1+0 & 0+1+0
\end{array}\right]\)
=\(\left[\begin{array}{ccc}
5 & 8 & 14 \\
0 & -1 & 1 \\
-1 & 0 & 1
\end{array}\right]\) and
⇒ \(\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
-1+0+0 & -2+1+0 & -3+0+0 \\
0+0+1 & 0-1+1 & 0+0+0 \\
2+0+4 & 4+3+4 & 6+0+0
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
-1 & -1 & -3 \\
1 & 0 & 0 \\
6 & 11 & 6
\end{array}\right]\)
⇒ \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right] \neq\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]\)
Hence, the required result is verified.
Question 15. Find \(A^2-5 \)A+6 I if A=\(\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]\)
Solution:
Here, \(A^2=A \cdot A\)=\(\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
4+0+1 & 0+0-1 & 2+0+0 \\
4+2+3 & 0+1-3 & 2+3+0 \\
2-2+0 & 0-1+0 & 1-3+0
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
5 & -1 & 2 \\
9 & -2 & 5 \\
0 & -1 & -2
\end{array}\right]\)
⇒ \(\mathrm{A}^2-5 \mathrm{~A}+6 \mathrm{I}=\left[\begin{array}{ccc}
5 & -1 & 2 \\
9 & -2 & 5 \\
0 & -1 & -2
\end{array}\right]-5\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]+6\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
5 & -1 & 2 \\
9 & -2 & 5 \\
0 & -1 & -2
\end{array}\right]-\left[\begin{array}{ccc}
10 & 0 & 5 \\
10 & 5 & 15 \\
5 & -5 & 0
\end{array}\right]+\left[\begin{array}{lll}
6 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 6
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
5-10+6 & -1-0+0 & 2-5+0 \\
9-10+0 & -2-5+6 & 5-15+0 \\
0-5+0 & -1+5+0 & -2-0+6
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
1 & -1 & -3 \\
-1 & -1 & -10 \\
-5 & 4 & 4
\end{array}\right]\)
Question 16. If A=\(\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]\) , prove that \(A^3-6 A^2+7\) A+2 I=0
Solution:
⇒ \(A^2=A \times A=\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]\)
= \(\left[\begin{array}{lll}
1+0+4 & 0+0+0 & 2+0+6 \\
0+0+2 & 0+4+0 & 0+2+3 \\
2+0+6 & 0+0+0 & 4+0+9
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
5 & 0 & 8 \\
2 & 4 & 5 \\
8 & 0 & 13
\end{array}\right]\)
⇒ \(\mathrm{A}^3=\mathrm{A}^2-\mathrm{A}=\left[\begin{array}{ccc}
5 & 0 & 8 \\
2 & 4 & 5 \\
8 & 0 & 13
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
5+0+16 & 0+0+0 & 10+0+24 \\
2+0+10 & 0+8+0 & 4+4+15 \\
8+0+26 & 0+0+0 & 16+0+39
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
21 & 0 & 34 \\
12 & 8 & 23 \\
34 & 0 & 55
\end{array}\right]\)
⇒\(\mathrm{A}^3-6 \mathrm{~A}^2+7 \mathrm{~A}+2 \mathrm{I}=\left[\begin{array}{lll}
21 & 0 & 34 \\
12 & 8 & 23 \\
34 & 0 & 55
\end{array}\right]-6\left[\begin{array}{ccc}
5 & 0 & 8 \\
2 & 4 & 5 \\
8 & 0 & 13
\end{array}\right]+7\left[\begin{array}{ccc}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]+2\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{lll}
21 & 0 & 34 \\
12 & 8 & 23 \\
34 & 0 & 55
\end{array}\right]-\left[\begin{array}{ccc}
30 & 0 & 48 \\
12 & 24 & 30 \\
48 & 0 & 78
\end{array}\right]+\left[\begin{array}{ccc}
7 & 0 & 14 \\
0 & 14 & 7 \\
14 & 0 & 21
\end{array}\right]+\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
21-30+7+2 & 0-0+0+0 & 34-48+14+0 \\
12-12+0+0 & 8-24+14+2 & 23-30+7+0 \\
34-48+14+0 & 0-0+0+0- & 55-78+21+2
\end{array}\right]\)
= \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)=0
Question 17. If A=\(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\) and I=\(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\), then find k so that \(A^2\)=k A-2 I.
Solution:
Given; \(A^2-k A-2 I \Rightarrow A \cdot A=k A-2 I\)
⇒ \(\Rightarrow\left[\begin{array}{cc}
3 & -2 \\
4 & -2
\end{array}\right]\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\)
= \(k\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]-2\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
9-8 & -6+4 \\
12-8 & -8+4
\end{array}\right]\)
= \(\left[\begin{array}{ll}
3 k & -2 k \\
4 k & -2 k
\end{array}\right]-\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right] \Rightarrow\left[\begin{array}{cc}
1 & -2 \\
4 & -4
\end{array}\right]\)
= \(\left[\begin{array}{cc}
3 k-2 & -2 k \\
4 k & -2 k-2
\end{array}\right]\)
By definition of equality of matrix, as the given matrices are equal, their corresponding elements are also equal. Comparing the corresponding elements, we get.
3 \(\mathrm{k}-2=1 \Rightarrow \mathrm{k}=1 ;-2 \mathrm{k}=-2 \Rightarrow \mathrm{k}=1 ; 4 \mathrm{k}=4 \Rightarrow \mathrm{k}=1 ;-4=-2 \mathrm{k}-2 \Rightarrow \mathrm{k}\)=1
Hence, k=1
Question 18. If A=\(\left[\begin{array}{cc}
0 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 0
\end{array}\right]\) And I is the Identity matrix Of order 2, show that I+A=(I-A) \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
Solution:
Here, \(\mathrm{A}=\left[\begin{array}{cc}
0 & -\mathrm{t} \\
\mathrm{t} & 0
\end{array}\right]\) where \(\mathrm{t}=\tan \left(\frac{\alpha}{2}\right)\)
Now, \(\cos \alpha=\frac{1-\tan ^2\left(\frac{\alpha}{2}\right)}{1+\tan ^2\left(\frac{\alpha}{2}\right)}=\frac{1-t^2}{1+t^2} and \sin \alpha=\frac{2 \tan \left(\frac{\alpha}{2}\right)}{1+\tan ^2\left(\frac{\alpha}{2}\right)}=\frac{2 t}{1+t^2}\)
RHS =(I-A)\(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
=\(\left[\left(\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right)-\left(\begin{array}{cc}
0 & -t \\
t & 0
\end{array}\right)\right]\left[\begin{array}{cc}
\frac{1-t^2}{1+t^2} & \frac{-2 t}{1+t^2} \\
\frac{2 t}{1+t^2} & \frac{1-t^2}{1+t^2}
\end{array}\right]\)
= \(\left[\begin{array}{cc}
1 & t \\
-t & 1
\end{array}\right]\left[\begin{array}{cc}
\frac{1-t^2}{1+t^2} & \frac{-2 t}{1+t^2} \\
\frac{2 t}{1+t^2} & \frac{1-t^2}{1+t^2}
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\frac{1-t^2+2 t^2}{1+t^2} & \frac{-2 t+t\left(1-t^2\right)}{1+t^2} \\
\frac{-t\left(1-t^2\right)+2 t}{1+t^2} & \frac{2 t^2+1-t^2}{1+t^2}
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\frac{1+t^2}{1+t^2} & \frac{-2 t+t-t^3}{1+t^2} \\
\frac{-t+t^3+2 t}{1+t^2} & \frac{2 t^2+1-t^2}{1+t^2}
\end{array}\right]\)
=\(\left[\begin{array}{cc}
\frac{1+t^2}{1+t^2} & \frac{-t\left(1+t^2\right)}{1+t^2} \\
\frac{t\left(1+t^2\right)}{r+t^2} & \frac{1+t^2}{1+t^2}
\end{array}\right]\)
= \(\left[\begin{array}{cc}
1 & -t \\
t & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 1
\end{array}\right]\)
and L H S=I+A=\(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]+\left[\begin{array}{cc}
0 & -t \\
t & 0
\end{array}\right]\)
=\(\left[\begin{array}{cc}
1 & -t \\
t & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 1
\end{array}\right]\)=R H S
Question 19. A trust fund has (30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year and the second bond pays 7% interest per year. Using matrix multiplication. determine how to divide ₹30,000 among the two types of bonds, if the trust fund must obtain an annual total interest of (a) ₹1800 and (b) ₹2000.
Solution:
Let the amount invested in the first type of bond is x, and then that invested in the second type of bonds
will he ₹(30000 – x).
1. According to given condition,
⇒ \(\left[\begin{array}{ll}
\mathrm{x} & 30000-\mathrm{x}
\end{array}\right]\left[\begin{array}{c}
\frac{5}{100} \\
\frac{7}{100}
\end{array}\right]=[1800] \Rightarrow\left[\frac{5 \mathrm{x}}{100}+\frac{(30000-\mathrm{x}) 7}{100}\right]=[1800]\)
⇒ \(\left[\frac{5 x+210000-7 x}{100}\right]=[1800] \Rightarrow 210000-2 x=180000 \Rightarrow 30000=2 x \Rightarrow x\)=15000
⇒ Hence, the amounts invested in the two types of bonds are respectively 15000 and ₹(30000 – 15000) = ₹15000
2. According to the given condition.
⇒ \({[\mathrm{x} 30000-\mathrm{x}]\left[\begin{array}{c}
\frac{5}{100} \\
\frac{7}{100}
\end{array}\right]=[2000] }\)
⇒ \({\left[\frac{5 \mathrm{x}}{100}+\frac{(30000-\mathrm{x}) 7}{100}\right]=[2000] \Rightarrow\left[\frac{5 \mathrm{x}+210000-7 \mathrm{x}}{100}\right]=\{2000] }\)
⇒ \(-2 \mathrm{x}+210000=200000 \Rightarrow-2 \mathrm{x}=-10000 \Rightarrow \mathrm{x}\)=5000
Hence, the amounts invested in two types of bonds are respectively ₹ 5000 and ₹(30000-5000)=₹ 25000.
Question 20. The bookshop of a particular school has 10 dozen chemistry books, I dozen physics books, and 10 dozen economics books. Their selling prices are ₹80, ₹60, and ₹40 for one book of, each subject respectively. Find the total amount that the bookshop will receive from selling the books, using matrix algebra.
Solution:
Let A=\(\left[\begin{array}{lll}10 \times 12 & 8 \times 12 & 10 \times 12\end{array}\right]\) and B=\(\left[\begin{array}{l}80 \\ 60 \\ 40\end{array}\right]\)
The amount received by the bookseller on selling these types of tools can be computed by evaluating products A and B.
Now, A B =\([\begin{array}{lll}
120 & 96 & 120
\end{array}][\begin{array}{l}
80 \\
60 \\
40
\end{array}\)
=\([120 \times 80+96 \times 60+120 \times 40]_{1 \times 1}\)
=[9600+5760+4800]\(_{|\times|}=[20160]_{1 \times 1}\)
The amount received by the bookseller =₹ 20160
Question 21. The restrictions on n, k, and p so that PY + WY will be defined are
- k=3,p=n
- k is arbitrary, p=2
- p is arbitrary, k=3
- k = 2, p = 3
Solution:
1. Matrices P and Y are of the orders p x k and 3 x k respectively.
Therefore, matrix PY will be defined if k=3
Consequently, P Y will be of the order p x k. Matrices W and Y are of the orders n x 3 and 3 x k respectively.
Since the number of columns in W is equal to the number of rows in Y, matrix WY is well-defined and is of the order n x k.
Matrices PY and WY can be added only when their orders are the same.
However, PY is of the order p x k, therefore we must have p=n.
Thus, k=3 and p= n are the restrictions on n, k and p so that PY+WY will be defined.
Question 22. If \(\mathrm{n}=\mathrm{p}\) then the order of the matrix 7 \(\mathrm{X}\)-52 is
- \(\mathrm{p} \times 2\)
- \(2 \times \mathrm{n}\)
- \(n \times 3\)
- \(\mathrm{p} \times \mathrm{n}\)
Solution: 2. \(2 \times \mathrm{n}\)
Matrix \(\mathrm{X}\) is of the order 2 \(\times \mathrm{n}\).
Therefore, matrix 7 \(\mathrm{X}\) is also of the same order.
Matrix Z is of the order 2 \(\times pi.e. 2 \times n\)
[Since, \(\mathrm{n}=\mathrm{p} ]\)
Therefore, matrix 5 \(\mathrm{Z}\) is also of the same order.
Now, both the matrices 7 X and 5 Z are of the order 2 \(\times\) n
Thus, matrix 7 X-5 Z is well-defined and is of the order 2 \(\times\)n.
Matrices Exercise 3.3
Question 1. Find the transpose of Each of the following matrices.
- \(\left[\begin{array}{c}5 \\ 1 / 2 \\ -1\end{array}\right]\)
- \(\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]\)
- \(\left[\begin{array}{ccc}-1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1\end{array}\right]\)
Solution:
Let A=\(\left[\begin{array}{c}5 \\ 1 / 2 \\ -1\end{array}\right]_{3-1}, then A^{\prime}=\left[\begin{array}{lll}5 & 1 / 2 & -1\end{array}\right]_{1-3}\)
Let A=\(\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]_{2 \times 2}, then A^{\prime}=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]_{2 \times 2}\)
Let A=\(\left[\begin{array}{ccc}-1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1\end{array}\right]_{3 \ldots 5}, then A^{\prime}=\left[\begin{array}{ccc}-1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1\end{array}\right]_{3 \times 1}\)
Question 2. If A=\(\left[\begin{array}{ccc}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{array}\right]\) and B=\(\left[\begin{array}{ccc}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{array}\right]\), then verify that
- \((A+B)^{\prime}=A^{\prime}+B^{\prime}\)
- \((\mathrm{A}-\mathrm{B})^{\prime}=\mathrm{A}^{\prime}-\mathrm{B}^{\prime}\)
Solution:
1. Here, A+B=\(\left[\begin{array}{ccc}
-1 & 2 & 3 \\
5 & 7 & 9 \\
-2 & 1 & 1
\end{array}\right]+\left[\begin{array}{ccc}
-4 & 1 & -5 \\
1 & 2 & 0 \\
1 & 3 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
-5 & 3 & -2 \\
6 & 9 & 9 \\
-1 & 4 & 2
\end{array}\right]\)
⇒ \((A+B)^{\prime}=\left[\begin{array}{ccc}
-5 & 6 & -1 \\
3 & 9 & 4 \\
-2 & 9 & 2
\end{array}\right]\)
Also \(A^{\prime}+B^{\prime}=\left[\begin{array}{ccc}
-1 & 2 & 3 \\
5 & 7 & 9 \\
-2 & 1 & 1
\end{array}\right]^{\prime}+\left[\begin{array}{ccc}
-4 & 1 & -5 \\
1 & 2 & 0 \\
1 & 3 & 1
\end{array}\right]^{\prime}\)
=\(\left[\begin{array}{ccc}
-1 & 5 & -2 \\
2 & 7 & 1 \\
3 & 9 & 1
\end{array}\right]+\left[\begin{array}{ccc}
-4 & 1 & 1 \\
1 & 2 & 3 \\
-5 & 0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
-5 & 6 & -1 \\
3 & 9 & 4 \\
-2 & 9 & 2
\end{array}\right]\)
From Eq. (1) and (2), it is verified that }\((A+B)^{\prime}=A^{\prime}+B^{\prime}\).
2. \( Here, A-B=\left[\begin{array}{ccc}
-1 & 2 & 3 \\
5 & 7 & 9 \\
-2 & 1 & 1
\end{array}\right]-\left[\begin{array}{ccc}
-4 & 1 & -5 \\
1 & 2 & 0 \\
1 & 3 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
-1+4 & 2-1 & 3+5 \\
5-1 & 7-2 & 9-0 \\
-2-1 & 1-3 & 1-1
\end{array}\right]=\left[\begin{array}{ccc}
3 & 1 & 8 \\
4 & 5 & 9 \\
-3 & -2 & 0
\end{array}\right]\)
⇒ \(\Rightarrow(A-B)^{\prime}=\left[\begin{array}{ccc}
3 & 4 & -3 \\
1 & 5 & -2 \\
8 & 9 & 0
\end{array}\right]\)
Also, \(A^{\prime}-B^{\prime}=\left[\begin{array}{ccc}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{array}\right]-\left[\begin{array}{ccc}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{array}\right]\)
= \(\left[\begin{array}{ccc}-1+4 & 5-1 & -2-1 \\ 2-1 & 7-2 & 1-3 \\ 3-(-5) & 9-0 & 1-1\end{array}\right]=\left[\begin{array}{ccc}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0\end{array}\right]\)
From Eqs. (1) and (2), it is verified that \((A-B)^{\prime}=A^{\prime}-B^{\prime}\).
Question 3. If \(A^{\prime}=\left[\begin{array}{cc}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]\) and B=\(\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]\), then verify that
1.\((\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime}\)
2.\((\mathrm{A}-\mathrm{B})^{\prime}=\mathrm{A}^{\prime}-\mathrm{B}^{\prime}\)
Solution:
A=\(\left(A^{\prime}\right)^{\prime}=\left[\begin{array}{cc}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right]^{\prime}=\left[\begin{array}{ccc}
3 & -1 & 0 \\
4 & 2 & 1
\end{array}\right]\)
Here,A+B=\(\left[\begin{array}{ccc}
3 & -1 & 0 \\
4 & 2 & 1
\end{array}\right]+\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]=\left[\begin{array}{lll}
2 & 1 & 1 \\
5 & 4 & 4
\end{array}\right]\)
⇒ \((A+B)^{\prime}=\left[\begin{array}{lll}
2 & 1 & 1 \\
5 & 4 & 4
\end{array}\right]^{\prime}=\left[\begin{array}{ll}
2 & 5 \\
1 & 4 \\
1 & 4
\end{array}\right]\) → Equation 1
⇒ \(B^{\prime}=\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]=\left[\begin{array}{cc}
-1 & 1 \\
2 & 2 \\
1 & 3
\end{array}\right], A^{\prime}=\left[\begin{array}{cc}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right]\)
\(A^{\prime}+B^{\prime}=\left[\begin{array}{cc}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right]+\left[\begin{array}{cc}
-1 & 1 \\
2 & 2 \\
1 & 3
\end{array}\right]=\left[\begin{array}{ll}
2 & 5 \\
1 & 4 \\
1 & 4
\end{array}\right]\) → Equation 2
From equations (1) and (2), it is verified that \((A+B)^{\prime}=A^{\prime}+B^{\prime}\).
RHS =\(A^{\prime}-B^{\prime}=\left[\begin{array}{cc}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right]-\left[\begin{array}{cc}
-1 & 1 \\
2 & 2 \\
1 & 3
\end{array}\right]\)
=\(\left[\begin{array}{cc}
3+1 & 4-1 \\
-1-2 & 2-2 \\
0-1 & 1-3
\end{array}\right]\)
=\(\left[\begin{array}{cc}
4 & 3 \\
-3 & 0 \\
-1 & -2
\end{array}\right]\)
LHS =\((A-B)^{\prime}=\left(\left[\begin{array}{ccc}
3 & -1 & 0 \\
4 & 2 & 1
\end{array}\right]-\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]\right)\)
=\(\left[\begin{array}{ccc}
4 & -3 & -1 \\
3 & 0 & -2
\end{array}\right]=\left[\begin{array}{cc}
4 & 3 \\
-3 & 0 \\
-1 & -2
\end{array}\right]\)
From equation (1) and (2), it is verified that \((A-B)^{\prime}=A^{\prime}-B^{\prime}\)
Question 4. If \(A^{\prime}=\left[\begin{array}{cc}
-2 & 3 \\
1 & 2
\end{array}\right]\) and B=\(\left[\begin{array}{cc}
-1 & 0 \\
1 & 2
\end{array}\right]\) then find A+2 \(B)^{\prime}\)
Solution:
⇒ \( Given,A^{\prime}=\left[\begin{array}{cc}
-2 & 3 \\
1 & 2
\end{array}\right]\)
A=\(\left(A^{\prime}\right)^{\prime}=\left[\begin{array}{cc}
-2 & 3 \\
1 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
-2 & 1 \\
3 & 2
\end{array}\right]\)
and \(2 B=2\left[\begin{array}{cc}
-1 & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{cc}
-2 & 0 \\
2 & 4
\end{array}\right]\)
A+2 B=\(\left[\begin{array}{cc}
-2 & 1 \\
3 & 2
\end{array}\right]+\left[\begin{array}{cc}
-2 & 0 \\
2 & 4
\end{array}\right]=\left[\begin{array}{cc}
-4 & 1 \\
5 & 6
\end{array}\right]\)
⇒ \((A+2 B)^{\prime}=\left[\begin{array}{cc}
-4 & 1 \\
5 & 6
\end{array}\right]=\left[\begin{array}{cc}
-4 & 5 \\
1 & 6
\end{array}\right]\)
Question 5. For the matrices, A and B, verify that \((A B)^{\prime}=B^{\prime} A^{\prime}\), where
- \(\mathrm{A}=\left[\begin{array}{c}1 \\ -4 \\ 3\end{array}\right], \mathrm{B}=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]\)
- \(\mathrm{A}=\left[\begin{array}{l}0 \\ 1 \\ 2\end{array}\right], \mathrm{B}=\left[\begin{array}{lll}1 & 5 & 7\end{array}\right]\)
Solution:
1. Here, A B=\(\left[\begin{array}{c}1 \\ -4 \\ 3\end{array}\right]\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3\end{array}\right]\)
\((\mathrm{AB})^{\prime}=\left[\begin{array}{ccc}
-1 & 2 & 1 \\
4 & -8 & -4 \\
-3 & 6 & 3
\end{array}\right]^{\prime}=\left[\begin{array}{ccc}
-1 & 4 & -3 \\
2 & -8 & 6 \\
1 & -4 & 3
\end{array}\right]\) →Equation 1
Also, \(\mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]^{\prime}\left[\begin{array}{c}1 \\ -4 \\ 3\end{array}\right]^{\prime}=\left[\begin{array}{c}-1 \\ 2 \\ 1\end{array}\right]\left[\begin{array}{lll}1 &-4 & 3\end{array}\right]\)
= \(\left[\begin{array}{ccc}-1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3\end{array}\right]\) → Equation 2
From Eqs. (1) and (2), we find that \((A B)^{\prime}=B^{\prime} A^{\prime}\)
2. Here, \((\mathrm{AB})=\left[\begin{array}{l}0 \\ 1 \\ 2\end{array}\right]\left[\begin{array}{lll}1 & 5 & 7\end{array}\right]=\left[\begin{array}{ccc}0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14\end{array}\right]\)
⇒ \((\mathrm{AB})^{\prime}=\left[\begin{array}{ccc}0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14\end{array}\right]^{\prime}=\left[\begin{array}{ccc}0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14\end{array}\right]\) → Equation 1
Also, \(B’\mathrm{A}^{\prime}=\left[\begin{array}{lll}1 & 5 & 7\end{array}\right]^{\prime}\left[\begin{array}{l}0 \\ 1 \\ 2\end{array}\right]^{\prime}=\left[\begin{array}{l}1 \\ 5 \\ 7\end{array}\right]\left[\begin{array}{lll}0 & 1 & 2\end{array}\right]=\left[\begin{array}{lll}0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14\end{array}\right]\) → Equation 2
From Eq, (1) and (2), it is verified that \((A B)^{\prime}=B^{\prime} A^{\prime}\).
Question 6. 1. If A=\(\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\), then verify that \(A^{\prime}\) A=I.
2. If A=\(\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{array}\right]\), then verify that \(A^{\prime}\) A=1.
Solution:
1. Here, \(\mathrm{A}=\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right] \)
⇒ \(\mathrm{A}^{\prime}=\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
⇒ \(\mathrm{A}^{\prime} \mathrm{A}=\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right] \)
= \(\left[\begin{array}{cc}
(\cos \alpha)(\cos \alpha)+(\sin \alpha)(\sin \alpha) & (\cos \alpha)(\sin \alpha)+(-\sin \alpha)(\cos \alpha) \\
(\sin \alpha)(\cos \alpha)+(\cos \alpha)(-\sin \alpha) & (\sin \alpha)(\sin \alpha)+(\cos \alpha)(\cos \alpha)
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\cos ^2 \alpha+\sin ^2 \alpha & 0 \\
0 & \sin ^2 \alpha+\cos ^2 \alpha
\end{array}\right]\)
=\(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=I\left[\sin ^2 \alpha+\cos ^2 \alpha=1\right]\)
2. Here, A=\(\left[\begin{array}{cc}
\sin \alpha & \cos \alpha \\
-\cos \alpha & \sin \alpha
\end{array}\right] \)
⇒ \(A^{\prime}=\left[\begin{array}{cc}
\sin \alpha & \cos \alpha \\
-\cos \alpha & \sin \alpha
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
\sin \alpha & -\cos \alpha \\
\cos \alpha & \sin \alpha
\end{array}\right]\)
⇒ \(\mathrm{A}^{\prime} \mathrm{A}=\left[\begin{array}{cc}
\sin \alpha & -\cos \alpha \\
\cos \alpha & \sin \alpha
\end{array}\right]\left[\begin{array}{cc}
\sin \alpha & \cos \alpha \\
-\cos \alpha & \sin \alpha
\end{array}\right]\)
=\(\left[\begin{array}{cc}
(\sin \alpha)(\sin \alpha)+(-\cos \alpha)(-\cos \alpha) & (\sin \alpha)(\cos \alpha)+(-\cos \alpha)(\sin \alpha) \\
(\sin \alpha)(\cos \alpha)+(\cos \alpha)(-\sin \alpha) & (\cos \alpha)(\cos \alpha)+(\sin \alpha)(\sin \alpha)
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\sin ^2 \alpha+\cos ^2 \alpha & \sin \alpha \cos \alpha-\cos \alpha \sin \alpha \\
\sin \alpha \cos \alpha-\sin \alpha \cos \alpha & \cos ^2 \alpha+\sin ^2 \alpha
\end{array}\right]\)
=\(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=1\left[\sin ^2 \alpha+\cos ^2 \alpha=1\right]\)
Hence, we verified that \(A^{\prime}\) A=1
Question 7. 1. Show that the matrix, A=\(\left[\begin{array}{ccc}1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3\end{array}\right]\) is a Symmetric mâtrix.
2. Show that the matrix, A=\(\left[\begin{array}{ccc}0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0\end{array}\right]\) is a Skew-symmetric matrix.
Solution:
1. Here, A=\(\left[\begin{array}{ccc}1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3\end{array}\right]\)
⇒ \(A^{\prime}=\left[\begin{array}{ccc}1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3\end{array}\right]=\left[\begin{array}{ccc}1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3\end{array}\right]\)=A
⇒ \(\mathrm{A}^{\prime}=\mathrm{A}\). Hence, A is a symmetric matrix.
2. Here, \(\mathrm{A}=\left[\begin{array}{ccc}0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0\end{array}\right]\)
\(\mathrm{A}^{\prime}=\left[\begin{array}{ccc}0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0\end{array}\right]=\left[\begin{array}{ccc}0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0\end{array}\right]\)=-\(\left[\begin{array}{ccc}0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0\end{array}\right]=-\mathrm{A}\)
∴ \(\mathrm{A}^{+}=-\mathrm{A}\). Hence, \(\mathrm{A}\) is skew-symmetric matrix.
Question 8. For the matrix A=\(\left[\begin{array}{ll}1 & 5 \\ 6 & 7\end{array}\right]\), verify that:
1. \(\left(\mathrm{A}+\mathrm{A}^{\prime}\right)\) is a symmetric matrix.
2.\(\left(\mathrm{A}-\mathrm{A}^{\prime}\right)\) is a skew-symmetric matrix.
Solution:
Given, A=\(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]\)
⇒ \(A+A^{\prime}=\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]+\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]^{\prime}\)
=\(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]+\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right]=\left[\begin{array}{cc}
2 & 11 \\
11 & 14
\end{array}\right]\)
⇒ \(\Rightarrow A+A^{\prime}=\left[\begin{array}{ll}
2 & 11 \\
11 & 14
\end{array}\right] and \left(A+A^{\prime}\right)^{\prime}=\left[\begin{array}{cc}
2 & 11 \\
11 & 14
\end{array}\right]=\left[\begin{array}{cc}
2 & 11 \\
11 & 14
\end{array}\right]=A+A^{\prime}\)
⇒ \((\mathrm{A}+\mathrm{A}^{\prime})^{\prime}=\mathrm{A}+\mathrm{A}^{\prime}\), So,\((\mathrm{A}+\mathrm{A}^{\prime})\) is a symmetric matrix.
2. \(A-A^{\prime}=\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]-\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]^{\prime}\)
= \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]-\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right]=\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]\)
⇒ \(A-A^{\prime}=\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right] and \left(A-A^{\prime}\right)^{\prime}=\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array}\right]\)
=-\(\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]=-\left(A-A^{\prime}\right)\)
⇒ \((A-A)^{\prime}=-\left(A-A^{\prime}\right)\)
Hence, \(\left(\mathrm{A}-\mathrm{A}^{\prime}\right)\) is a skew-symmetric matrix.
Question 9. Find \(\frac{1}{2}\left(A+A^{\prime}\right) and \frac{I}{2}\left(A-A^{\prime}\right)\), when A=\(\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]\)
Solution:
Given A=\(\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]\)
Now,\(\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left(\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]+\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]\right)\)
=\(\frac{1}{2}\left(\left[\begin{array}{ccc}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right]+\left[\begin{array}{ccc}
0 & -a & -b \\
a & 0 & -c \\
b & c & 0
\end{array}\right]\right)\)
=\(\left[\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)
and \(\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left(\left[\begin{array}{ccc}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right]-\left[\begin{array}{ccc}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right]\right)\)
= \(\frac{1}{2}\left(\left[\begin{array}{ccc}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right]-\left[\begin{array}{ccc}
0 & -a & -b \\
a & 0 & -c \\
b & c & 0
\end{array}\right]\right)\)
=\(\frac{1}{2}\left[\begin{array}{ccc}
0 & 2 a & 2 b \\
-2 a & 0 & 2 c \\
-2 b & -2 c & 0
\end{array}\right]=\left[\begin{array}{ccc}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right]\)
Question 10. Express the following matrices as the sum of a symmetric and a skew-symmetric matrix.
- \(\left[\begin{array}{cc}3 & 5 \\ 1 & -1\end{array}\right]\)
- \(\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]\)
- \(\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]\)
- \(\left[\begin{array}{cc}1 & 5 \\ -1 & 2\end{array}\right]\)
Solution:
1. A square matrix A can be expressed as the sum of a symmetric and skew-symmetric matrix.
⇒ \(\mathrm{A}=\frac{1}{2}\left[\mathrm{~A}+\mathrm{A}^{\prime}\right]+\frac{1}{2}\left[\mathrm{~A}-\mathrm{A}^{\prime}\right]\)
Let A=\(\left[\begin{array}{cc}3 & 5 \\ 1 & -1\end{array}\right], A^{\prime}=\left[\begin{array}{cc}3 & 1 \\ 5 & -1\end{array}\right]\)
Now,\(\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left(\left[\begin{array}{cc}3 & 5 \\ 1 & -1\end{array}\right]+\left[\begin{array}{cc}3 & 1 \\ 5 & -1\end{array}\right]\right)\)
= \(\frac{1}{2}\left[\begin{array}{cc}6 & 6 \\ 6 & -2\end{array}\right]=\left[\begin{array}{cc}3 & 3 \\ 3 & -1\end{array}\right]=P( let )\)
⇒ \(P^{\prime}=\left[\begin{array}{cc}
3 & 3 \\
3 & -1
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
3 & 3 \\
3 & -1
\end{array}\right]=\mathrm{P}\)
Thus,\(\mathrm{P}=\frac{1}{2}\left(\mathrm{~A}+\mathrm{A}^{\prime}\right)\) is a symmetric matrix.
Again,\(\frac{1}{2}\left(\mathrm{~A}-\mathrm{A}^{\prime}\right)=\frac{1}{2}\left(\left[\begin{array}{cc}3 & 5 \\ 1 & -1\end{array}\right]-\left[\begin{array}{cc}3 & 1 \\ 5 & -1\end{array}\right]\right)\)
=\(\frac{1}{2}\left[\begin{array}{cc}0 & 4 \\ -4 & 0\end{array}\right]=\left[\begin{array}{cc}0 & 2 \\ -2 & 0\end{array}\right]=\mathrm{Q}((let)\)
\(Q^{\prime}=\left[\begin{array}{cc}0 & 2 \\ -2 & 0\end{array}\right]^{\prime}=\left[\begin{array}{cc}0 & -2 \\ 2 & 0\end{array}\right]=-\left[\begin{array}{cc}0 & 2 \\ -2 & 0\end{array}\right]\)=-Q
Thus,\(\mathrm{Q}=\frac{1}{2}\left(\mathrm{~A}-\mathrm{A}^{\prime}\right)\) is a skew – symmetric matrix.
Now, P+Q=\(\left[\begin{array}{cc}3 & 3 \\ 3 & -1\end{array}\right]+\left[\begin{array}{cc}0 & 2 \\ -2 & 0\end{array}\right]=\left[\begin{array}{cc}3 & 5 \\ 1 & -1\end{array}\right]=\mathrm{A}\).
Hence, matrix A is the sum of a symmetric matrix and a Skew-Symmetric matrix.
2. A square matrix A can be expressed as the sum of a symmetric and skew symmetric matrices.
Let A=\(\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]\), then \(A^{\prime}=\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]=\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]\)
Now, \(A+A^{\prime}=\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]+\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]\)
=\(\left[\begin{array}{ccc}12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6\end{array}\right]\)
Let,P=\(\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{ccc}12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6\end{array}\right]=\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]\)
⇒ \(P^{\prime}=\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]\)=P
Thus,\(\mathrm{P}=\frac{1}{2}\left(\mathrm{~A}+\mathrm{A}^{\prime}\right)\) is a symmetric matrix.
Now, \(A-A^{\prime}=\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]-\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]\)
= \(\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\)
LetQ=\(\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\)
⇒ \(Q^{\prime}=\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)=-Q
Thus,\(\mathrm{Q}=\frac{1}{2}\left(\mathrm{~A}-\mathrm{A}^{\prime}\right)\) is a skew-symmetric matrix.
Now, P+Q=\(\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]+\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=\left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]\)=A
Hence, matrix A is the sum of a symmetric matrix and a Skew-Symmetric matrix”
3. A square matrix A can be expressed as the sum of symmetric and skew-symmetric matrices
\(\mathrm{A}=\frac{1}{2}\left[\mathrm{~A}+\mathrm{A}^{\prime}\right]+\frac{1}{2}\left[\mathrm{~A}-\mathrm{A}^{\prime}\right]\)Let A=\(\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right], then A^{\prime}=\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]\)
Now, \(A+A^{\prime}=\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]+\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]=\left[\begin{array}{ccc}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right]\)
⇒ \(\mathrm{P}=\frac{1}{2}\left(\mathrm{~A}+\mathrm{A}^{\prime}\right)=\frac{1}{2}\left[\begin{array}{ccc}
6 & 1 & -5 \\
1 & -4 & -4 \\
-5 & -4 & 4
\end{array}\right]\)
=\(\left[\begin{array}{ccc}
3 & 1 / 2 & -5 / 2 \\
1 / 2 & -2 & -2 \\
-5 / 2 & -2 & 2
\end{array}\right]\)
⇒ \(\mathrm{P}^{\prime}=\left[\begin{array}{ccc}
3 & 1 / 2 & -5 / 2 \\
1 / 2 & -2 & -2 \\
-5 / 2 & -2 & 2
\end{array}\right]=\left[\begin{array}{ccc}
3 & 1 / 2 & -5 / 2 \\
1 / 2 & -2 & -2 \\
-5 / 2 & -2 & 2
\end{array}\right]=\mathrm{P}\)
Thus,\(\mathrm{P}=\frac{1}{2}\left(\mathrm{~A}+\mathrm{A}^{\prime}\right)\) is a symmetric matrix.
Now,\(A-A^{\prime}=\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]-\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]=\left[\begin{array}{ccc}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{array}\right]\)
Let,Q=\(\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{ccc}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{array}\right]=\left[\begin{array}{ccc}0 & 5 / 2 & 3 / 2 \\ -5 / 2 & 0 & 3 \\ -3 / 2 & -3 & 0\end{array}\right]\)
⇒ \(Q^{\prime}=\left[\begin{array}{ccc}
0 & 5 / 2 & 3 / 2 \\
-5 / 2 & 0 & 3 \\
-3 / 2 & -3 & 0
\end{array}\right]^{\prime}=\left[\begin{array}{ccc}
0 & -5 / 2 & -3 / 2 \\
5 / 2 & 0 & -3 \\
3 / 2 & 3 & 0
\end{array}\right]=-\mathrm{Q}\)
Thus, Q=\(\frac{1}{2}\left(A-A^{\prime}\right)\) is a skew – symmetric matrix.
Now, \(\mathrm{P}+\mathrm{Q}=\left[\begin{array}{ccc}3 & 1 / 2 & -5 / 2 \\ 1 / 2 & -2 & -2 \\ -5 / 2 & -2 & 2\end{array}\right]+\left[\begin{array}{ccc}0 & 5 / 2 & 3 / 2 \\ -5 / 2 & 0 & 3 \\ -3 / 2 & -3 & 0\end{array}\right]=\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]=\mathrm{A}\)
Hence, matrix A is the sum of a symmetric matrix and a Skew-Symmetric matrix.
A square matrix A can be expressed as the sum of symmetric and skew-symmetric matrices
\(\mathrm{A}=\frac{1}{2}\left[\mathrm{~A}+\mathrm{A}^{\prime}\right]+\frac{1}{2}\left[\mathrm{~A}-\mathrm{A}^{\prime}\right]\)Let A=\(\left[\begin{array}{cc}1 & 5 \\ -1 & 2\end{array}\right]\) Then,\( A^{\prime}=\left[\begin{array}{cc}1 & 5 \\ -1 & 2\end{array}\right]^{\prime}=\left[\begin{array}{cc}1 & -1 \\ 5 & 2\end{array}\right]\)
Now, \(A+A^{\prime}=\left[\begin{array}{cc}1 & 5 \\ -1 & 2\end{array}\right]+\left[\begin{array}{cc}1 & -1 \\ 5 & 2\end{array}\right]=\left[\begin{array}{ll}2 & 4 \\ 4 & 4\end{array}\right]\)
Let P=\(\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{ll}2 & 4 \\ 4 & 4\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]\)
⇒ \(P^{\prime}=\left[\begin{array}{ll}
1 & 2 \\
2 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{ll}
1 & 2 \\
2 & 2
\end{array}\right]\)=P
Thus, P=\(\frac{1}{2}\left(A+A^{\prime}\right)\) is a symmetric matrix.
Now, \(A-A^{\prime}=\left[\begin{array}{cc}1 & 5 \\ -1 & 2\end{array}\right]-\left[\begin{array}{cc}1 & -1 \\ 5 & 2\end{array}\right]=\left[\begin{array}{cc}0 & 6 \\ -6 & 0\end{array}\right]\)
Let\(Q=\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{cc}0 & 6 \\ -6 & 0\end{array}\right]=\left[\begin{array}{cc}0 & 3 \\ -3 & 0\end{array}\right]\)
⇒ \(Q^{\prime}=\left[\begin{array}{cc}
0 & 3 \\
-3 & 0
\end{array}\right]=\left[\begin{array}{cc}
0 & -3 \\
3 & 0
\end{array}\right]=-Q\)
Thus,\(\mathrm{Q}=\frac{1}{2}\left(\mathrm{~A}-\mathrm{A}^{\prime}\right)\) is a skew – symmetric matrix.
⇒ \(\mathrm{P}+\mathrm{Q}=\left[\begin{array}{ll}
1 & 2 \\
2 & 2
\end{array}\right]+\left[\begin{array}{cc}
0 & 3 \\
-3 & 0
\end{array}\right]=\left[\begin{array}{cc}
1 & 5 \\
-1 & 2
\end{array}\right]=\mathrm{A}\)
Hence, matrix A is the sum of a symmetric matrix and a Skew-symmetric matrix,
Question 11. If A and B are symmetric matrices of the same order, then AB-BA is a:
- Skew-symmetric matrix
- Symmetric matrix
- Zero matrix
- Identity matrix
Solution 1. Skew-symmetric matrix
1. Given, that A and B are symmetric matrices.
⇒ \(A^{\prime}\)=A and \(B^{\prime}\) =B
⇒ \((A B-B A)^{\prime} =(A B)^{\prime}-(B A)^{\prime}{\left[(A-B)^{\prime}=A^{\prime}-B^{\prime}\right]}\)
=\(B^{\prime} A^{\prime}-A^{\prime} B^{\prime} {\left[(A B)^{\prime}=B^{\prime} A^{\prime}\right]}\)
= B A-A B \({\left[ A^{\prime}=A and B^{\prime}=B\right]} \)
= -(A B-B A)
⇒ \((A B-B A)^{\prime}\)=-(A B-B A)
Thus, \((\mathrm{AB}-\mathrm{BA})\) is a skew – symmetric matrix.
Question 12. If \(\mathrm{A}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]\), then \(\mathrm{A}+\mathrm{A}^{\prime}=\mathrm{I}\), if the value of \(\alpha\) is ?
- \(\pi / 6\)
- \(\pi / 3\)
- \(\pi\)
- 3 \(\pi / 2\)
Solution: 2. \(\pi / 3\)
Here, \(\mathrm{A}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]\)
and \(\mathrm{A}^{\prime}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]^{\prime}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\)
Given. \(\mathrm{A}+\mathrm{A}^{\prime}=\mathrm{I}\)
⇒ \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]+\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\)
= \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \Rightarrow\left[\begin{array}{cc}
2 \cos \alpha & 0 \\
0 & 2 \cos \alpha
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Comparing the corresponding elements of the above matrices, we have
⇒ \(2 \cos \alpha=1 \Rightarrow \cos \alpha=\frac{1}{2}=\cos \frac{\pi}{3} \Rightarrow \alpha=\frac{\pi}{3}\)
Matrices Exercise 3.4
Question 1. Matrices A and B will be inverse of each other only if
- AB=BA
- AB=0,BA=I
- AB=BA=0
- AB=BA=I
Solution: 4. AB=BA=I
We know that if A is a square matrix of order ‘n’ and if there exists another square matrix B
of the same order ‘n’, such that AB = BA = I., then B is said to be the inverse of A’
In this case, it is clear that A is the inverse of B’
Thus, matrices A and B will be inverse of each other only if AB = BA = I
Matrices Miscellaneous Exercise
Question 1. If A and B are symmetric matrices prove that AB – BA is a skew-symmetric matrix.
Solution:
Here, A and B are symmetric matrices” then \(A^{\prime}-A \) and \( B^{\prime}=B\)
Now,\((A B-B A)^{\prime} =(A B)^{\prime}-(B A)^{\prime}\) \(((A-B)^{\prime}=A^{\prime}-B^{\prime}[(A B)^{\prime}=B^{\prime} A^{\prime}]\)
=\(B^{\prime} A^{\prime}-A^{\prime} B^{\prime}=B A-A B( B^{\prime}=B\) and \(A^{\prime}=A)\)
=-(A B-B A)
⇒ \((A B-B A)^{\prime}\)=-(A B-B A)
Thus, \((\mathrm{AB}-\mathrm{BA})\) is a skew-symmetric matrix.
Question 2. Show that the matrix B’AB is symmetric or skew-symmetric according as A is symmetric or skewsymmetric.
Solution:
We suppose that A is a symmetric matrix, then \(A^{\prime}=A\)
Consider \(\left(B^{\prime} A B\right)^{\prime}=\left\{B^{\prime}(A B)\right\}^{\prime}=(A B)^{\prime}\left(B^{\prime}\right)^{\prime} \)
⇒ \({\left[(\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime}\right]} \)
=\(\left(\mathrm{B}^{\prime} \mathrm{A}^{\prime}\right) \mathrm{B}\)
⇒ \({\left[\left(B^{\prime}\right)^{\prime}=\mathrm{B}, \mathrm{A}^{\prime}=\mathrm{A}\right]}\)
=\(B^{\prime}\left(A^{\prime} B\right)=B^{\prime}(A B) \)
⇒ \(\left(B^{\prime} A B\right)^{\prime}=B^{\prime}\) A B
Which shows that \(B^{\prime}\) A B is a symmetric matrix.
Now, we suppose that A is a skew-symmetric matrix.
Then, \(\mathrm{A}^{+}=-\mathrm{A}\)
Consider, \(\left(B^{\prime} A B\right)^{\prime} =\left[B^{\prime}(A B)\right]^{\prime}=(A B)^{\prime}\left(B^{\prime}\right)^{\prime} \left[\left(A B^{\prime}\right)=B^{\prime} A^{\prime}\right]\)
=\(\left(B^{\prime} A^{\prime}\right) B=B^{\prime}(-A) B=-B^{\prime} A B \left[\left(A^{\prime}\right)^{\prime}=A\right]\left[ A^{\prime}=-A\right]\)
∴ \(\left(B^{\prime} A B\right)^{\prime}=-B^{\prime} A B\), which shows that \(B^{\prime}\) A B is a skew – symmetric matrix.
Question 3. Find the values of x, y and z if the matrix A= \(\left[\begin{array}{ccc}
0 & 2 y & z \\
x & y & -z \\
x & -y & z
\end{array}\right]\) satisfies the equation \(A^{\prime}\) A=1
Solution:
⇒ Given, \(A^{\prime} A=I\)
⇒ \(\left[\begin{array}{ccc}
0 & x & x \\
2 y & y & -y \\
z & -z & z
\end{array}\right]\left[\begin{array}{ccc}
0 & 2 y & z \\
x & y & -z \\
x & -y & z
\end{array}\right]\)
=\(\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
⇒ \(\left[\begin{array}{ccc}
0+x^2+x^2 & 0+x y-x y & 0-x z+x z \\
0+y x-y x & 4 y^2+y^2+y^2 & 2 y z-y z-y z \\
0-z x+z x & 2 y z-y z-y z & z^2+z^2+z^2
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
⇒ \(\left[\begin{array}{ccc}
2 x^2 & 0 & 0 \\
0 & 6 y^2 & 0 \\
0 & 0 & 3 z^2
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
On comparing the corresponding elements, we have
2. \(\mathrm{x}^2=1,6 \mathrm{y}^2=1,3 \mathrm{z}^2=1 \Rightarrow \mathrm{x}^2=\frac{1}{2}, \mathrm{y}^2=\frac{1}{6}, \mathrm{z}^2=\frac{1}{3}\)
\(\mathrm{x}= \pm \frac{1}{\sqrt{2}}, \mathrm{y}= \pm \frac{1}{\sqrt{6}}, \mathrm{z}= \pm \frac{1}{\sqrt{3}}\)Question 4. For what value of x ;\(\left[\begin{array}{lll}1 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]\)=0
Solution:
Given \(\left[\begin{array}{lll}1 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]\)=0
⇒ \(\left[\begin{array}{lll}1 & 2 & 1\end{array}\right]\left[\begin{array}{c}0+4+0 \\ 0+0+x \\ 0+0+2 x\end{array}\right]\)=0
⇒ \(\left[\begin{array}{lll}1 & 2 & 1\end{array}\right]\left[\begin{array}{c}4 \\ x \\ 2 x\end{array}\right]\)=0
[4+2 x+2 x]=[0] \(\Rightarrow 4+4 x=0 \Rightarrow\) x=-1
Question 5. If A=\(\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\), show that \(A^2-5 \)A+71=0
Solution:
Given, A=\(\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\)
Now, \(A^2=A. A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\)
=\(\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]\)
⇒ \(A^2-5\) A+7 =\(\left[\begin{array}{cc}
8 & 5 \\
-5 & 3
\end{array}\right]-5\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]+7\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
=\(\left[\begin{array}{cc}
8 & 5 \\
-5 & 3
\end{array}\right]-\left[\begin{array}{cc}
15 & 5 \\
-5 & 10
\end{array}\right]+\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right]\)
= \(\left[\begin{array}{cc}
8-15+7 & 5-5+0 \\
-5+5+0 & 3-10+7
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)=0
Question 6. Find x, if \(\left[\begin{array}{lll}x & -5 & -1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]\)=0
Solution:
Here, \(\left[\begin{array}{lll}x & -5 & -1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]\)=0
⇒ \(\left[\begin{array}{lll}x & -5 & -1\end{array}\right]\left(\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]\right)\)=0
⇒ \(\left[\begin{array}{lll}
x & -5 & -1
\end{array}\right]\left[\begin{array}{c}
x+0+2 \\
0+8+1 \\
2 x+0+3
\end{array}\right]\)=0
⇒ \([x(x+2)+(-5)(9)+(-1)(2 x+3)]=[0]\)
⇒ \(\left[x^2-48\right]=[0]\)
⇒ \(x^2-48=0 \Rightarrow x= \pm \sqrt{48}= \pm 4 \sqrt{3}\)
Question 7. A manufacturer produces three products x,y, and z which he sells in two markets, and the annual sales are indicated below.
Market: 1 2
Products: 10,000 2,000 18,000 6,000 20,000 8,000
1. If unit sale prices of x, and are ₹2.50, ₹1.50, and ₹1.00 respectively; find the total revenue in each market with the help of matrix algebra.
2. If the unit costs of the above three commodities are ₹2.00, ₹1.00, and 50 paise respectively; find the gross profit.
Solution:
Matrix representing the sales is A=\(\left[\begin{array}{ccc}10000 & 2000 & 18000 \\ 6000 & 20000 & 8000\end{array}\right]\)
1. Matrix representing the sale price per unit is \(\mathrm{B}=\left[\begin{array}{B}
5 / 2 \\
3 / 2 \\
1
\end{array}\right]\)
Total revenue in each market is given by the product;
⇒ \(\mathrm{AB} =\left[\begin{array}{ccc}
10000 & 2000 & 18000 \\
6000 & 20000 & 8000
\end{array}\right]\left[\begin{array}{B}
5 / 2 \\
3 / 2 \\
1
\end{array}\right]\)
= \(\left[\begin{array}{l}
\left(10000 \times \frac{5}{2}\right)+\left(2000 \times \frac{3}{2}\right)+(18000 \times 1) \\
\left(6000 \times \frac{5}{2}\right)+\left(20000 \times \frac{3}{2}\right)+(8000 \times 1)
\end{array}\right]\)
= \(\left[\begin{array}{c}
46000 \\
53000
\end{array}\right]\)
Hence, the total revenue in Market 1 is ₹ 46000, and that in Market 2 is ₹ 53000.
The matrix representing the cost price per unit is \(\mathrm{C}=\left[\begin{array}{l}2.00 \\ 1.00 \\ 0.50\end{array}\right]\)
The total cost in the two markets is given by the product
⇒ \(\mathrm{AC}=\left[\begin{array}{ccc}
10000 & 2000 & 18000 \\
6000 & 20000 & 8000
\end{array}\right]\left[\begin{array}{c}
2 \\
1 \\
1 / 2
\end{array}\right]\)
= \(\left[\begin{array}{l}
(10000 \times 2)+(2000 \times 1)+\left(18000 \times \frac{1}{2}\right) \\
(6000 \times 2)+(20000 \times 1)+\left(8000 \times \frac{1}{2}\right)
\end{array}\right]=\left[\begin{array}{l}
31000 \\
36000
\end{array}\right]\)
Profit in market \(\mathrm{I}\)=₹(46000-31000)=₹ 15000
and profit in market 2 =₹(53000-36000)=₹ 17000
The gross profit =₹(15000+17000)=₹ 32000.
Question 8. Find the matrix X so that \(X\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]\).
Solution:
Here, \(X\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]\)
The matrix given on the RHS of the equation is a 2 \(\times\) 3 matrix and the one given on the LHS of the equation is also a 2 \(\times 3\) matrix. Therefore, X has to be a 2 \(\times\) 2 matrix. Now, Let X=\(\left[\begin{array}{ll}a & c \\ b & d\end{array}\right]\)
Therefore; we have \(\left[\begin{array}{ll}\mathrm{a} & \mathrm{c} \\ \mathrm{b} & \mathrm{d}\end{array}\right]\left[\begin{array}{lll}\mathrm{l} & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]\)
⇒ \(\left[\begin{array}{ccc}
a+4 c & 2 a+5 c & 3 a+6 c \\
b+4 d & 2 b+5 d & 3 b+6 d
\end{array}\right]=\left[\begin{array}{ccc}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right]\)
Equating the corresponding elements of the two matrices, we leave
a+4 c=-7, 2 a+5 c=-8, 3 a+6 c=-9
b+4 d=2, 2 b+5 d=4, 3 b+6 d=6
Now, a+4 c=-7 \(\Rightarrow\) a=-7-4 c
2 \(\mathrm{a}+5 \mathrm{c}=-8 \Rightarrow-14-8 \mathrm{c}+5 \mathrm{c}=-8 \Rightarrow-3 \mathrm{c}=6 \Rightarrow \mathrm{c}\)=-2
a=-7-4(-2)=-7+8=1 also; 3(1)+6(-2)=3-12=-9
Also, b+4 d=2 \(\Rightarrow\) b=2-4 d and 2 b+5 d=4 \(\Rightarrow\) 4-8 d+5 d=4
-3 d=0 \(\Rightarrow\) d=0
b=2-4(0)=2 Also; 3(2)+6 \(\times\) 0=6
Thus, a=1, b=2, c=-2 and d=0
Hence, the required matrix X is \(\left[\begin{array}{cc}1 & -2 \\ 2 & 0\end{array}\right]\)
Question 9. If A=\(\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right] \)is such that \(A^2=\)1, then
- 1+\(\alpha^2+\beta \gamma\)=0
- 1-\(\alpha^2+\beta \gamma\)=0
- 1-\(\alpha^2-\beta \gamma\)=0
- 1+\(\alpha^2-\beta \gamma\)=0
Solution:
Given, \(A^2\)=1
⇒ \(A \cdot A=T \Rightarrow\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]\left[\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
\alpha^2+\beta \gamma & \alpha \beta-\alpha \beta \\
\alpha \gamma-\gamma \alpha & \gamma \beta+\alpha^2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
On comparing the corresponding elements, we have
∴ \(\alpha^2+\beta \gamma=1 \Rightarrow \alpha^2+\beta \gamma-1=0 \Rightarrow 1-\alpha^2-\beta \gamma\)=0
Question 10. If a matrix n is both a symmetric and skew-symmetric matrix, then
- A is a diagonal matrix
- A is a zero matrix
- A is a square matrix
- None of these
Solution: 2. A is a zero matrix
Let A be a square matrix such that A is both a symmetric and skew-symmetric matrix.
⇒ \(\mathrm{A}^{\prime}=\mathrm{A}\) {A is symmetric }
and \(A^{\prime}\)=-A {[ A is skew- symmetric] }
A=-A
A+A=O \(\Rightarrow 2 A=O \Rightarrow\) A=0
Question 11. If A is square matrix such that \(\mathrm{A}^2=\mathrm{A}\), then \((\mathrm{I}+\mathrm{A})^3-7 \mathrm{~A}\) is equal to
- A
- I – A
- 1
- 3 A
Solution: 3. 1
Here, \(\mathrm{A}^2\)= A
⇒ \((I+A)^2\)=(I+A)(I+A) [A I=A=\([A, A^2=A]\).
⇒ \((I+A)^3 =(I+A)^2 \cdot(I+A)=(I+3 A) \cdot(I+A)\)
= I+I A+3(A I)+3(A A)
= I+A+3 A+3 A=I+7 A \(\left[ A^2=A\right]\)
∴ \((I+A)^3\)-7 A=I