Determinants Exercise 4.1
Question 1. \(\left|\begin{array}{cc}
2 & 4 \\
-5 & -1
\end{array}\right|\)
Solution:
Let A=\(\left|\begin{array}{cc}2 & 4 \\ -5 & -1\end{array}\right| \Rightarrow|A|=2 \times(-1)-(-5) \times\) 4=-2+20=18
Question 2. 1.\(\left|\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right|\)
2. \(\left|\begin{array}{cc}x^2-x+1 & x-1 \\ x+1 & x+1\end{array}\right|\)
Solution:
⇒ \(\left|\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)
=\((\cos \theta)(\cos \theta)-(\sin \theta)(-\sin \theta) \)
=\(\cos ^2 \theta+\sin ^2 \theta=1\left (\sin ^2 \theta+\cos ^2 \theta=1\right)\)
⇒ \(\left|\begin{array}{cc}
x^2-x+1 & x-1 \\
x+1 & x+1
\end{array}\right|=\left(x^2-x+1\right)(x+1)-(x+1)(x-1)\)
= \(x^3-x^2+x+x^2-x+1-\left(x^2-1\right)=x^3+1-\left(x^2-1\right)=x^3-x^2+2\)
Question 3. If A= \(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right] \), then show that |2 A|=4|A|
Solution:
Given, A=\(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\)
2 A=2\(\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]=\left[\begin{array}{ll}
2 \times 1 & 2 \times 2 \\
2 \times 4 & 2 \times 2
\end{array}\right]=\left[\begin{array}{ll}
2 & 4 \\
8 & 4
\end{array}\right] \)
LHS =|2 A|=\(\left[\begin{array}{ll}
2 & 4 \\
8 & 4
\end{array}\right]=(2 \times 4)-(8 \times 4)\)=-24
Read and Learn More Class 12 Maths Chapter Wise with Solutions
Now, |A|=\(\left|\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right|=(1 \times 2)-(4 \times 2)\)=-6
RHS =4|A|=4 \(\times(-6)\)=-24
LHS = RHS
Hence; |2A|=4|A|
Question 4. If A=\(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right]\) , then show that |3 A|=27|A|
Solution:
Given, A=\(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right]\)
then 3 A=\(\left[\begin{array}{ccc}
3 & 0 & 3 \\
0 & 3 & 6 \\
0 & 0 & 12
\end{array}\right]\)
Now; \(|\mathrm{A}|=\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right|\)=1(4-0)-0(0)+1(0)=4
⇒ \(|3 \mathrm{~A}|=\left|\begin{array}{lll}
3 & 0 & 3 \\
0 & 3 & 6 \\
0 & 0 & 12
\end{array}\right|\)=3(36-0)-0(0)+3(0)=108
⇒ \(|3 \mathrm{~A}|=108=27 \times 4=27|\mathrm{~A}|\),Hence proved.
Question 5. Evaluate the determinants.
- \(\left|\begin{array}{ccc}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|\)
- \(\left|\begin{array}{ccc}3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1\end{array}\right|\)
- \(\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|\)
- \(\left|\begin{array}{ccc}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right|\)
Solution:
1. Let |A|=\(\left|\begin{array}{ccc}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|\)
By expanding along \(R_2\)
⇒ \(|\mathrm{A}|=-0\left|\begin{array}{cc}
-1 & -2 \\
-5 & 0
\end{array}\right|+0\left|\begin{array}{cc}
3 & -2 \\
3 & 0
\end{array}\right|-(-1)\left|\begin{array}{cc}
3 & -1 \\
3 & -5
\end{array}\right|=\{3 \times(-5)-3 \times(-1)\}\)=-15+3=-12
2. Let |A|=\(\left|\begin{array}{ccc}3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1\end{array}\right|\)
By expanding along \(\mathrm{R}_1\) (first row), we get
⇒ \(|A| =3\left|\begin{array}{cc}
1 & -2 \\
3 & 1
\end{array}\right|-(-4)\left|\begin{array}{cc}
1 & -2 \\
2 & 1
\end{array}\right|+5\left|\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right|\)
=3(1+6)+4(1+4)+5(3-2)=3(7)+4(5)+5(1)=21+20+5=46
3. Let |A|=\(\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right|\)
By expanding along \(R_1\) (first row), we get
⇒ \(|A|=0\left|\begin{array}{cc}
0 & -3 \\
3 & 0
\end{array}\right|-1\left|\begin{array}{cc}
-1 & -3 \\
-2 & 0
\end{array}\right|+2\left|\begin{array}{ll}
-1 & 0 \\
-2 & 3
\end{array}\right|\)
=-1(0-6)+2(-3-0)=-1(-6)+2(-3)=6-6=0
4. Let \(|A|=\left|\begin{array}{ccc}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right|\)
By expanding along \(R_2\) (second row), we get
|A|=\(-0\left|\begin{array}{cc}
-1 & -2 \\
-5 & 0
\end{array}\right|+2\left|\begin{array}{cc}
2 & -2 \\
3 & 0
\end{array}\right|-(-1)\left|\begin{array}{cc}
2 & -1 \\
3 & -5
\end{array}\right|\)
=0+2(0+6)+(-10+3)=12-7=5
Question 6. If A=\(\left[\begin{array}{lll}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right]\), find|A| ?
Solution:
Given, A=\(\left[\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right]\)
By expanding along R_1 (first row), we get
⇒ \(|A|=1\left|\begin{array}{ll}
1 & -3 \\
4 & -9
\end{array}\right|-1\left|\begin{array}{ll}
2 & -3 \\
5 & -9
\end{array}\right|+(-2)\left|\begin{array}{ll}
2 & 1 \\
5 & 4
\end{array}\right|\)
=1(-9+12)-1(-18+15)-2(8-5)
=1(3)-1(-3)-2(3)=3+3-6=0
Question 7. Find the values of x, if:
1. \(\left|\begin{array}{cc}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|\)
2. \(\left|\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right|=\left|\begin{array}{cc}x & 3 \\ 2 x & 5\end{array}\right|\)
Solution:
1. Given, \(\left|\begin{array}{cc}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 \mathrm{x} & 4 \\ 6 & \mathrm{x}\end{array}\right|\)
On expanding both determinants, we get
⇒ \((2 \times 1)-(5 \times 4)=(2 x \times x)-(6 \times 4) \Rightarrow 2-20=2 x^2-24\)
⇒ \(2 x^2=-18+24 \Rightarrow x^2=\frac{6}{2}=3 \Rightarrow x= \pm \sqrt{3}\)
2. Given, \(\left|\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right|=\left|\begin{array}{cc}x & 3 \\ 2 x & 5\end{array}\right|\)
On expanding both determinants, we get
⇒ \((2 \times 5)-(4 \times 3)=(5 \times x)-(3 \times 2 x)\)
10-12=5 x-6 x \(\Rightarrow-2=-x \Rightarrow\) x=2
Question 8. If \(\left|\begin{array}{cc}
x & 2 \\
18 & x
\end{array}\right|=\left|\begin{array}{cc}
6 & 2 \\
18 & 6
\end{array}\right|\), then x is equal to ?
- 6
- \(\pm\) 6
- -6
- 0
Solution: 2. \(\pm\) 6
On expanding both determinants, we get
⇒ \((\mathrm{x} \times \mathrm{x})-(18 \times 2)=(6 \times 6)-(18 \times 2) \Rightarrow \mathrm{x}^2-36=36-36\)
\(\mathrm{x}^2-36=0 \Rightarrow \mathrm{x}^2=36 \Rightarrow \mathrm{x}= \pm 6\)So, (B) is the correct option.
Determinants Exercise 4.2
Question 1. Find the area of the triangle with vertices at the points in each of the following:
(1,0), (6,0), (4,3)
(2,7). (1,1), (10,8)
(-2,-3), (3,2), (-1,-8)
Solution:
Area of triangle =\(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|\)
1. Required area =\(\frac{1}{2}\left|\begin{array}{lll}
1 & 0 & 1 \\
6 & 0 & 1 \\
4 & 3 & 1
\end{array}\right|=\frac{1}{2}|1(0-3)-0(6-4)+1(18-0)|\)
⇒ \(\frac{1}{2}|{-3+18}|=\frac{15}{2}\)..sq. units
2. Required area =\(\frac{1}{2}\left|\begin{array}{lll}
2 & 7 & 1 \\
1 & 1 & 1 \\
10 & 8 & 1
\end{array}\right|\)
= \(\frac{1}{2}|2(1-8)-7(1-10)+1(8-10)|\)
= \(\frac{1}{2}|2(-7)-7(-9)+1(-2)|=\frac{1}{2}|-14+63-2|=\frac{47}{2}\) sq. units
3. Required area =\(\frac{1}{2}\left|\begin{array}{ccc}
-2 & -3 & 1 \\
3 & 2 & 1 \\
-1 & -8 & 1
\end{array}\right|\)
= \(\frac{1}{2}|-2(2+8)+3(3+1)+1(-24+2)|\)
= \(\frac{1}{2}|-20+12-22|=\frac{1}{2}|-30|\)=15 sq. units
(Since the area of the triangle is always positive)
Question 2. Show that the points A(a,b + c), B(b,c +a), C(e, a + b) are collinear
Solution:
Area of \(\triangle A B C =\frac{1}{2}\left|\begin{array}{lll}
a & b+c & 1 \\
b & c+a & 1 \\
c & a+b & 1
\end{array}\right|\)
=\(\frac{1}{2}|a\{(c+a) \times 1-(a+b) \times 1\}-(b+c)\{b \times 1-1 \times c\}+1\{b \times(a+b)-(c+a) \times c\}|\)
=\(\frac{1}{2}\left|a(c+a-a-b)-(b+c)(b-c)+1\left(a b+b^2-c^2-a c\right)\right|\)
=\(\frac{1}{2}\left|a c-a b-b^2+c^2+a b+b^2-c^2-a c\right|=\frac{1}{2} \times \)0=0
Since, the area of AABC = 0.
Hence, points A(a, b + c), … C(c, a + b) are collinear,
Question 3. Find the value of k, the area of a triangle is 4 sq. units and the vertices are:
1. (k,0), (4,0),(0,2)
2. (-2,0),(0,4),(0, k)
Solution:
Area of triangle =\(\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|\)
1. Given, \(\frac{1}{2}\left|\begin{array}{lll}
\mathrm{k} & 0 & 1 \\
4 & 0 & 1 \\
0 & 2 & 1
\end{array}\right|=4\)
⇒ \(|\mathrm{k}(0-2)+1(8-0)|=8 \Rightarrow \mathrm{k}(0-2)+1(8-0)= \pm 8\)
On taking positive sign;
-2k + 8 = 8 ⇒ -2k = 0 ⇒ k = 0
On taking negative sign;
-2k + 8 = 8 ⇒-2k = -16⇒ k = 8
k = 0,8
2. Given, \(\frac{1}{2}\left|\begin{array}{ccc}-2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & \mathrm{k} & 1\end{array}\right|=4 \Rightarrow|-2(4-\mathrm{k})+1(0-0)|\)=8
⇒ –\(2(4-\mathrm{k})+1(0-0)= \pm 8 \Rightarrow(-8+2 \mathrm{k})= \pm\) 8
On taking positive sign, 2 k-8=8 \(\Rightarrow\) 2 \(\mathrm{k}=16 \Rightarrow \)k=8
On taking negative sign, 2 k-8=-8 \(\Rightarrow 2 \mathrm{k}\)=0 \(\Rightarrow\) k=0
k =0,8
Question 4. 1. Find the equation of the line joining (1, 2) and (3, 6) using determinants.
2. Find the equation of the line joining (3, 1) and (9, 3) using determinants.
Solution.
1. Let P(x,y) beany point on the line joining A(1,2)andB(3,6) .
⇒ Area of triangle =\(\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|\)=0
⇒ \(\frac{1}{2}\left|\begin{array}{lll}
1 & 2 & 1 \\
3 & 6 & 1 \\
x & y & 1
\end{array}\right|=0 \Rightarrow \frac{1}{2}[1(6-y)-2(3-x)+1(3 y-6 x)]\)=0
6-y-6+2x+3y-6x = 0 ⇒ 2y-4x = 0 ⇒y = 2x
Fluence, the equation of the line joining the given points is y = 2x.
2. Let P(x, y) be any point on the line joining A(3,i)and B(9.3)
Area of triangle =\(\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right|\)=0
⇒ \(\frac{1}{2}\left|\begin{array}{lll}
3 & 1 & 1 \\
9 & 3 & 1 \\
x & y & 1
\end{array}\right|\)=0
⇒ \(\frac{1}{2}|3(3-y)-1(9-x)+1(9 y-3 x)\)|=0
9-3 y-9+x+9 y-3 x=0 \(\Rightarrow 6 y-2 x=0 \Rightarrow\) x-3 y=0
Hence, the equation of the line joining the given points is x-3 y=0.
Question 5. If the area of a triangle is 35 sq. units with vertices (2,-6),(5,4), and (k,4), then k is.
- 12
- -2
- -1 2, -2
- 12,-2
Solution: 4. 12,-2
Given, \(\frac{1}{2}\left|\begin{array}{ccc}
2 & -6 & 1 \\
5 & 4 & 1 \\
k & 4 & 1
\end{array}\right|\)=35
|2(4-4)+6(5-k)+1(20-4 k)|=70
2(4-4)+6(5-k)+1(20-4 k)= \(\pm\) 70
30-6 k+20-4 k= \(\pm 70\)
On taking positive sign, -10 k +50=70 \(\Rightarrow\)-10 k =20 \(\Rightarrow\) k=-2
On taking negative sign, -10 k +50=-70 \(\Rightarrow\)-10 \(\mathrm{k}=-120 \Rightarrow\) k =12
k=12,-2 Hence, the correct option is 4.
Determinants Exercise 4.3
Question 1. \(\left|\begin{array}{cc}2 & -4 \\ 0 & 3\end{array}\right|\)
2. \(\left|\begin{array}{ll}a & c \\ b & d\end{array}\right|\)
Solution:
Here, \(\left|\begin{array}{cc}2 & -4 \\ 0 & 3\end{array}\right|\)is given
Minors, \(M_{11}=3, M_{12}=0, M_{21}=-4 and M_{22}\)=2
Also cofactors, \(A_{11}=(-1)^{1+1} \mathrm{M}_{11}=1 \times 3=3\)
⇒ \(A_{12}=(-1)^{1+2} M_{12}=(-1) \times\) 0=0
⇒ \(A_{21}=(-1)^{2+1} M_{21}=(-1) \times(-4)\)=4
and \(\mathrm{A}_{22}=(-1)^{2+2} \mathrm{M}_{22}=1 \times \)2=2
2. Here, \(\left|\begin{array}{ll}\mathrm{a} & \mathrm{c} \\ \mathrm{b} & \mathrm{d}\end{array}\right|\) is given
Minors, \(\mathrm{M}_{11}=\mathrm{d}, \mathrm{M}_{12}=\mathrm{b}, \mathrm{M}_{21}=\mathrm{c}\) and \(\mathrm{M}_{22}=\mathrm{a}\)
Also cofactors, \(A_{11}=(-1)^{1+1} M_{11}=1 \times\) d=d
⇒ \(A_{12}=(-1)^{1+2} M_{12}=(-1) \times \)b=-b
⇒ \(A_{21}=(-1)^{2+1} M_{21}=(-1) \times c\)=-c
and \(A_{22}=(-1)^{2+2} \mathrm{M}_{22}=1 \times \mathrm{a}=\mathrm{a}\)
Question 2. \(\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|\)
\(\left|\begin{array}{ccc}1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2\end{array}\right|\)
Solution:
1. Here, \(\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\) is given
Minors of elements of the first row are:
⇒ \(M_{11}=\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|=1-0=1, M_{12}=\left|\begin{array}{ll}
0 & 0 \\
0 & 1
\end{array}\right|\)=0-0=0
and \(M_{13}=\left|\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right|\)=0-0=0
Minors of elements of the second row are:
\(M_{21}=\left|\begin{array}{ll}
0 & 0 \\
0 & 1
\end{array}\right|=0-0=0, M_{22}=\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\)=1-0=1
and \(M_{23}=\left|\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right|\)=0-0=0
Minors of elements of the third row are:
⇒ \(M_{31}=\left|\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right|=0-0=0, M_{32}=\left|\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right|\)=0-0=0
and \(M_{33}=\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\)=1-0=1
Hence, the cofactors of elements of the first row are:
⇒ \(A_{11}=(-1)^{1+1} M_{11}=1 \times 1=1, A_{12}\)
=\((-1)^{162} M_{12}=(-1) \times 0=0, A_{13}=(-1)^{1+1} M_{13}=1 \times\) 0=0
The cofactors of elements of the second row are:
⇒ \(A_{21}=(-1)^{2+1} M_{21}=(-1) \times 0=0, A_{22}=(-1)^{2+2} M_{21}=1 \times \)1=1,
⇒ \(A_{23}=(-1)^{2 \times 3} M_3=(-1) \times \)0=0
The cofactors of elements of the third row are:
⇒ \(A_{31}=(-1)^{3+1} M_{34}=1 \times 0=0, A_{32}=(-1)^{3.2} M_{12}=(-1) \times\) 0=0
⇒ \(A_{33}=(-1)^{3,3} M_{33}=1 \times \)1=1
2. Here, \(\left|\begin{array}{ccc}
1 & 0 & 4 \\
3 & 5 & -1 \\
0 & 1 & 2
\end{array}\right|\) is given
Minors of elements of the first row are:
⇒ \(\mathrm{M}_{11}=\left|\begin{array}{cc}
5 & -1 \\
1 & 2
\end{array}\right|=10+1=11, \mathrm{M}_{13}=\left|\begin{array}{cc}
3 & -1 \\
0 & 2
\end{array}\right|=6-0=6\) and
⇒ \(\mathrm{M}_{13}=\left|\begin{array}{ll}
3 & 5 \\
⇒ 0 & 1
\end{array}\right|\)=3-0=3
Minors of elements of the second row are:
⇒ \(M_{31}=\left|\begin{array}{ll}
0 & 4 \\
1 & 2
\end{array}\right|\)=0-4=-4,
⇒ \(M_{23}=\left|\begin{array}{ll}
1 & 4 \\
0 & 2
\end{array}\right|\)=2-0=2 and
\(M_{23}=\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\)=1-0=1
Minors of elements of the third row are:
⇒ \(M_{31}=\left|\begin{array}{cc}
0 & 4 \\
5 & -1
\end{array}\right|\)=0-20=-20,
⇒ \(M_{32}=\left|\begin{array}{cc}
1 & 4 \\
3 & -1
\end{array}\right|\)=-1-12=-13 and
⇒ \(M_{33}=\left|\begin{array}{ll}
1 & 0 \\
3 & 5
\end{array}\right|\)=5-0=5
Hence, the cofactors of elements of the first row are:
⇒ \(A_{11}=(-1)^{1+1} M_{14}=1 \times 11=11, A_{12}=(-1)^{1+2} M_{12}=(-1) \times 6=-6, A_{13}\)
=\((-1)^{1+3} M_{13}=1 \times\) 3=3
The cofactors of elements of the second row are:
\(A_{21}=(-1)^{2+1} M_{21}=(-1) \times(-4)=4, A_{22}=(-1)^{2+2} M_{22}=1 \times \)2=2
⇒ \(A_{21}=(-1)^{2+3} M_{21}=(-1) \times\) 1=-1
The cofactors of elements of the third row are:
⇒ \(A_{11}=(-1)^{3+1} M_{31}=1 \times(-20)=-20, A_{12}=(-1)^{3+2} M_{12}=(-1) \times(-13)=13\)
⇒ \(A_{31}=(-1)^{3-3} M_{33}=1 \times\) 5=5
Question 3. Using cofactors of the elements of the second row, evaluate \(\Delta=\left|\begin{array}{lll}
5 & 3 & 8 \\
2 & 0 & 1 \\
1 & 2 & 3
\end{array}\right|\)
Solution:
Given, \(\Delta=\left|\begin{array}{lll}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{array}\right|\)
The cofactors of the elements of the second row are:
⇒ \(\mathrm{A}_{21}=(-1)^{2+1}\left|\begin{array}{ll}
3 & 8 \\
2 & 3
\end{array}\right|\)=-(9-16)=7,
⇒ \([\mathrm{A}_4=(-1)^{* / 1} \mathrm{M}_4]\)
⇒ \(\mathrm{A}_{22}=(-1)^{2+2}\left|\begin{array}{ll}
5 & 8 \\
1 & 3
\end{array}\right|\)=(15-8)=7
and \(A_y=(-1)^{2+3}\left|\begin{array}{ll}5 & 3 \\ 1 & 2\end{array}\right|=-(10-3)=-7\)
Now, expansion of \(\Delta\) using cofactors of elements of second row is given by
⇒ \(\Delta=a_{y 1} A_{y t}+a_{y 2} A_{y z}+a_y A_{y y}=(2 \times 7)+(0 \times 7)+(1) \times(-7)=14-7=7\)
Question 4. Using cofactors of elements of the third column, evaluate \(\Delta=\left|\begin{array}{lll}
1 & x & y z \\
1 & y & z x \\
1 & z & x y
\end{array}\right|\)
Solution:
Given, \(\Delta=\left|\begin{array}{lll}1 & x & y z \\ 1 & y & x \\ 1 & z & x y\end{array}\right|\)
The cofactors of the elements of the third column are:
⇒ \(A_{13}=(-1)^{1+3}\left|\begin{array}{ll}
1 & y \\
1 & z
\end{array}\right|=1(z-y)\)=z-y,
⇒ \(A_{23}=(-1)^{2+3}\left|\begin{array}{ll}
1 & x \\
1 & z
\end{array}\right|\)=-1(z-x)=x-z,
⇒ \(A_{x 3}=(-1)^{3+3}\left|\begin{array}{ll}
1 & x \\
1 & y
\end{array}\right|\)=1(y-x)=y-x
Now, the expansion of A using cofactors of elements of the third column is given by
\(\Delta =a_{13} A_{13}+a_{39} A_{23}+a_{33} A_{31}=y z(z-y)+z x(x-z)+x y(y-x)\)=\(y z^2-y^2 z+z x^2-z^2 x+x y^2-x^2 \)
y=\(x^2(z-y)+x\left(y^2-z^2\right)+y z(z-y)\)
= \((z-y)\left\{x^2-x(y+z)+y z\right\}=(z-y)\left\{x^2-x y-x z+y z\right\}\)
= (z-y)[x(x-y)-z(x-y)]=(y-z)(x-y)(z-x)=(x-y)(y-z)(z-x)
Question 5. If \(\Delta=\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|\) and \(A_1\) is cofactor of \(a_{i j}\), then value of \(\Delta\) is given by ?
- \(a_{11} A_{31}+a_{12} A_{32}+a_{19} A_{30}\)
- \(a_{11} A_{11}+a_{12} A_{21}+a_{10} A_n\)
- \(a_{31} A_{11}+a_{23} A_{12}+a_{2 y} A_{13}\)
- \(a_{11} \mathbf{A}_{11}+a_{2 /} A_{24}+a_n A_3\)
Solution: 4. \(a_{11} \mathbf{A}_{11}+a_{2 /} A_{24}+a_n A_3\)
⇒ \(\Delta\) is equal to the sum of the products of the elements of a row (or a column) with their corresponding cofactors.
⇒ \(\Delta=a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13} \text { or } a_{21} A_{21}+a_{22} A_{22}+a_{23} A_{23}\)
or \(a_{12} A_{12}+a_{22} A_{22}+a_{12} A_{32}\) or \(a_{13} A_{13}+a_{23} A_{23}+a_{33} A_{33}\)
Hence, the sum of the products of the elements of the first column with their corresponding cofactors is \(\Delta=a_{11} A_{11}+a_{21} A_{21}+a_{31} A_{31}\)
Hence, the correct option is 4.
Determinants Exercise 4.4
Question 1. \(\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\)
Solution :
Let A=\(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
⇒ \(A_{11}=4, A_{12}=-3, A_{21}\)=-2 and \(A_{22}\)=1
⇒ \({adj} A=\left[A_4\right]=\left[\begin{array}{ll}
A_{11} & A_{22} \\
A_{21} & A_{22}
\end{array}\right]^{\prime}\)
= \(\left[\begin{array}{cc}
4 & -3 \\
-2 & 1
\end{array}\right]=\left[\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right]\)
Question 2. \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
2 & 3 & 5 \\
-2 & 0 & 1
\end{array}\right]\)
Solution:
Let A=\(\left[\begin{array}{ccc}
1 & -1 & 2 \\
2 & 3 & 5 \\
-2 & 0 & 1
\end{array}\right]\)
The cofactors of elements of the first row are:
\(A_{11}=\left|\begin{array}{ll}
3 & 5 \\
0 & 1
\end{array}\right|\)=3-0=3,
⇒ \(A_{12}=-\left|\begin{array}{cc}
2 & 5 \\
-2 & 1
\end{array}\right|=-(2+10)=-12 \text { and } A_{13}=\left|\begin{array}{cc}
2 & 3 \\
-2 & 0
\end{array}\right|\)=0-(-6)=6
The cofactors of elements of the second row are:
⇒ \(A_{21}=-\left|\begin{array}{cc}
-1 & 2 \\
0 & 1
\end{array}\right|=-(-1-0)=1, A_2=\left|\begin{array}{cc}
1 & 2 \\
-2 & 1
\end{array}\right|\)
= \((1+4)=5 and A_{23}=-\left|\begin{array}{cc}
1 & -1 \\
-2 & 0
\end{array}\right|\)=-(0-2)=2
The cofactors of elements of the third row are:
⇒ \(A_{11}=\left|\begin{array}{cc}
-1 & 2 \\
3 & 5
\end{array}\right|=(-5-6)=-11,\)
⇒ \(A_{93}=-\left|\begin{array}{ll}
1 & 2 \\
2 & 5
\end{array}\right|\)=-(5-4)=-1
and \(A_{33}=\left|\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right|\)=3+2=5
Hence,\({adj}(A)=\left[A_0\right]^{\prime}=\left[\begin{array}{ccc}3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5\end{array}\right]^{\prime}=\left[\begin{array}{ccc}3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5\end{array}\right]\)
Question 3. \(\left[\begin{array}{cc}
2 & 3 \\
-4 & -6
\end{array}\right]\)
Solution:
Let A=\(\left[\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right],|A|=\left|\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right|=-12-(-12)\)=-12+12=0
⇒ \(|\mathrm{A}| \mathrm{I}=0\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=\mathrm{O}\)
Cofactors of A are \(A_{11}=-6, A_{12}=4, A_{21}=-3, A_2=2\)
⇒ \({adj}(A)=\left[A_{i 1}\right]=\left[\begin{array}{ll}
-6 & 4 \\
-3 & 2
\end{array}\right]=\left[\begin{array}{cc}
-6 & -3 \\
4 & 2
\end{array}\right]\)
Now, \((adj A) \mathrm{A}=\left[\begin{array}{cc}-6 & -3 \\ 4 & 2\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right]=\left[\begin{array}{cc}-12+12 & -18+18 \\ 8-8 & 12-12\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=\mathrm{0}\)
Also, \(A(adj A)=\left[\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right]\left[\begin{array}{cc}-6 & -3 \\ 4 & 2\end{array}\right]=\left[\begin{array}{cc}-12+12 & -6+6 \\ 24-24 & 12-12\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\)=0
Hence, \(\mathrm{A}({adj} \mathrm{A})=({adj} \mathrm{A}) \mathrm{A}=|\mathrm{A}| \mathrm{I}_2\).
Question 4. \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right]\)
Solution:
Let A=\(\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]\)
Now, \(|A|=\left|\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right|\)=1(0-0)-(-1)(9+2)+2(0-0)=0+11+0=11
⇒ \(|A| I=11\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
11 & 0 & 0 \\
0 & 11 & 0 \\
0 & 0 & 11
\end{array}\right]\)
Cofactors of A are:
⇒ \(A_{11}=0, A_{12}=-(9+2)=-11, A_{13}=0, A_{21}=-(-3-0)=3, A_{22}=3-2=1, A_{33}\)
=-(0+1)=-1,
⇒ \(A_{31}=2-0=2, A_{32}=-(-2-6)=8, A_{39}=0+3=3 \)
⇒ \({adj}(A)=\left[A_7\right]^{\prime}=\left[\begin{array}{ccc}
0 & -11 & 0 \\
3 & 1 & -1 \\
2 & 8 & 3
\end{array}\right]=\left[\begin{array}{ccc}
0 & 3 & 2 \\
-11 & 1 & 8 \\
0 & -1 & 3
\end{array}\right]\)
Now, \(A({adj} A)=\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right]\left[\begin{array}{ccc}
0 & 3 & 2 \\
-11 & 1 & 8 \\
0 & -1 & 3
\end{array}\right]\)
= \(\left[\begin{array}{lll}
0+11+0 & 3-1-2 & 2-8+6 \\
0+0+0 & 9+0+2 & 6+0-6 \\
0+0+0 & 3+0-3 & 2+0+9
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
11 & 0 & 0 \\
0 & 11 & 0 \\
0 & 0 & 11
\end{array}\right]\)
Also, \(({adj} A) A=\left[\begin{array}{ccc}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right]\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]\)
= \(\left[\begin{array}{ccc}
0+9+2 & 0+0+0 & 0-6+6 \\
-11+3+8 & 11+0+0 & -22-2+24 \\
0-3+3 & 0+0+0 & 0+2+9
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
11 & 0 & 0 \\
0 & 11 & 0 \\
0 & 0 & 11
\end{array}\right]\)
=\(11\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Hence, \(\mathrm{A}({adj} \mathrm{A})=({adj} \mathrm{A}) \mathrm{A}=|\mathrm{A}| \mathrm{I}\)
Question 5. \(\left[\begin{array}{cc}
2 & -2 \\
4 & 3
\end{array}\right]\)
Solution:
Let A=\(\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right].\)
We have, \(|\mathrm{A}|=\left|\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right|=6-(-8)=14 \neq 0 \mathrm{A}^{-1}\) exists
Cofactors of \(\mathrm{A}_{11} are \mathrm{A}_{11}\)=3,
⇒ \(\mathrm{~A}_{12}=-4, \mathrm{~A}_{21}=2, \mathrm{~A}_{22}=2\)
⇒ \({adj}(\mathrm{A})=\left[\mathrm{A}_{i j}\right]^{\prime}=\left[\begin{array}{cc}
3 & -4 \\
2 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
3 & 2 \\
-4 & 2
\end{array}\right]\)
Now, \(A^{-1}=\frac{1}{|A|}({adj} A)=\frac{1}{14}\left[\begin{array}{cc}
3 & 2 \\
-4 & 2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\frac{3}{14} & \frac{2}{14} \\
-\frac{4}{14} & \frac{2}{14}
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\frac{3}{14} & \frac{1}{7} \\
-\frac{2}{7} & \frac{1}{7}
\end{array}\right]\)
Question 6. \(\left[\begin{array}{ll}
-1 & 5 \\
-3 & 2
\end{array}\right]\)
Solution:
Let A=\(\left[\begin{array}{ll}-1 & 5 \\ -3 & 2\end{array}\right]\).
We have, \(|A|==-2-(-15)=13 \neq\) 0
\(\mathrm{A}^{-1}\) exists
Now, cofactors of A are \(A_{11}=2, A_{12}=3, A_{21}=-5, A_{21}\)=-1
⇒ \({adj}(A)=\left[A_0\right]^{\prime}=\left[\begin{array}{cc}
2 & 3 \\
-5 & -1
\end{array}\right]^{\prime}=\left[\begin{array}{ll}
2 & -5 \\
3 & -1
\end{array}\right]\)
Now, \(A^{-1}=\frac{1}{|A|}({adj} A)=\frac{1}{13}\left[\begin{array}{cc}2 & -5 \\ 3 & -1\end{array}\right]\)
= \(\left[\begin{array}{cc}\frac{2}{13} & -\frac{5}{13} \\ \frac{3}{13} & -\frac{1}{13}\end{array}\right]\)
Question 7. \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
Solution:
Let A=\(\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]\)
We have, |\(\mathrm{A}|=1(10-0)-2(0-0)+3(0-0)=10 \neq 0\)
⇒ \(\mathrm{A}^{-1}\) exists
Now, cofactors of A are
⇒ \(A_{11}\)=10-0=10, \(A_{12}\)=-(0-0)=0, \(A_{13}\)=0-0=0,
⇒ \(A_{21}\)=-(10-0)=-10, \(A_{22}\)=5-0=5, \(A_{23}\)=-(0-0)=0 ,
⇒ \(\mathrm{A}_{31}=8-6=2, \mathrm{~A}_{32}=-(4-0)=-4, \quad \mathrm{~A}_{35}=2-0=2 \)
⇒ \({adj}(\mathrm{A})=\left[\mathrm{A}_{i j}\right]^{\prime}=\left[\begin{array}{ccc}
10 & 0 & 0 \\
-10 & 5 & 0 \\
2 & -4 & 2
\end{array}\right]^{\prime}\)
= \(\left[\begin{array}{ccc}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
\end{array}\right] \)
Now, \(A^{-1}=\frac{1}{|A|}({adj} A)=\frac{1}{10}\left[\begin{array}{ccc}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
1 & -1 & \frac{1}{5} \\
0 & \frac{1}{2} & -\frac{2}{5} \\
0 & 0 & \frac{1}{5}
\end{array}\right]\).
Question 8. \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
3 & 3 & 0 \\
5 & 2 & -1
\end{array}\right]\)
Solution:
Let A=\(\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1\end{array}\right]\)
⇒ \(|A|=1(-3-0)-0+0=-3 \neq\) 0
⇒ \(\mathrm{A}^{-1}\) exists
Cofactors of A are:
⇒ \(A_{11}=-3-0=-3, A_{12}=-(-3-0)=3, A_{13}=6-15=-9\),
⇒ \(A_{21}=-(0-0)=0, A_{22}=-1-0=-1, A_{23}=-(2-0)=-2\),
⇒ \(A_{31}=0-0=0, A_{12}=-(0-0)=0, A_{33}\)=3-0=3
⇒ \({adj}(A)=\left[A_{i 1}\right]^{\prime}=\left[\begin{array}{ccc}
-3 & 3 & -9 \\
0 & -1 & -2 \\
0 & 0 & 3
\end{array}\right]^{\prime}\)
= \(\left[\begin{array}{ccc}
-3 & 0 & 0 \\
3 & -1 & 0 \\
-9 & -2 & 3
\end{array}\right]\)
Now, \(A^{-1}=\frac{1}{|A|}({adj} A)=\frac{1}{-3}\left[\begin{array}{ccc}
-3 & 0 & 0 \\
3 & -1 & 0 \\
-9 & -2 & 3
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
-1 & \frac{1}{3} & 0 \\
3 & \frac{2}{3} & -1
\end{array}\right]\)
Question 9. \([\left[\begin{array}{ccc}
2 & 1 & 3 \\
4 & -1 & 0 \\
-7 & 2 & 1
\end{array}\right]\)
Solution:
Let A=\(\left[\begin{array}{ccc}2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1\end{array}\right]\)
|A|=2(-1-0)-1(4-0)+3(8-7)=-2-4+3=-3 \(\neq\) 0
⇒ \(\mathrm{A}^{-1}\) exists
Cofactors of A are:
⇒ \(A_{11}=-1-0=-1, A_{12}=-(4-0)=-4, A_{13}=8-7=1\)
⇒ \(A_{21}=-(1-6)=5, A_{22}=2+21=23, \quad A_{33}\)=-(4+7)=-11,
⇒ \(A_{31}=0+3=3, A_{32}=-(0-12), A_{33}\)=-2-4=-6
⇒ \({adj}(A)=\left[A_{61}\right]^{\prime}=\left[\begin{array}{ccc}
-1 & -4 & 1 \\
5 & 23 & -11 \\
3 & 12 & -6
\end{array}\right]\)
=\(\left[\begin{array}{ccc}
-1 & 5 & 3 \\
-4 & 23 & 12 \\
1 & -11 & -6
\end{array}\right]\)
Now, \(A^{-1}=\frac{1}{|A|}({adj}A)\)
= \(\frac{1}{-3}\left[\begin{array}{ccc}
-1 & 5 & 3 \\
-4 & 23 & 12 \\
1 & -11 & -6
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
\frac{1}{3} & -\frac{5}{3} & -1 \\
\frac{4}{3} & -\frac{23}{3} & -4 \\
-\frac{1}{3} & \frac{11}{3} & 2
\end{array}\right]\)
Question 10. \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]\)
Solution:
Let A=\(\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\)
We have, \(|A|=\left|\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right|\)
=1(8-6)-(-1)(0+9)+2(0-6)=2+9-12=-1 \(\neq\) 0
⇒ \(\mathrm{A}^{-1}\) exists
Cofactors of A are:
⇒ \(A_{11}=8-6=2, A_{12}=-(0+9)=-9, A_{12}=0-6=-6\),
⇒ \(A_{21}=-(-4+4)=0, A_{22}=4-6=-2, A_{23}=-(-2+3)=-1\)
⇒ \(A_{31}=3-4=-1, A_{22}=-(-3-0)=3, A_{33}=2-0=2\)
⇒ \({adj}(A)=\left[A_4\right]^{+}=\left[\begin{array}{ccc}
2 & -9 & -6 \\
0 & -2 & -1 \\
-1 & 3 & 2
\end{array}\right]^{\prime}=\left[\begin{array}{ccc}
2 & 0 & -1 \\
-9 & -2 & 3 \\
-6 & -1 & 2
\end{array}\right]\)
Now; \(A^{-1}=\frac{1}{|A|}({adj} A)=\frac{1}{-1}\left[\begin{array}{ccc}
2 & 0 & -1 \\
-9 & -2 & 3 \\
-6 & -1 & 2
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right]\)
Question 11. \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos \alpha & \sin \alpha \\
0 & \sin \alpha & -\cos \alpha
\end{array}\right]\)
Solution:
Let A=\(\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha\end{array}\right] \Rightarrow|A|\)
= \(\left|\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & -\cos \alpha\end{array}\right|=1\left(-\cos ^2 \alpha-\sin ^2 \alpha\right)\)
=-\(\left(\cos ^2 \alpha+\sin ^2 \alpha\right)=-1 \neq 0 \left[ \cos ^2 \theta+\sin ^2 \theta=1\right]\)
⇒ \(\mathrm{A}^{-1}\) exists
Cofactors of A are:
⇒ \(\mathrm{A}_{11}=-\cos ^2 \alpha-\sin ^2 \alpha\)=-1,
⇒ \(\mathrm{~A}_{12}=-(0-0)=0, \mathrm{~A}_{13}\)=0-0=0
⇒ \(\mathrm{~A}_{21}=-(0-0)=0, \mathrm{~A}_{22}=-\cos \alpha-0=-\cos \alpha, \mathrm{A}_{23}=-(\sin \alpha-0)=-\sin \alpha,\)
⇒ \(\mathrm{A}_{21}=0-0=0, \mathrm{~A}_{32}=-(\sin \alpha-0)=-\sin \alpha, \mathrm{A}_{35}=\cos \alpha-0=\cos \alpha \)
⇒ \({adj}(\mathrm{A})=\left[\mathrm{A}_{1 j}\right]^{\prime}=\left[\begin{array}{ccc}
-1 & 0 & 0 \\
0 & -\cos \alpha & -\sin \alpha \\
0 & -\sin \alpha & \cos \alpha
\end{array}\right]^{\prime}\)
= \(\left[\begin{array}{ccc}
-1 & 0 & 0 \\
0 & -\cos \alpha & -\sin \alpha \\
0 & -\sin \alpha & \cos \alpha
\end{array}\right]\)
Now, \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}({adj} \mathrm{A})=\frac{1}{-1}\left[\begin{array}{ccc}
-1 & 0 & 0 \\
0 & -\cos \alpha & -\sin \alpha \\
0 & -\sin \alpha & \cos \alpha
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos \alpha & \sin \alpha \\
0 & \sin \alpha & -\cos \alpha
\end{array}\right]\)
Question 12. Let A=\(\left[\begin{array}{ll}
3 & 7 \\
2 & 5
\end{array}\right]\) and\( \mathrm{B}=\left[\begin{array}{ll}
6 & 8 \\
7 & 9
\end{array}\right]\). Verify that \((A B)^{-1}=B^{-1} A^{-1}\).
Solution:
Given, A=\(\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right] ;|A|=\left|\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right|=15-14=1 \neq 0\)
⇒ \(\mathrm{A}^{-1}\) exists
Cofactors of A are \(\mathrm{A}_{11}\)=5,
⇒ \(\mathrm{~A}_{12}=-2, \mathrm{~A}_{21}=-7, \mathrm{~A}_{27}\)=3
⇒ \({adj}(A)=\left[A_{i j}\right]^{\prime}=\left[\begin{array}{cc}5 & -2 \\ -7 & 3\end{array}\right]^{\prime}=\left[\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right]\)
Now, \(A^{-1}=\frac{1}{|A|}({adj} A)=\frac{1}{1}\left[\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right]=\left[\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right] \)
Here, B=\(\left[\begin{array}{ll}6 & 8 \\ 7 & 9\end{array}\right] |B|=\left|\begin{array}{ll}6 & 8 \\ 7 & 9\end{array}\right|=54-56=-2 \neq \)
⇒ \(\mathrm{B}^{-1}\) exists
Cofactors of B are \(\mathrm{B}_{11}=9, \mathrm{~B}_{22}=-7, \mathrm{~B}_{21}=-8, \mathrm{~B}_{22}\)=6
⇒ \({adj}(B)=\left[B_{i j}\right]^{\prime}=\left[\begin{array}{cc}
9 & -7 \\
-8 & 6
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
9 & -8 \\
-7 & 6
\end{array}\right]\)
⇒ \(B^{-1}=\frac{1}{|B|}(\text { adj } B)=\frac{1}{-2}\left[\begin{array}{cc}
9 & -8 \\
-7 & 6
\end{array}\right]\)
Now,\(B^{-1} A^{-1} =\frac{1}{-2}\left[\begin{array}{cc}
9 & -8 \\
-7 & 6
\end{array}\right]\left[\begin{array}{cc}
5 & -7 \\
-2 & 3
\end{array}\right]\)
=\(\frac{1}{-2}\left[\begin{array}{cc}
45+16 & -63-24 \\
-35-12 & 49+18
\end{array}\right] \)
=\(\frac{1}{-2}\left[\begin{array}{cc}
61 & -87 \\
-47 & 67
\end{array}\right]\)
=\(\left[\begin{array}{cc}
-\frac{61}{2} & \frac{87}{2} \\
\frac{47}{2} & -\frac{67}{2}
\end{array}\right]\) Equation 1
Now, let \(\mathrm{C}=\mathrm{AB}=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]\left[\begin{array}{ll}6 & 8 \\ 7 & 9\end{array}\right]\)
=\(\left[\begin{array}{ll}18+49 & 24+63 \\ 12+35 & 16+45\end{array}\right]\)
=\(\left[\begin{array}{ll}67 & 87 \\ 47 & 61\end{array}\right]\)
⇒ \(|\mathrm{AB}|=\left|\begin{array}{ll}
67 & 87 \\
47 & 61
\end{array}\right|\)
=\((67 \times 61)-(47 \times 87)=4087-4089=-2 \neq 0\)
∴ \(\mathrm{C}^{-1} exists\),
Cofactors of C are \(C_{11}=61, C_{12}=-47, C_{21}=-87, C_{22}=67\)
⇒ \({adj}(\mathrm{AB})=\left[\mathrm{C}_{\mathrm{ii}}\right]^{+}=\left[\begin{array}{cc}
61 & -47 \\
-87 & 67
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
61 & -87 \\
-47 & 67
\end{array}\right]\)
⇒ \((\mathrm{AB})^{-1}=\frac{1}{|\mathrm{AB}|}(\mathrm{adj} \mathrm{AB})=\frac{1}{-2}\left[\begin{array}{cc}
61 & -87 \\
-47 & 67
\end{array}\right]\)
= \(\left[\begin{array}{cc}
-\frac{61}{2} & \frac{87}{2} \\
\frac{47}{2} & -\frac{67}{2}
\end{array}\right]\) → Equation 2
From Eq. (1) and (2), we get \((A B)^{-1}=B^{-1} A^{-1}\)
Hence, the given result is proved.
Question 13. If A=\(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\), show that \(A^2-5 A+7 I=0\). Hence, find \(A^{-1}\).
Solution:
Given, A=\(\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\)
⇒ \(A^2=A \cdot A=\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
9-1 & 3+2 \\
-3-2 & -1+4
\end{array}\right]=\left[\begin{array}{cc}
8 & 5 \\
-5 & 3
\end{array}\right]\)
Now,\( A^2-5 A+7 I\)
= \(\left[\begin{array}{cc}
8 & 5 \\
-5 & 3
\end{array}\right]-5\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]+7\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
8 & 5 \\
-5 & 3
\end{array}\right]-\left[\begin{array}{cc}
15 & 5 \\
-5 & 10
\end{array}\right]+\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right]\)
= \(\left[\begin{array}{cc}
8-15+7 & 5-5+0 \\
-5+5+0 & 3-10+7
\end{array}\right]\)
=\(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)=0
\(A^2-5 A+7 I\)=0
⇒ \(|A|=\left|\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right|=6+1=7 \neq \)0, \(A^{-1}\) exists.
Now, A \(\cdot\) A-5 A=-71
Post multiplying by \(\mathrm{A}^{-1}\) on both sides, we get
A \(\cdot A\left(A^{-1}\right)-5 A^{-1}=-7 A^{-1} \Rightarrow A I-5 I=-7 A^{-1}\)
Using \(A^{-1}=I\) and \(I^{-1}=A^{-1}\)
⇒ \(A^{-1}=-\frac{1}{7}(A-5 I) \Rightarrow A^{-1}=\frac{1}{7}(5 I-A)=\frac{1}{7}\left(\left[\begin{array}{ll}
5 & 0 \\
0 & 5
\end{array}\right]-\left[\begin{array}{cc}
3 & -1 \\
-1 & 2
\end{array}\right]\right)\)
=\(\frac{1}{7}\left[\begin{array}{cc}
2 & -1 \\
1 & 3
\end{array}\right] \)
⇒ \(A^{-1}=\frac{1}{7}\left[\begin{array}{cc}
2 & -1 \\
1 & 3
\end{array}\right]\)
Question 14. For the matrix A=\(\left[\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right]\), find the numbers ‘ a ‘ and ‘ b ‘ such that \(A^2+a A+b I\)=0.
Solution:
Given, A=\(\left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right]\)
⇒ \(A^2=A \cdot A=\left[\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right]\left[\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right]\)
=\(\left[\begin{array}{ll}
9+2 & 6+2 \\
3+1 & 2+1
\end{array}\right]=\left[\begin{array}{cc}
11 & 8 \\
4 & 3
\end{array}\right]\)
Given, \(\mathrm{A}^2+\mathrm{aA}+\mathrm{bI}\)=0
On putting the values of \(\mathrm{A}^2, \mathrm{~A}\) and 1 ; we get
⇒ \(\left[\begin{array}{ll}
11 & 8 \\
4 & 3
\end{array}\right]+\mathrm{a}\left[\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right]+\mathrm{b}\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)=0
⇒ \(\left[\begin{array}{cc}
11+3 a+b & 8+2 a+0 \\
4+a+0 & 3+a+b
\end{array}\right]=0 \)
⇒ \({\left[\begin{array}{cc}
11+3 a+b & 8+2 a \\
4+a & 3+a+b
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right] }\)
If two matrices are equal, then their corresponding elements are also equal.
11+3a+b=0 → Equation 1
8+2 a=0 → Equation 2
4+a=0 → Equation 3
and 3+a+b=0 → Equation 4
Solving Eq. (3) and (4), we get 4+\(\mathrm{a}=0 \Rightarrow \mathrm{a}\)=-4
And 3+a+b=0 \(\Rightarrow 3-4+b=0 \Rightarrow\) b=1
Thus, a=-4 and b=1
Question 15. For the matrix A=\(\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]\), show that \(A^3-6 A^2+5 A+111\)=0. Hence, find \(A^{-1}\),
Solution:
Given, A=\(\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right] ;|A|=1(6-3)-1(3+6)+1(-1-4)-3-9-5=-11 \neq 0\)
⇒ \(\mathrm{A}^{-1}\) exists
Now,\(\mathbf{A}^2=\mathbf{A} \cdot \mathrm{A} =\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]\)
=\(\left[\begin{array}{ccc}
1+1+2 & 1+2-1 & 1-3+3 \\
1+2-6 & 1+4+3 & 1-6-9 \\
2-1+6 & 2-2-3 & 2+3+9
\end{array}\right]=\left[\begin{array}{ccc}
4 & 2 & -1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]\)
And \(A^3=A^2 \cdot A =\left[\begin{array}{ccc}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]\)
=\(\left[\begin{array}{ccc}
4+2+2 & 4+4-1 & 4-6+3 \\
-3+8-28 & -3+16+14 & -3-24-42 \\
7-3+28 & 7-6-14 & 7+9+42
\end{array}\right]\)
=\(\left[\begin{array}{ccc}
8 & 7 & 1 \\
-23 & 27 & -69 \\
32 & -13 & 58
\end{array}\right]\)
⇒ \(A^3-6 A^2+ \text { A + } 111 \\
=\left[\begin{array}{ccc}
8 & 7 & 1 \\
-23 & 27 & -69 \\
32 & -13 & 58
\end{array}\right]-6\left[\begin{array}{ccc}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]+5\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]+11\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
=\(\left[\begin{array}{ccc}
8 & 7 & 1 \\
-23 & 27 & -69 \\
32 & -13 & 58
\end{array}\right]-\left[\begin{array}{ccc}
24 & 12 & 6 \\
-18 & 48 & -84 \\
42 & -18 & 84
\end{array}\right]+\left[\begin{array}{ccc}
5 & 5 & 5 \\
5 & 10 & -15 \\
10 & -5 & 15
\end{array}\right]+\left[\begin{array}{ccc}
11 & 0 & 0 \\
0 & 11 & 0 \\
0 & 0 & 11
\end{array}\right]\)
=\(\left[\begin{array}{ccc}
8-24+5+11 & 7-12+5+0 & 1-6+5+0 \\
-23+18+5+0 & 27-48+10+11 & -69+84-15+0 \\
32-42+10+0 & -13+18-5+0 & 58-84+15+11
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)=0
Now, \(A^3-6 A^2+5 A+111\)=0
⇒ \((A A A) A^{-1}-6(A A) A^{-1}+5 A A^{-1}+111 A^{-1}\)=0 (Post-multiplying by }\( A^{-t} as |A| \neq 0)\)
⇒ \(A A\left(A A^{-1}\right)-6 A\left(A^{-1}\right)+5\left(A^{-1}\right)+11\left(\mathrm{IA}^{-1}\right)\)=0
⇒ \(\mathrm{AAI}-6 \mathrm{AI}+5 \mathrm{I}+11 \mathrm{~A}^{-1}=0 (Using \mathrm{AA}^{-1}=\mathrm{I} and\mathrm{IA}^{-1}=\mathrm{A}^{-1}\)
⇒ \(\mathrm{A}^2-6 \mathrm{~A}+5 \mathrm{I}=-11 \mathrm{~A}^{-1} \)
⇒ \((Using \mathrm{AAI}=\mathrm{A}^2 and \mathrm{AI}=\mathrm{A}\)
⇒ \(A^{-1}=-\frac{1}{11}\left(A^2-6 A+51\right) \Rightarrow A^{-1}=\frac{1}{11}\left(-A^2+6 A-5 I\right)\)
=\(\frac{1}{11}\left\{-\left[\begin{array}{ccc}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]+6\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]-5\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\right\} \)
= \(\frac{1}{11}\left\{\left[\begin{array}{ccc}
-4 & -2 & -1 \\
3 & -8 & 14 \\
-7 & 3 & -14
\end{array}\right]+\left[\begin{array}{ccc}
6 & 6 & 6 \\
6 & 12 & -18 \\
12 & -6 & 18
\end{array}\right]-\left[\begin{array}{ccc}
5 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 5
\end{array}\right]\right\}\)
=\(\frac{1}{11}\left[\begin{array}{ccc}
-4+6-5 & -2+6-0 & -1+6-0 \\
3+6-0 & -8+12-5 & 14-18-0 \\
-7+12-0 & 3-6-0 & -14+18-5
\end{array}\right]\)
=\(\frac{1}{11}\left[\begin{array}{ccc}
-3 & 4 & 5 \\
9 & -1 & -4 \\
5 & -3 & -1
\end{array}\right]\)
Question 16. If A=\(\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]\) , verify that \(A^3-6 A^2+9 A-4\) I=O and hence, find \(A^{-1}\)
Solution:
Given, A=\(\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right] \)
⇒\(A^2=A \cdot A=\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
4+1+1 & -2-2-1 & 2+1+2 \\
-2-2-1 & 1+4+1 & -1-2-2 \\
2+1+2 & -1-2-2 & 1+1+4
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
6 & -5 & 5 \\
-5 & 6 & -5 \\
5 & -5 & 6
\end{array}\right]\)
and \(A^3=A^2 \cdot A =\left[\begin{array}{ccc}
6 & -5 & 5 \\
-5 & 6 & -5 \\
5 & -5 & 6
\end{array}\right]\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
12+5+5 & -6-10-5 & 6+5+10 \\
-10-6-5 & 5+12+5 & -5-6-10 \\
10+5+6 & -5-10-6 & 5+5+12
\end{array}\right]\)
=\(\left[\begin{array}{ccc}
22 & -21 & 21 \\
-21 & 22 & -21 \\
21 & -21 & 22
\end{array}\right]\)
⇒ \(A^3-6 A^2 +9 A-41\)
= \(\left[\begin{array}{ccc}
22 & -21 & 21 \\
-21 & 22 & -21 \\
21 & -21 & 22
\end{array}\right]-6\left[\begin{array}{ccc}
6 & -5 & 5 \\
-5 & 6 & -5 \\
5 & -5 & 6
\end{array}\right]+9\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]-4\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
22 & -21 & 21 \\
-21 & 22 & -21 \\
21 & -21 & 22
\end{array}\right]-\left[\begin{array}{ccc}
36 & -30 & 30 \\
-30 & 36 & -30 \\
30 & -30 & 36
\end{array}\right]+\left[\begin{array}{ccc}
18 & -9 & 9 \\
-9 & 18 & -9 \\
9 & -9 & 18
\end{array}\right]-\left[\begin{array}{ccc}
4 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 4
\end{array}\right]\)
=\(\left[\begin{array}{ccc}
22-36+18-4 & -21+30-9-0 & 21-30+9-0 \\
-21+30-9-0 & 22-36+18-4 & -21+30-9-0 \\
21-30+9-0 & -21+30-9-0 & 22-36+18-4
\end{array}\right]\)
= \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)=0
⇒ \(A^3-6 A^2+9 A-4 I=0 \Rightarrow(A A A) A^{-1}-6(A A) A^{-1}+9 A^{-1}-4 I A^{-1}\)=0
(Post-multiplying by } \(\mathrm{A}^{-1} \text { as }|\mathrm{A}| \neq 0 \)
⇒ \([|A|=|\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}|\)=2(4-1)+1(-2+1)+1(1-2)=6-1-1=4 \(\neq 0\)
⇒ \(\mathrm{AA}\left(\mathrm{AA}^{-1}\right)-6 \mathrm{~A}\left(\mathrm{AA}^{-1}\right)+9\left(\mathrm{AA}^{-1}\right)-4\left(\mathrm{IA}^{-1}\right)\)=0
⇒ \(\mathrm{AAI}-6 \mathrm{AI}+9 \mathrm{I}-4 \mathrm{~A}^{-1}=0\)
(Using }\(\mathrm{AA}^{-1}= I\) and \(1 \mathrm{~A}^{-1}=\mathrm{A}^{-1}\)
⇒ \(A^2-6 A+9 I=4 A^{-1}\)
(Using \(A^2 T=A^2\) and A\(\mathrm{~A}^{-1}\)=A )
⇒ \(A^{-1}=\frac{1}{4}\left(A^2-6 A+91\right)\)
⇒ \(\frac{1}{4}\left\{\left[\begin{array}{ccc}
6 & -5 & 5 \\
-5 & 6 & -5 \\
5 & -5 & 6
\end{array}\right]-6\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]+9\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\right\}\)
⇒ \(\frac{1}{4}\left\{\left[\begin{array}{ccc}
6 & -5 & 5 \\
-5 & 6 & -5 \\
5 & -5 & 6
\end{array}\right]-\left[\begin{array}{ccc}
12 & -6 & 6 \\
-6 & 12 & -6 \\
6 & -6 & 12
\end{array}\right]+\left[\begin{array}{ccc}
9 & 0 & 0 \\
0 & 9 & 0 \\
0 & 0 & 9
\end{array}\right]\right\}\)
=\(\frac{1}{4}\left[\begin{array}{ccc}
6-12+9 & -5+6+0 & 5-6+0 \\
-5+6+0 & 6-12+9 & -5+6+0 \\
5-6+0 & -5+6+0 & 6-12+9
\end{array}\right]\)
=\(\frac{1}{4}\left[\begin{array}{ccc}
3 & 1 & -1 \\
1 & 3 & 1 \\
-1 & 1 & 3
\end{array}\right]\)
Question 17. Let A be anon-singular square matrix of order 3×3, then |adj A| equal to?
- \(|\mathrm{A}|\)
- \(|\mathrm{A}|^2\)
- \(|\mathrm{A}|\)
- 3\(|\mathrm{~A}|\)
Solution: 2. \(|\mathrm{A}|^2\)
We know that \(({adj} \mathrm{A}) \mathrm{A}=|\mathrm{A}| \mathrm{I}\)
= \(|A|\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
|A| & 0 & 0 \\
0 & |A| & 0 \\
0 & 0 & |A|
\end{array}\right]\)
⇒ \(|({adj} A) A|=\left|\begin{array}{ccc}
A & 0 & 0 \\
0 & |A| & 0 \\
0 & 0 & |A|
\end{array}\right|\)
=\(|A|^3\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|=|A|^3|\mathrm{I}|\)
⇒ \(|{adj} A| A|=| A|^3 (|I|=1)\)
⇒ \(|{adj} A|=|A|^2\)
Hence, the correct option is 2
Question 18. If A is an invertible matrix of order 2, then det \(\left(\mathrm{A}^{-1}\right)\) is equal to?
- \({det}(\mathrm{A})\)
- \(\frac{1}{det}(\mathrm{A}\)
- 1
- zero
Solution:
We know that \(\mathrm{AA}^{-1}=\mathrm{I}\)
⇒ \(|\mathrm{AA}^{-1}|=|\mathrm{I}| \Rightarrow|\mathrm{A}|\mathrm{A}^{-1}|\)=1
⇒ {Using\(|\mathrm{AA}^{-1}|=|\mathrm{A}||\mathrm{A}^{-1}|\) and \(|\mathrm{I}|=1)\)
⇒ \(|\mathrm{A}^{-1}|=\frac{1}{|\mathrm{~A}|}=\frac{1}{{det}(\mathrm{A})}\) .
Hence, the correct option is (B).
Determinants Exercise 4.5
Question 1. x+2 y=2,2 x+3 y=3.
Solution:
The given system can be written as AX= B, where
⇒ \(\mathrm{A}=\left[\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right], \mathrm{X}=\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y}
\end{array}\right] \)
and \(\mathrm{B}=\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)
Here, \(|\mathrm{A}|=\left|\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right|=1(3)-2(2)=3-4=-1 \neq\) 0
A is non-singular. Therefore, \(\mathrm{A}^{-1}\) exists.
Hence, the given system of equations is consistent.
Question 2. 2 x-y=5, x+y=4
Solution:
The given system can be written as A X=B, where
A=\(\left[\begin{array}{cc}
2 & -1 \\
1 & 1
\end{array}\right], X=\left[\begin{array}{l}
x \\
y
\end{array}\right] \text { and } B=\left[\begin{array}{l}
5 \\
4
\end{array}\right]\)
Here, \(|A|=\left|\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right|=2(1)-(-1)(1)=2+1=3 \neq \)
A is non-singular. Therefore, \(\mathrm{A}^{-1}\) exists.
Hence, the given system of equations is consistent.
Question 3. x+3 y=5,2 x+6 y=8
Solution:
The given system can be written as AX = B, where
A=\(\left[\begin{array}{ll}
1 & 3 \\
2 & 6
\end{array}\right], X=\left[\begin{array}{l}
x \\
y
\end{array}\right] \text { and } B=\left[\begin{array}{l}
5 \\
8
\end{array}\right]\)
Here, \(|A|=\left|\begin{array}{ll}1 & 3 \\ 2 & 6\end{array}\right|=1(6)-3(2)=6-6=0\)
A is a singular matrix.
Nothing can be said about consistency as yet. We compute
⇒ \(({adj} \mathrm{A}) \mathrm{B} =\left[\begin{array}{cc}
6 & -2 \\
-3 & 1
\end{array}\right]\left[\begin{array}{l}
5 \\
8
\end{array}\right]\)
=\(\left[\begin{array}{cc}
6 & -3 \\
-2 & 1
\end{array}\right]\left[\begin{array}{l}
5 \\
8
\end{array}\right]=\left[\begin{array}{c}
30-24 \\
-10+8
\end{array}\right]\)
= \(\left[\begin{array}{c}
6 \\
-2
\end{array}\right] \neq 0\)
Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.
Question 4. x+y+z=1,2 x+3 y+2 z=2, a x+a y+2 a z=4
Solution:
The given system can be written as AX – B, where
A=\(\left[\begin{array}{ccc}
1 & 1 & 1 \\
2 & 3 & 2 \\
a & a & 2 a
\end{array}\right], X=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and
B=\(\left[\begin{array}{l}
1 \\
2 \\
4
\end{array}\right]\)
Here,|A| =\(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 3 & 2 \\
a & a & 2 a
\end{array}\right|=1(6 a-2 a)-1(4 a-2 a)+1(2 a-3 a)\)
=4 a-2 a-a=4 a-3 a=a \(\neq 0\)
A is non-singular. Therefore, \(\mathrm{A}^{-1}\) exists.
Hence, the given system of equations is consistent.
Question 5. 3 x-y-2 z=2,2 y-z=-1,3 x-5 y=3
Solution:
The given system is 3 x-y-2 z=2,0 x+2 y-z=-1 and 3 x-5 y+0 z=3 which can be written as AX =B, where
A=\(\left[\begin{array}{ccc}
3 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{array}\right], X=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and
B=\(\left[\begin{array}{c}
2 \\
-1 \\
3
\end{array}\right]\)
Here, \(|A|=\left|\begin{array}{ccc}3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right|=3(0-5)+1(0+3)-2(0-6)\)=-15+3+12=0
A is a singular matrix.
Therefore, nothing can be said about consistency as yet. So, we compute (adj A)B.
Cofactors of A are:
⇒ \(A_{11}=-5, A_{12}=-3, A_{13}\)=-6,
⇒ \(A_{21}=10, A_{22}=6, A_{23}=12\),
⇒ \(A_{91}=5, A_{32}=3, A_{33}\)=6
⇒ \({adj}(A)=\left[A_4\right]^{\prime}=\left[\begin{array}{ccc}
-5 & -3 & -6 \\
10 & 6 & 12 \\
5 & 3 & 6
\end{array}\right]=\left[\begin{array}{ccc}
-5 & 10 & 5 \\
-3 & 6 & 3 \\
-6 & 12 & 6
\end{array}\right]\)
⇒ \(({adj} A) B=\left[\begin{array}{ccc}
-5 & 10 & 5 \\
-3 & 6 & 3 \\
-6 & 12 & 6
\end{array}\right] \cdot\left[\begin{array}{c}
2 \\
-1 \\
3
\end{array}\right]\)
= \(\left[\begin{array}{c}
-10-10+15 \\
-6-6+9 \\
-12-12+18
\end{array}\right]=\left[\begin{array}{c}
-5 \\
-3 \\
-6
\end{array}\right] \neq 0\)
Thus, the solution of the given system of equations does not exist.
Hence, the system of equations is inconsistent.
Question 6. 5 x-y+4 z=5,2 x+3 y+5 z=2,5 x-2 y+6 z=-1
Solution:
The given system can be written as A X=B, where
⇒ \(\mathbf{A}=\left[\begin{array}{ccc}
5 & -1 & 4 \\
2 & 3 & 5 \\
5 & -2 & 6
\end{array}\right]\),
X=\(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
and \(B=\left[\begin{array}{c}
5 \\
2 \\
-1
\end{array}\right]\)
Here, \(|A|=\left|\begin{array}{ccc}5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6\end{array}\right|\)
=5(18+10)-(-1)(12-25)+4(-4-15)=140-13-76=51 \(\neq 0\)
A is non-singular.
Therefore, \(\mathrm{A}^{-1}\) exists.
Hence, the given system of equations is consistent.
Question 7. 5 x+2 y=4,7 x+3 y=5
Solution:
The given system can be written as AX = B, where
⇒ \(\mathrm{A}=\left[\begin{array}{ll}5 & 2 \\ 7 & 3\end{array}\right], X=\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y}\end{array}\right] \)
and \(\mathrm{B}=\left[\begin{array}{l}4 \\ 5\end{array}\right]\)
Here, \(|\mathrm{A}|=\left|\begin{array}{ll}5 & 2 \\ 7 & 3\end{array}\right|=15-14=1 \neq 0\)
Thus, A is non-singular.
Therefore, its \(\mathrm{A}^{-1}\) exists.
Therefore, the given system is consistent and has a unique solution given by
⇒ \(\mathrm{A}^{-1}(\mathrm{AX})=\mathrm{A}^{-1} \mathrm{~B} \Rightarrow \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}\)
Cofactors of A are, \(A_{11}=3, A_{12}=-7, A_{21}=-2, A_{22}=5\)
⇒ \({adj}(A)=\left[A_{i j}\right]^{\prime}=\left[\begin{array}{cc}
3 & -7 \\
-2 & 3
\end{array}\right]^{\prime}=\left[\begin{array}{cc}
3 & -2 \\
-7 & 5
\end{array}\right]\)
Now, \(A^{-1}=\frac{1}{|A|}({adj} A)=\frac{1}{1}\left[\begin{array}{cc}3 & -2 \\ -7 & 5\end{array}\right]=\left[\begin{array}{cc}3 & -2 \\ -7 & 5\end{array}\right]\)
⇒ \(\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}=\left[\begin{array}{cc}
3 & -2 \\
-7 & 5
\end{array}\right]\left[\begin{array}{l}
4 \\
5
\end{array}\right]=\left[\begin{array}{l}
3 \times 4+(-2) \times 5 \\
(-7) \times 4+5 \times 5
\end{array}\right]\)
= \(\left[\begin{array}{c}
2 \\
-3
\end{array}\right] \Rightarrow\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
2 \\
-3
\end{array}\right]\)
Hence, x=2 and y=-3.
Question 8. 2 x-y=-2,3 x+4 y=3
Solution:
The given system can be written as AX = B, where
A=\(\left[\begin{array}{cc}
2 & -1 \\
3 & 4
\end{array}\right], X=\left[\begin{array}{l}
x \\
y
\end{array}\right]\)
and \(B=\left[\begin{array}{c}
-2 \\
3
\end{array}\right]\)
Here, |A|=\(\left|\begin{array}{cc}2 & -1 \\ 3 & 4\end{array}\right|=2 \times 4-(-3)=11 \neq 0\)
Thus, A is non-singular.
Therefore, its \(\mathrm{A}^{-1}\) exists.
Therefore, the given system is consistent and has a unique solution given by
⇒ \(\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}\)
Cofactors of A are, \(A_{11}=4, A_{12}=-3, A_{21}=1, A_{21}=2\)
⇒ \({adj}(A)=\left[A_i\right]^{\prime}=\left[\begin{array}{cc}
4 & -3 \\
1 & 2
\end{array}\right]^{\prime}-\left[\begin{array}{cc}
4 & 1 \\
-3 & 2
\end{array}\right]\)
⇒ \(A^{-1}=\frac{1}{|A|}({adj} A)=\frac{1}{11}\left[\begin{array}{cc}
4 & 1 \\
-3 & 2
\end{array}\right]\)
Now, \(X=A^{-1} B=\frac{1}{11}\left[\begin{array}{cc}4 & 1 \\ -3 & 2\end{array}\right]\left[\begin{array}{c}-2 \\ 3\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}-8+3 \\ 6+6\end{array}\right]\)
= \(\frac{1}{11}\left[\begin{array}{c}-5 \\ 12\end{array}\right]=\left[\begin{array}{c}-\frac{5}{11} \\ \frac{12}{11}\end{array}\right]\)
⇒ \(\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}\frac{-5}{11} \\ \frac{12}{11}\end{array}\right]\)
Hence, x=\(\frac{-5}{11}\) and y=\(\frac{12}{11}\).
Question 9. 4 x-3 y=3,3 x-5 y=7
Solution:
The given system can be written as AX = B, where
A=\(\left[\begin{array}{ll}
4 & -3 \\
3 & -5
\end{array}\right]\),
X=\(\left[\begin{array}{l}
x \\
y
\end{array}\right] \)and B=\(\left[\begin{array}{l}
3 \\
7
\end{array}\right]\)
Here, \(|A|=\left|\begin{array}{ll}4 & -3 \\ 3 & -5\end{array}\right|=4(-5)-3(-3)=-20+9=-11 \neq 0\)
Thus, A is non-singular.
Therefore, its \(\mathrm{A}^{-1}\) exists,
Therefore, the given system is consistent and has a unique solution given by
⇒ \(\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}\)
Cofactors of A are, \(A_{11}=-5, A_{12}=-3, A_{21}=3, A_{22}=4\)
⇒ \({adj}(A)=\left[A_{i j}\right]^{\prime}=\left[\begin{array}{cc}
-5 & -3 \\
3 & 4
\end{array}\right]^{\prime}=\left[\begin{array}{ll}
-5 & 3 \\
-3 & 4
\end{array}\right]\)
⇒ \(A^{-1}=\frac{1}{|A|}({adj} A)=\frac{1}{-11}\left[\begin{array}{ll}
-5 & 3 \\
-3 & 4
\end{array}\right] \)
Now, \(X=A^{-1} B=\frac{1}{-11}\left[\begin{array}{ll}
-5 & 3 \\
-3 & 4
\end{array}\right]\left[\begin{array}{l}
3 \\
7
\end{array}\right]\)
=\(\frac{1}{-11}\left[\begin{array}{c}
-15+21 \\
-9+28
\end{array}\right]=\frac{1}{-11}\left[\begin{array}{c}
6 \\
19
\end{array}\right] \)
⇒ \(\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
\frac{-6}{11} \\
\frac{-19}{11}
\end{array}\right]\)
Hence, x=-\(\frac{6}{11}\) and y=-\(\frac{19}{11}\)
Question 10. 5 x+2 y=3,3 x+2 y=5
Solution:
The given system can be written as AX = B, where
⇒ \(\mathrm{A}=\left[\begin{array}{ll}5 & 2 \\ 3 & 2\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y}\end{array}\right] and \mathrm{B}=\left[\begin{array}{l}3 \\ 5\end{array}\right]\)
Here, |A|=\(\left|\begin{array}{ll}5 & 2 \\ 3 & 2\end{array}\right|=10-6=4 \neq 0\)
Thus, A is non-singular.
Therefore, its \(\mathrm{A}^{-1}\) exists.
Therefore, the given system is consistent and has a unique solution given by \(X=A^{-1} B\)
Cofactors of A are, \(A_{11}=2, \mathrm{~A}_{12}=-3, \mathrm{~A}_{21}=-2, \mathrm{~A}_{22}\)=5
adj \((A)=\left[A_4\right]^{\prime}=\left[\begin{array}{cc}2 & -3 \\ -2 & 5\end{array}\right]^*=\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]\)
⇒ \(\mathrm{A}^{-1}=\frac{1}{\mid \mathrm{A}}({adj} \mathrm{A})=\frac{1}{4}\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]\)
Now, X=\(\mathrm{A}^{-1}
\mathrm{~B}=\frac{1}{4}\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]\left[\begin{array}{l}3 \\ 5\end{array}\right]\)
=\(\frac{1}{4}\left[\begin{array}{c}6-10 \\ -9+25\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}-4 \\ 16\end{array}\right]\)
=\(\left[\begin{array}{c}-1 \\ 4\end{array}\right] \Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}-1 \\ 4\end{array}\right]\)
Hence, x=-1 and y=4
Question 11. 2 x+y+z=1, x-2 y-z=\(\frac{3}{2}\), 3 y-5 z=9
Solution:
The given system can be written as AX=B where
A=\(\left[\begin{array}{ccc}
2 & 1 & 1 \\
2 & -4 & -2 \\
0 & 3 & -5
\end{array}\right], X=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
and B=\(\left[\begin{array}{l}
1 \\
3 \\
9
\end{array}\right]\)
Here, \(|A|=\left|\begin{array}{ccc}2 & 1 & 1 \\ 2 & -4 & -2 \\ 0 & 3 & -5\end{array}\right|=2(20+6)-1(-10-0)+1(6-0)=52+10+6=68 \neq 0\)
Thus, A is non-singular.
Therefore, its \(\mathrm{A}^{-1}\) exists.
Therefore, the given system is consistent and has a unique solution given by X=\(\mathrm{A}^{-1} \mathrm{~B}\)
Cofactors of A are,
⇒ \(A_{11}=20+6=26, A_{12}=-(-10-0)=10, A_{13}\)=6-0=6
⇒ \(A_{21}=-(-5-3)=8, A_{22}=-10-0=-10, A_{23}=-(6-0)\)=-6
⇒ \(A_{31}=(-2+4)=2, A_{32}=-(-4-2)=6, A_{33}\)=-8-2=-10
⇒ \({adj}(A)=\left[A_1\right]^{\prime}=\left[\begin{array}{ccc}
26 & 10 & 6 \\
8 & -10 & -6 \\
2 & 6 & -10
\end{array}\right]=\left[\begin{array}{ccc}
26 & 8 & 2 \\
10 & -10 & 6 \\
6 & -6 & -10
\end{array}\right]\)
⇒ \(A^{-1}=\frac{1}{|A|}({adj} A)=\frac{1}{68}\left[\begin{array}{ccc}
26 & 8 & 2 \\
10 & -10 & 6 \\
6 & -6 & -10
\end{array}\right]\)
Now, \(X=A^{-1} B \Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{68}\left[\begin{array}{ccc}26 & 8 & 2 \\ 10 & -10 & 6 \\ 6 & -6 & -10\end{array}\right]\left[\begin{array}{l}1 \\ 3 \\ 9\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
=\(\frac{1}{68}\left[\begin{array}{c}
26+24+18 \\
10-30+54 \\
6-18-90
\end{array}\right]=\frac{1}{68}\left[\begin{array}{c}
68 \\
34 \\
-102
\end{array}\right]\)
⇒ \(\Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
1 \\
\frac{1}{2} \\
\frac{-3}{2}
\end{array}\right]\)
Hence; x=1, y=\(\frac{1}{2}\) and z=\(\frac{-3}{2}\)
Question 12. x-y+z=4, 2 x+y-3 z=0, x+y+z=2
Solution:
The given system can be written as AX=B, where
A=\(\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right], X=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
and B=\(\left[\begin{array}{l}
4 \\
0 \\
2
\end{array}\right]\)
Here, \(|\mathrm{A}|=\left|\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right|=1(1+3)-(-1)(2+3)+1(2-1)=4+5+1=10 \neq 0\)
Thus, A is non-singular.
Therefore, its \(\mathrm{A}^{-1}\) exists.
Therefore, the given system is consistent and has a unique solution given by \(\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}\)
Cofactors of A are.
⇒ \(A_{11}=1+3=4, A_{13}=-(2+3)=-5, A_{13}=2-1=1 \)
⇒ \(A_{21}=-(-1-1)=2, A_{22}=1-1=0, A_{25}=-(1+1)=-2\)
⇒ \(A_{71}=3-1=2, A_{22}=-(-3-2)=5, A_{23}=1+2=3\)
⇒ \({adj}(A)=\left[A_{i j}\right]^{\prime}=\left[\begin{array}{ccc}
4 & -5 & 1 \\
2 & 0 & -2 \\
2 & 5 & 3
\end{array}\right]=\left[\begin{array}{ccc}
4 & 2 & 2 \\
-5 & 0 & 5 \\
1 & -2 & 3
\end{array}\right]\)
⇒ \(A^{-1}=\frac{1}{|A|}(\text { adj } A)=\frac{1}{10}\left[\begin{array}{ccc}
4 & 2 & 2 \\
-5 & 0 & 5 \\
1 & -2 & 3
\end{array}\right]\)
Now, \(X=A^{-1} B \Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{10}\left[\begin{array}{c}
16+0+4 \\
-20+0+10 \\
4+0+6
\end{array}\right] \Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
=\(\frac{1}{10}\left[\begin{array}{c}
20 \\
-10 \\
10
\end{array}\right]=\left[\begin{array}{c}
2 \\
-1 \\
1
\end{array}\right]\)
Hence; x=2, y=-1 and z=1
Question 13. 2 x+3 y+3 z=5, x-2 y+z=-4, 3 x-y-2 z=3
Solution:
The given system can be written as AX = B, where
A=\(\left[\begin{array}{ccc}
2 & 3 & 3 \\
1 & -2 & 1 \\
3 & -1 & -2
\end{array}\right], X=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
and B=\(\left[\begin{array}{c}
5 \\
-4 \\
3
\end{array}\right]\)
Here, \(|\mathrm{A}|=\left|\begin{array}{ccc}2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2\end{array}\right|=2(4+1)-3(-2-3)+3(-1+6)=10+15+15=40 \neq 0\)
Thus, A is non-singular.
Therefore, its \(\mathrm{A}^{-4}\) exists.
Therefore, the given system is consistent and has a unique solution given by X =\(\mathrm{A}^{-1}\) B Cofictors of A are,
⇒ \(A_{11}=4+1=5, A_{12}=-(-2-3)=5, A_{13}=(-1+6)=5\)
⇒ \(A_{21}=-(-6+3)=3, A_{27}=(-4-9)=-13, A_{33}=-(-2-9)=11\)
⇒ \(A_{31}=3+6=9, A_{21}=-(2-3)=1, A_{32}=-4-3=-7\)
⇒ \({adj}(\mathrm{A})=\left[\mathrm{A}_1\right]^{\prime}=\left[\begin{array}{ccc}
5 & 5 & 5 \\
3 & -13 & 11 \\
9 & 1 & -7
\end{array}\right]^{\prime}=\left[\begin{array}{ccc}
5 & 3 & 9 \\
5 & -13 & 1 \\
5 & 11 & -7
\end{array}\right]\)
⇒ \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}({adj} \mathrm{A})=\frac{1}{40}\left[\begin{array}{ccc}
5 & 3 & 9 \\
5 & -13 & 1 \\
5 & 11 & -7
\end{array}\right]\)
⇒ \(\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B} \Rightarrow\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\frac{1}{40}\left[\begin{array}{ccc}
5 & 3 & 9 \\
5 & -13 & 1 \\
5 & 11 & -7
\end{array}\right] \cdot\left[\begin{array}{c}
5 \\
-4 \\
3
\end{array}\right]\)
= \(\frac{1}{40}\left[\begin{array}{c}
25-12+27 \\
25+52+3 \\
25-44-21
\end{array}\right]=\frac{1}{40}\left[\begin{array}{c}
40 \\
80 \\
-40
\end{array}\right]\)
= \(\left[\begin{array}{c}
1 \\
2 \\
-1
\end{array}\right]\)
Hence, x=1, y=2 and z=-1
Question 14. x-y+2 z=7,3 x+4 y-5 z=-5,2 x-y+3 z=12
Solution:
The given system can be written as A X=B, where
⇒ \(\mathrm{A}=\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right]\),
X=\(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right],\)
and B=\(\left[\begin{array}{c}
7 \\
-5 \\
12
\end{array}\right]\)
Here, \(|A|=\left|\begin{array}{ccc}1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3\end{array}\right|=1(12-5)-(-1)(9+10)+2(-3-8)=7+19-22=4 \neq 0\)
Thus, A is non-singular.
Therefore, its \(\mathrm{A}^{-1}\) exists.
Therefore, the given system is consistent and has a unique solution given by
⇒ \(\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}\)
Cofactors of A are,
⇒ \(A_{11}=12-5=7, A_{22}=-(9+10)=-19, A_{13}\)=-3-8=-11
⇒ \(A_{21}=-(-3+2)=1, A_{22}=3-4=-1, A_{23}=-(-1+2)\)=-1
⇒ \(A_{31}=5-8=-3, A_{32}=-(-5-6)=11, A_{39}\)=4+3=7
⇒ \({adj}(\mathrm{A})=\left[\mathrm{A}_{\mathrm{ij}}\right]^{\prime}=\left[\begin{array}{ccc}
7 & -19 & -11 \\
1 & -1 & -1 \\
-3 & 11 & 7
\end{array}\right]^{\prime}=\left[\begin{array}{ccc}
7 & 1 & -3 \\
-19 & -1 & 11 \\
-11 & -1 & 7
\end{array}\right] \)
⇒ \(\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}({adj} \mathrm{A})=\frac{1}{4}\left[\begin{array}{ccc}
7 & 1 & -3 \\
-19 & -1 & 11 \\
-11 & -1 & 7
\end{array}\right]\)
⇒ \(\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B} \Rightarrow\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]\)
=\(\frac{1}{4}\left[\begin{array}{ccc}
7 & 1 & -3 \\
-19 & -1 & 11 \\
-11 & -1 & 7
\end{array}\right] \cdot\left[\begin{array}{c}
7 \\
-5 \\
12
\end{array}\right]\)
= \(\frac{1}{4}\left[\begin{array}{c}
49-5-36 \\
-133+5+132 \\
-77+5+84
\end{array}\right] \Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\)
= \(\frac{1}{4}\left[\begin{array}{l}
8 \\
4 \\
12
\end{array}\right]=\left[\begin{array}{l}
2 \\
1 \\
3
\end{array}\right]\)
Hence, x=2, y=1 and z=3
Question 15. If A=\(\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right]\), find \(A^{-1}\). Using \(A^{-1}\), solve the system of equations 2 x-3 y+5 z=11, 3 x+2 y-4 z=-5, x+y-2 z=-3.
Solution:
The given system can written as AX = R,where A = =\(\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\),
X=\(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] and B=\left[\begin{array}{c}
11 \\
-5 \\
-3
\end{array}\right]\)
Here, \(|A|=\left|\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right|\)
=2(-4+4)-(-3)(-6+4)+5(3-2)=0-6+5=-1 \(\neq 0\)
Thus, A is non-singular.
Therefore, its \(\mathrm{A}^{-1}\) exists.
Therefore, the given system is consistent and has a unique solution given by X=\(A^{-1}\) B Cofactors of A are,
⇒ \(A_{11}=-4+4=0, A_{12}=-(-6+4)=2 \cdot A_{13}=3-2=1\)
⇒ \(A_{21}=-(6-5)=-1, A_{22}=-4-5=-9 \cdot A_{33}=-(2+3)=-5\)
⇒ \(A_{13}=(12-10)=2, A_{12}=-(-8-15)=23, A_{73}=4+9=13\)
⇒ \({adj}(\mathrm{A})=\left[\mathrm{A}_{\mathrm{ii}}\right]^{\prime}=\left[\begin{array}{ccc}
0 & 2 & 1 \\
-1 & -9 & -5 \\
2 & 23 & 13
\end{array}\right]=\left[\begin{array}{ccc}
0 & -1 & 2 \\
2 & -9 & 23 \\
1 & -5 & 13
\end{array}\right]\)
⇒ \(A^{-1}=\frac{1}{|A|}({adj} \mathrm{A})=\frac{1}{-1}\left[\begin{array}{ccc}
0 & -1 & 2 \\
2 & -9 & 23 \\
1 & -5 & 13
\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 & -2 \\
-2 & 9 & -23 \\
-1 & 5 & -13
\end{array}\right]\)
Now, \(X=A^{-1} \mathrm{~B} \Rightarrow\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 & -2 \\
-2 & 9 & -23 \\
-1 & 5 & -13
\end{array}\right]\left[\begin{array}{c}
11 \\
-5 \\
-3
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
0-5+6 \\
-22-45+69 \\
-11-25+39
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
Hence, x=1, y=2 and z=3
Question 16. The cost of 4 kg onion, 3 kg wheat, and 2kg rice is ₹ 60. The cost of 2 kg} onion, 4 kg wheat, and 6kg rice is ₹ 90. The cost of 6kg onion, 2 kg wheat, and 3 kg rice is ₹ 70. Find the cost of each item per kg by matrix method.
Solution:
Let the prices (per kg ) of onion, wheat, and rice be ₹ x,₹ y, and ₹ z respectively. Then 4 x+3 y+2 z=60,2 x+4 y+6 z=90,6 x+2 y+3 z=70
The given system can be written as AX = B, where A =\(\left[\begin{array}{lll}4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]\)
and B =\(\left[\begin{array}{l}60 \\ 90 \\ 70\end{array}\right]\)
Here, \(|A|=\left|\begin{array}{lll}4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3\end{array}\right|=4(12-12)-3(6-36)+2(4-24)=0+90-40=50 \neq 0\)
Thus, A is non-singular.
Therefore, its \(A^{-1}\) exists.
Therefore, the given system is consistent and has a unique solution given by X=\(\mathrm{A}^{-1} B\)
Cofactors of A are,
⇒ \(A_{11}=12-12=0, A_{12}=-(6-36)=30, A_{13}=4-24=-20,\)
⇒ \(A_{21}=-(9-4)=-5, A_{22}=12-12=0, A_{23}=-(8-18)=10\),
⇒ \(A_{31}=(18-8)=10, A_{92}=-(24-4)=-20, A_{31}=16-6\)=10
⇒ \({adj}(A)=\left[A_{11}\right]^{\prime}=\left[\begin{array}{ccc}
0 & 30 & -20 \\
-5 & 0 & 10 \\
10 & -20 & 10
\end{array}\right]=\left[\begin{array}{ccc}
0 & -5 & 10 \\
30 & 0 & -20 \\
-20 & 10 & 10
\end{array}\right]\)
⇒ \(A^{-1}=\frac{1}{|A|}(\text { adj } A)=\frac{1}{50}\left[\begin{array}{ccc}
0 & -5 & 10 \\
30 & 0 & -20 \\
-20 & 10 & 10
\end{array}\right]\)
Now, \(X=A^{-1} B \Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{50}\left[\begin{array}{ccc}
0 & -5 & 10 \\
30 & 0 & -20 \\
-20 & 10 & 10
\end{array}\right]\)
=\(\left[\begin{array}{l}
60 \\
90 \\
70
\end{array}\right]=\frac{1}{50}\left[\begin{array}{c}
\sigma-450+700 \\
1800+0-1400 \\
-1200+900+700
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{50}\left[\begin{array}{l}
250 \\
400 \\
400
\end{array}\right]\)
= \(\left[\begin{array}{l}
5 \\
8 \\
8
\end{array}\right] \Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
5 \\
8 \\
8
\end{array}\right]\)
x=5, y=8 and z=8 .
Hence, the price of onion per kg is ₹ 5, the price of wheat per kg is ₹ 8 and that of rice per kg is ₹ 8.
Determinants Miscellaneous Exercise
Question 1. Prove that the determinant \(\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|\) is independent of \theta.
Solution:
Let \(|A|=\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|\)
Expanding to the corresponding first row, we get
⇒ \(|A| =x\left|\begin{array}{cc}
-x & 1 \\
1 & x
\end{array}\right|-\sin \theta\left|\begin{array}{cc}
-\sin \theta & 1 \\
\cos \theta & x
\end{array}\right|+\cos \theta\left|\begin{array}{cc}
-\sin \theta & -x \\
\cos \theta & 1
\end{array}\right|\)
= x\(\left(-x^2-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta(-\sin \theta+x \cos \theta)\)
=-\(x^3-x+x \sin ^2 \theta+\sin \theta \cos \theta-\sin \theta \cos \theta+x \cos ^2 \theta \quad\left(\sin ^2 \theta+\cos ^2 \theta=1\right)\)
=-\(x^3-x+x\left(\sin ^2 \theta+\cos ^2 \theta\right)=-x^3-x+x\)
= \(-x^3\)
Hence, A is independent of \(\theta\).
Question 2. Evaluate \(\left|\begin{array}{ccc}
\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
-\sin \beta & \cos \beta & 0 \\
\sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha
\end{array}\right|\).
Solution:
Given, \(|A|=\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{array}\right|\)
Expanding corresponding to \(R_1\), we get
|A| =\(\cos \alpha \cos \beta(\cos \alpha \cos \beta-0)-\cos \alpha \sin \beta(-\cos \alpha \sin \beta-0)\)
–\(\sin \alpha\left(-\sin ^2 \beta \sin \alpha-\cos ^2 \beta \sin \alpha\right)\)
= \(\cos ^2 \alpha(\cos ^2 \beta+\sin ^2 \beta)+\sin ^2 \alpha(\sin ^2 \beta+\cos ^2 \beta) [\sin ^2 \theta+\cos ^2 \theta=1]\)
= \(\cos ^2 \alpha(1)+\sin ^2 \alpha(1)=\cos ^2 \alpha+\sin ^2 \alpha=1\) .
Question 3. If \(A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]\) and
B=\(\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]\), find \((A B)^{-1}\).
Solution:
We know that \((\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}\) and \(\mathrm{A}^{-1}\) is known, therefore we proceed to find \(\mathrm{B}^{-1}\).
Here, \(|B|=\left|\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right|=1(3-0)-2(-1-0)-2(2-0)=3+2-4=1 \neq 0\)
⇒ \(\mathrm{B}^{-1}\) exists. Cofactors of B are:
⇒ \(\mathrm{B}_{11}=(3-0)=3, \mathrm{~B}_{12}=-(-1-0)=1\),
⇒ \(\mathrm{~B}_{13}\)=(2-0)=2,
⇒ \(\mathrm{~B}_{11}=-(2-4)=2, \mathrm{~B}_{21}=(1-0)\)=1,
⇒ \(\mathrm{~B}_{23}\)=-(-2-0)=2,
⇒ \(\mathrm{~B}_{31}\)=(0+6)=6,
⇒ \(\mathrm{~B}_{32}\)=-(0-2)=2,
⇒ \(\mathrm{~B}_{33}=(3+2)\)=5,
⇒ \({adj}(\mathrm{B})=\left[\mathrm{B}_6\right]^{+}=\left[\begin{array}{lll}
3 & 1 & 2 \\
2 & 1 & 2 \\
6 & 2 & 5
\end{array}\right]=\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]\)
⇒ \(B^{-1}=\frac{1}{|B|} {adj}(B)=\frac{1}{1}\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]\)
= \(\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]\)
Now,\((A B)^{-1}=B^{-1} A^{-1}=\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
9-30+30 & -3+12-12 & 3-10+12 \\
3-15+10 & -1+6-4 & 1-5+4 \\
6-30+25 & -2+12-10 & 2-10+10
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
9 & -3 & 5 \\
-2 & 1 & 0 \\
1 & 0 & 2
\end{array}\right]\)
Question 4. Evaluate \(\left|\begin{array}{ccc}x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{array}\right|\)
Solution:
Let \(\Delta=\left|\begin{array}{ccc}x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{array}\right|\)
Expand along \(R_1\)
⇒ \(\Delta=x\left(x y+y^2-x^2\right)-y\left(y^2-x^2-x y\right)+(x+y)\left(x y-x^2-2 x y-y^2\right)\)
⇒ \(\Delta=x^2 y+x y^2-x^3-y^3+x^2 y+x y^2+x^2 y-x^3-2 x^2 y-x y^2+x y^2-x^2 y-2 x y^2-y^2\)
⇒ \(\Delta=-2\left(x^3+y^3\right)\)
Question 5. Evaluate \(\left|\begin{array}{ccc}1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y\end{array}\right|\),
Solution:
Let \(\Delta=\left|\begin{array}{ccc}1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y\end{array}\right|\)
Expand along C
⇒ \(\Delta=1\left(x^2+2 x y+y^2-x y\right)-1\left(x^2+x y-x y\right)+1\left(x y-x y-y^2\right)\)
⇒ \(\Delta-x^2+x y+y^2-x^2-y^2\)=x y
Question 6. If x, y, z are non-zero real numbers, then the inverse of matrix A=\(\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]\) is ?
- \(\left[\begin{array}{ccc}x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1}\end{array}\right]\)
- \(x y z\left[\begin{array}{ccc}x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1}\end{array}\right]\)
- \(\frac{1}{x y z}\left[\begin{array}{ccc}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]\)
- \(\frac{1}{x y z}\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\)
Solution: 1. \(\left[\begin{array}{ccc}x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1}\end{array}\right]\)
Given, A=\(\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]\)
⇒ \(|A|=\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]=x(y z-0)=x y z \neq\) 0(because x, y and z are non-zero )
⇒\(\mathrm{A}^{-1}\) exists
Cofactors of A are:
⇒ \(A_{11}=(y z-0)=y z_1 A_{12}=-(0-0)=0, A_{13}=0-0=0\),
⇒ \(A_{21}=-(0-0)=0, A_{22}=x z-0=x z_{23}\)=-(0-0)=0,
⇒ \(A_{31}=0-0=0, A_{31}=-(0-0)=0, A_{13}=(x y-0)\)=x y
⇒ \({adj}(A)=\left[A_{i t}\right]^{\prime}=\left[\begin{array}{ccc}
y z & 0 & 0 \\
0 & x z & 0 \\
0 & 0 & x y
\end{array}\right]^{\prime}=\left[\begin{array}{ccc}
y z & 0 & 0 \\
0 & x z & 0 \\
0 & 0 & x y
\end{array}\right] \)
Now, \(A^{-1}=\frac{1}{|A|}({adj} A) \Rightarrow A^{-1}=\frac{1}{x y z}\left[\begin{array}{ccc}
y z & 0 & 0 \\
0 & x z & 0 \\
0 & 0 & x y
\end{array}\right]=\left[\begin{array}{ccc}
\frac{1}{x} & 0 & 0 \\
0 & \frac{1}{y} & 0 \\
0 & 0 & \frac{1}{z}
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
x^{-1} & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z^{-1}
\end{array}\right]\)
Hence, the correct option is 1.
Question 7. Let A=\(\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]\), where 0 \(\leq \theta \leq 2 \pi\), then:
- \({\det} \mathrm{A}\)=0
- \({\det} \mathrm{A} \in(2, \infty)\)
- \({\det} \mathrm{A} \in(2,4)\)
- \({\det} \mathrm{A} \in[2,4]\)
Solution: 4. \({det} \mathrm{A} \in[2,4]\)
Given, A=\(\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]\)
⇒ \(|A|=\left|\begin{array}{ccc}
1 & \sin \theta & 1 \\
-\sin \theta & 1 & \sin \theta \\
-1 & -\sin \theta & 1
\end{array}\right|\)
=1\(\left(1+\sin ^2 \theta\right)-\sin \theta(-\sin \theta+\sin \theta)+1\left(\sin ^2 \theta+1\right)\)
⇒ \(|A|=2+2 \sin ^2 \theta\)
For 0 \(\leq \theta \leq 2 \pi,-1 \leq \sin \theta \leq 1 \Rightarrow 0 \leq \sin ^2 \theta \leq 1\)
1 \(\leq 1+\sin ^2 \theta \leq 2 \Rightarrow 2 \leq 2\left(1+\sin ^2 \theta\right) \leq 4 {det}(A) \in[2,4]\)
Hence, the correct option is 4.