Login/Sign Up
Home » Find the inverse of each of the following matrices.
Find the inverse of each of the following matrices.
\begin{aligned} &\text { (i) }\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{array}\right] \\ \end{aligned}
\begin{aligned} &\text { (ii) }\left[\begin{array}{ccc} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{array}\right] \\ \end{aligned}
Adjoint and Inverse of a Matrix | CBSE | Class 12 | Exercise 7.1 | Maths | RD Sharma
Find the inverse of each of the following matrices.
\begin{aligned} &\text { (i) }\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{array}\right] \\ \end{aligned}
\begin{aligned} &\text { (ii) }\left[\begin{array}{ccc} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{array}\right] \\ \end{aligned}

Solution:

(i) The criteria of existence of inverse matrix is the determinant of a given matrix should not be equal to zero.

|\mathrm{A}|=1\left|\begin{array}{ll}3 & 1 \\ 1 & 2\end{array}\right|-2\left|\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right|+3\left|\begin{array}{ll}2 & 3 \\ 3 & 1\end{array}\right|

=1(6-1)-2(4-3)+3(2-9)

=5-2-21

=-18 \neq 0

Thus, \mathrm{A}^{-1} exists

Cofactors of A are

C_{11}=5

C_{21}=-1

C_{31}=-7

C_{12}=-1

C_{22}=-7

C_{32}=5

C_{13}=-7

C_{23}=5

C_{33}=-1

It is known that adi A=\left[\begin{array}{lll}\mathrm{C}_{11} & \mathrm{C}_{12} & \mathrm{C}_{13} \\ \mathrm{C}_{21} & \mathrm{C}_{22} & \mathrm{C}_{23} \\ \mathrm{C}_{31} & \mathrm{C}_{32} & \mathrm{C}_{33}\end{array}\right]^{\mathrm{T}}

=\left[\begin{array}{ccc}5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1\end{array}\right]^{\mathrm{T}}

Therefore, adj A=\left[\begin{array}{ccc}5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1\end{array}\right] Now, A^{-1}=\frac{1}{|A|} adj A

Therefore, A^{-1}=\frac{1}{(-18)}\left[\begin{array}{ccc}5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1\end{array}\right]

Thus,

A^{-1}=\left[\begin{array}{ccc} \frac{-5}{18} & \frac{1}{18} & \frac{7}{18} \\ \frac{1}{18} & \frac{7}{18} & \frac{-5}{18} \\ \frac{7}{18} & \frac{-5}{18} & \frac{1}{18} \end{array}\right]

(ii) The criteria of existence of inverse matrix is the determinant of a given matrix should not be equal to zero.

|A|=1\left|\begin{array}{cc}-1 & -1 \\ 3 & -1\end{array}\right|-2\left|\begin{array}{cc}1 & -1 \\ 2 & -1\end{array}\right|+5\left|\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right|

=1(1+3)-2(-1+2)+5(3+2)

=4-2+25

=27 \neq 0

Thus, \mathrm{A}^{-1} exists

Cofactors of A are

C_{11}=4

C_{21}=17

C_{31}=3

C_{12}=-1

C_{22}=-11

C_{32}=6

C_{13}=5

C_{23}=1

C_{33}=-3

\operatorname{adj} A=\left[\begin{array}{lll}C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33}\end{array}\right]^{T} =\left[\begin{array}{ccc}4 & -1 & 5 \\ 17 & -11 & 1 \\ 3 & 6 & -3\end{array}\right]^{\mathrm{T}}

Therefore, adj A=\left[\begin{array}{ccc}4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3\end{array}\right]

Now, A^{-1}=\frac{1}{|A|} \operatorname{adj} A

Therefore, A^{-1}=\frac{1}{(27)}\left[\begin{array}{ccc}4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3\end{array}\right]

Thus, \mathrm{A}^{-1}=\left[\begin{array}{ccc}\frac{4}{27} & \frac{17}{27} & \frac{3}{27} \\ \frac{-1}{27} & \frac{-11}{27} & \frac{6}{27} \\ \frac{5}{27} & \frac{1}{27} & \frac{-3}{27}\end{array}\right]=\left[\begin{array}{ccc}\frac{4}{27} & \frac{17}{27} & \frac{1}{9} \\ \frac{-1}{27} & \frac{-11}{27} & \frac{2}{9} \\ \frac{5}{27} & \frac{1}{27} & \frac{-1}{9}\end{array}\right]

i

More Material