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Find the inverse of each of the following matrices.
\begin{aligned} &\text { (i) } \left.\begin{array}{ccc} -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right] \\ \end{aligned}
\begin{aligned} &(ii)\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \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}{ccc} -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right] \\ \end{aligned}
\begin{aligned} &(ii)\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \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}|=2\left|\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right|+1\left|\begin{array}{cc}-1 & -1 \\ 1 & 2\end{array}\right|+1\left|\begin{array}{cc}-1 & 2 \\ 1 & -1\end{array}\right|

=2(4-1)+1(-2+1)+1(1-2)

=6-2

=-4 \neq 0

Thus, A^{-1} exists

Cofactors of A are

C_{11}=3

\mathrm{C}_{21}=1

C_{31}=-1

\mathrm{C}_{12}=+1

\mathrm{C}_{22}=3

\mathrm{C}_{32}=1

C_{13}=-1

C_{23}=1 \mathrm{C}_{33}=3

It is known that 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}3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3\end{array}\right]^{\mathrm{T}}

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

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

Therefore, A^{-1}=\frac{1}{4}\left[\begin{array}{ccc}3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3\end{array}\right]

Thus, A^{-1}=\left[\begin{array}{ccc}\frac{3}{4} & \frac{1}{4} & \frac{-1}{4} \\ \frac{1}{4} & \frac{3}{4} & \frac{1}{4} \\ \frac{-1}{4} & \frac{1}{4} & \frac{1}{4}\end{array}\right]

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

|\mathrm{A}|=2\left|\begin{array}{ll}1 & 0 \\ 1 & 3\end{array}\right|-0\left|\begin{array}{ll}5 & 0 \\ 0 & 3\end{array}\right|-1\left|\begin{array}{ll}5 & 1 \\ 0 & 1\end{array}\right|

=2(3-0)-0-1(5)

=6-5

=1 \neq 0

Thus, A^{-1} exists

Cofactors of A are

C_{11}=3

C_{21}=-1

\mathrm{C}_{31}=1

C_{12}=-15

\begin{array}{l} C_{22}=6 \\ C_{32}=-5 \\ C_{13}=5 \\ C_{23}=-2 \\ C_{33}=2 \\ \text { It is known that adj } A=\left[\begin{array}{ccc} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{array}\right]^{T} \\ \text { Therefore, } \operatorname{adj} A=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right] \\ \text { Now, } A^{-1}=\frac{1}{|A|} \operatorname{adj} A \\ \text { Therefore, } A^{-1}=\frac{1}{1}\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right] \\ =\left[\begin{array}{ccc} 3 & -15 & 5 \\ -1 & 6 & -2 \\ 1 & -5 & 2 \end{array}\right]^{\mathrm{T}} \\ =\left[\begin{array}{ccc} -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right] \end{array}

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