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Adjoint and Inverse of a Matrix — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAdjoint and Inverse of a Matrix5 Marks
Question
Given $\text{A}=\begin{bmatrix}5 & 0 & 4 \\2 & 3 & 2 \\ 1 & 2 & 1\end{bmatrix},\text{B}^{-1}=\begin{bmatrix}1 & 3 & 3 \\1 & 4 & 3 \\ 1 & 3 & 4\end{bmatrix}.$ Compute $(AB)^{-1}.$
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Answer

We have,$\text{A}=\begin{bmatrix}5 & 0 & 4 \\2 & 3 & 2 \\ 1 & 2 & 1\end{bmatrix}$
$\text{B}^{-1}=\begin{bmatrix}1 & 3 & 3 \\1 & 4 & 3 \\ 1 & 3 & 4\end{bmatrix}$
We have $(AB)^{-1} = B^{-1}A^{-1} $For matrix A,$\text{C}_{11}=\begin{vmatrix}3 & 2 \\2 & 1 \end{vmatrix}=-1,\ \text{C}_{12}=-\begin{vmatrix}2 & 2 \\1 & 1 \end{vmatrix}=0$
$\text{and C}_{13}=\begin{vmatrix}2 & 3 \\1 & 2 \end{vmatrix}=1$
$\text{C}_{21}=-\begin{vmatrix}0 & 4 \\2 & 1 \end{vmatrix}=8,\ \text{C}_{22}=\begin{vmatrix}5 & 4 \\1 & 1 \end{vmatrix}=1$
$\text{and C}_{23}=-\begin{vmatrix}5 & 0 \\1 & 2 \end{vmatrix}=-10$
$\text{C}_{31}=\begin{vmatrix}0 & 4 \\3 & 1 \end{vmatrix}=-12,\ \text{C}_{32}=-\begin{vmatrix}5 & 4 \\2 & 2 \end{vmatrix}=-2$
$\text{and C}_{33}=\begin{vmatrix}5 & 0 \\2 & 3 \end{vmatrix}=15$
Now, $\text{adj (A)}=\begin{bmatrix}-1 & 0 & 1 \\8 & 1 & -2 \\ -12 & -2 & 15 \end{bmatrix}^\text{T}$$=\begin{bmatrix} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{bmatrix}$
and |A| = -1$\therefore\text{A}^{-1}=-\begin{bmatrix} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{bmatrix}$
$=\begin{bmatrix} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{bmatrix}$
So, $\text{B}^{-1}\text{A}^{-1}=\begin{bmatrix} 1 & 3 & 3 \\1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{bmatrix}$$=\begin{bmatrix} -2 & 19 & -27 \\ -2 & 18 & -25 \\ -3 & 29 & -42 \end{bmatrix}$

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