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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS3 Marks
Question
Given $\text{A}=\begin{bmatrix}2 & -3 \\-4 & 7 \end{bmatrix},$ compute $A^{-1}$ and show that $2A^{-1} = 9I - A$.
✓

Answer

We have, $\text{A}=\begin{bmatrix}2 & -3 \\-4 & 7 \end{bmatrix}$
Now, $\text{adj (A)}=\begin{bmatrix}7 & 3 \\ 4 & 2 \end{bmatrix}$
and $|A| = 2$
$\therefore\ \text{A}^{-1}=\frac{1}{2}\begin{bmatrix}7 & 3 \\ 4 & 2 \end{bmatrix}$
Now, $2A^{-1} = 9I - A$
$\text{L.H.S}=2\text{A}^{-1}=\begin{bmatrix}7 & 3 \\ 4 & 2 \end{bmatrix}$
$\text{R.H.S}=9\text{I}-\text{A}=9\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}-\begin{bmatrix}2 & -3 \\-4 & 7 \end{bmatrix}$
$=\begin{bmatrix}7 & 3 \\4 & 2 \end{bmatrix}=\text{L.H.S}$
Hence proved.

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