Solution:
Given that
![A=\left[\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4ad68f18868d9bbabc14b6bff4990edd_l3.png "Rendered by QuickLaTeX.com")
![\operatorname{adj} A=\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-62759f5dae15e5781c0845a3d3d532e8_l3.png "Rendered by QuickLaTeX.com")
![A^{-1}=\frac{1}{2}\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fe05dc530902df7d2264122db9697b38_l3.png "Rendered by QuickLaTeX.com")
We need to Show: 
We have
Left Hand Show ![=2 \mathrm{~A}^{-1}=2 . \frac{1}{2}\left[\begin{array}{ll}7 & 3 \\ 4 & 2\end{array}\right]=\left[\begin{array}{ll}7 & 3 \\ 4 & 2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-63363e0e52af44e73737965b75e180ab_l3.png "Rendered by QuickLaTeX.com")
Right Hand Side![=9 I-A=\left[\begin{array}{ll}9 & 0 \\ 0 & 9\end{array}\right]-\left[\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ccf31299163f7b1a90e0824070c28f0f_l3.png "Rendered by QuickLaTeX.com")
![=\left[\begin{array}{ll}7 & 3 \\ 4 & 2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fd38f5abc0f5f928206f74be3f8aa1d5_l3.png "Rendered by QuickLaTeX.com")
As a result,
Given ![A=\left[\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d8efe0d51dfb800dd8cf461c191b701a_l3.png "Rendered by QuickLaTeX.com")
Solution:
Given that
We need to Show:
We have
Left Hand Show
Right Hand Side
As a result,