!!DON'T SEND ANY LINKS"!!Answer 38th question.
Q.38. If $A=\left[\begin{array}{cc}3& -4\\ -1& 2\end{array}\right]$, find matrix B such that BA = I, where I is unity matrix of order 2.

Dear Student,

Please find below the solution to the asked query:

Given :  A  = $\left[\begin{array}{cc}3& - 4\\ - 1& 2\end{array}\right]$

And

BA = I  , So

B = A^-1I

We know if $\mathrm{M} = \left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right] , \mathrm{Then} {\mathrm{M}}^{-1} = \frac{1}{\mathrm{ad} - \mathrm{bc}}\left[\begin{array}{cc}\mathrm{d}& -\mathrm{b}\\ -\mathrm{c}& \mathrm{a}\end{array}\right]$ , So

A^-1 = $\frac{1}{3 \times 2 -\left(-4\right) \times \left(-1\right)}\left[\begin{array}{cc}2& - \left( -4\right)\\ -\left(-1\right)& 3\end{array}\right]= \frac{1}{6 -4}\left[\begin{array}{cc}2& 4\\ 1& 3\end{array}\right]= \frac{1}{2}\left[\begin{array}{cc}2& 4\\ 1& 3\end{array}\right]= \left[\begin{array}{cc}\mathbf{1}& \mathbf{ }\mathbf{2}\\ \frac{\mathbf{1}}{\mathbf{2}}& \frac{\mathbf{3}}{\mathbf{2}}\end{array}\right]$

So,

B = $\left[\begin{array}{cc}1& 2\\ \frac{1}{2}& \frac{3}{2}\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] =\left[\begin{array}{cc}1\times 1+2\times 0& 1\times 0+2\times 1\\ \frac{1}{2}\times 1 +\frac{3}{2}\times 0& \frac{1}{2}\times 0+\frac{3}{2}\times 1\end{array}\right] = \left[\begin{array}{cc}1+0& 0+2\\ \frac{1}{2} +0& 0+\frac{3}{2}\end{array}\right]=\mathbf{ }\left[\begin{array}{cc}\mathbf{1}& \mathbf{2}\\ \frac{\mathbf{1}}{\mathbf{2}}& \frac{\mathbf{3}}{\mathbf{2}}\end{array}\right]\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)}$
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