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Q.(i). If $A=\left[\begin{array}{cc}5& -8\\ -5& 8\end{array}\right]$, show by elementary row transformations that $A^{-1}$ does not exist.

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{A}=\left[\begin{array}{cc}5& -8\\ -5& 8\end{array}\right] \\ {\mathrm{R}}_1\to {\mathrm{R}}_1+{\mathrm{R}}_2 \\ \mathrm{A}=\left[\begin{array}{cc}5-5& -8+8\\ -5& 8\end{array}\right] \\ \mathrm{A}=\left[\begin{array}{cc}0& 0\\ -5& 8\end{array}\right] \\ \mathrm{As} {\mathrm{R}}_1 \mathrm{is} 0, \mathrm{hence} \mathrm{determinant} \mathrm{is} 0 , \mathrm{hence} \mathrm{matrix} \mathrm{will} \mathrm{be} \mathrm{non}-\mathrm{invertible}. \end{gathered}$$

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