If A is a 3 × 3 invertible matrix, show that for any scalar k ≠ 0, ( kA )^-1=1/k A^-1. Hence calculate (3 A )^-1, where A = [2, -1, 1; -1, 2, -1; 1,…

Consider \(\displaystyle (\mathrm{k} \mathrm{A})\left(\frac{1}{\mathrm{k}} \mathrm{A}^{-1}\right)=\mathrm{k} \cdot \frac{1}{\mathrm{k}}\left(\mathrm{A} \cdot \mathrm{A}^{-1}\right)=1 . I=I\) \[\therefore(3 \mathrm{A})^{-1}=\frac{1}{3} \mathrm{A}^{-1} \] Here, \(\displaystyle |\mathrm{A}|=4 \neq 0 \therefore \mathrm{A}^{-1}\) exists. \[\operatorname{adj} \mathrm{A}=\left[\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right] \]