Question

If x, y, z are non-zero real numbers, then the inverse of the matrix $A=\left[\begin{array}{ccc}x& 0& 0\\ 0& y& 0\\ 0& 0& z\end{array}\right]$, is
(a) $\left[\begin{array}{ccc}{x}^{-1}& 0& 0\\ 0& y^{-1}& 0\\ 0& 0& z^{-1}\end{array}\right]$

(b) $xyz\left[\begin{array}{ccc}{x}^{-1}& 0& 0\\ 0& y^{-1}& 0\\ 0& 0& z^{-1}\end{array}\right]$

(c) $\frac{1}{xyz}\left[\begin{array}{ccc}x& 0& 0\\ 0& y& 0\\ 0& 0& z\end{array}\right]$

(d) $\frac{1}{xyz}\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Answer 1

(a) $\left[\begin{array}{ccc}{x}^{-1}& 0& 0\\ 0& y^{-1}& 0\\ 0& 0& z^{-1}\end{array}\right]$


$$\begin{gathered} A=IA \\ \Rightarrow \left[\begin{array}{ccc}x& 0& 0\\ 0& y& 0\\ 0& 0& z\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A \\ \Rightarrow \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}{x}^{-1}& 0& 0\\ 0& y^{-1}& 0\\ 0& 0& z^{-1}\end{array}\right]A\left[\mathrm{Applying}{R}_1=\frac{1}{x}{R}_1,R_2=\frac{1}{y}{R}_2\mathrm{and}{R}_3=\frac{1}{z}{R}_3\right] \\ \Rightarrow A^{-1}=\left[\begin{array}{ccc}{x}^{-1}& 0& 0\\ 0& y^{-1}& 0\\ 0& 0& z^{-1}\end{array}\right] \end{gathered}$$