Question

If x,y, z are non-zero real numbers, then the inverse of 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

Given, $A=\left[\begin{array}{ccc}x& 0& 0\\ 0& y& 0\\ 0& 0& z\end{array}\right]$
$|A|=\left[\begin{array}{ccc}x& 0& 0\\ 0& y& 0\\ 0& 0& z\end{array}\right]=x(yz=0)=xyz\ne 0$ ($\because$ x, y and z are non - zero )
Cofactors of A are
$\begin{array}{ccc}{A}_{11}=(yz-0)=yz,& A_{12}=-(0-0)=0& A_{13}=0-0=0\\ A_{21}=-(0-0)=0,& A_{22}=xz-0=xz& A_{23}=-(0-0)=0\\ A_{31}=0-0=0& A_{32}=-(0-0)=0& A_{33}=(xy-0)=xy\end{array}$
$\therefore adj\left(A\right)=\left[\begin{array}{ccc}yz& 0& 0\\ 0& xz& 0\\ 0& 0& xy\end{array}\right]=\left[\begin{array}{ccc}yz& 0& 0\\ 0& xz& 0\\ 0& 0& xy\end{array}\right]$
Now, $A^{-1}=\frac{1}{|A|}\left(adjA\right)$
$=\frac{1}{xyz}\left[\begin{array}{ccc}yz& 0& 0\\ 0& xz& 0\\ 0& 0& xy\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{x}& 0& 0\\ 0& \frac{1}{y}& 0\\ 0& 0& \frac{1}{z}\end{array}\right]=\left[\begin{array}{ccc}{x}^{-1}& 0& 0\\ 0& y^{-1}& 0\\ 0& 0& z^{-1}\end{array}\right]$
Hence, the correct option is(a)