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If $\mathbf{x}, \mathbf{y}, \mathbf{z}$ are nonzero real numbers, then the inverse of $\operatorname{matrix} A=\left[\begin{array}{ccc}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]$ is
(a) $\left[\begin{array}{ccc}x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1}\end{array}\right]$
(b) $\operatorname{xyz}\left[\begin{array}{ccc}x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1}\end{array}\right]$
(c) $\frac{1}{x y z}\left[\begin{array}{ccc}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]$
(d) $\frac{1}{x y z}\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
If $\mathbf{x}, \mathbf{y}, \mathbf{z}$ are nonzero real numbers, then the inverse of $\operatorname{matrix} A=\left[\begin{array}{ccc}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]$ is
(a) $\left[\begin{array}{ccc}x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1}\end{array}\right]$
(b) $\operatorname{xyz}\left[\begin{array}{ccc}x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1}\end{array}\right]$
(c) $\frac{1}{x y z}\left[\begin{array}{ccc}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]$
(d) $\frac{1}{x y z}\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
Solution:
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Verified Answer
(a) $A=\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]$
The cofactors of the elements are
$\begin{array}{ll}\mathrm{A}_{11}=\mathrm{yz}, & \mathrm{A}_{12}=0 \\ \mathrm{~A}_{13}=0, & \mathrm{~A}_{21}=0 \\ \mathrm{~A}_{22}=\mathrm{xz}, & \mathrm{A}_{23}=0 \\ \mathrm{~A}_{31}=0, & \mathrm{~A}_{32}=0 \\ \mathrm{~A}_{33}=\mathrm{xy} & \end{array}$
$|A|=x y z$
$\operatorname{adj}(A)=\left[\begin{array}{ccc}y z & 0 & 0 \\ 0 & z x & 0 \\ 0 & 0 & x y\end{array}\right]$
$\therefore \mathrm{A}^{-1}=\frac{1}{\mathrm{xyz}}\left[\begin{array}{ccc}\mathrm{yz} & 0 & 0 \\ 0 & \mathrm{zx} & 0 \\ 0 & 0 & \mathrm{xy}\end{array}\right]=\left[\begin{array}{ccc}\mathrm{x}^{-1} & 0 & 0 \\ 0 & \mathrm{y}^{-1} & 0 \\ 0 & 0 & \mathrm{z}^{-1}\end{array}\right]$
The cofactors of the elements are
$\begin{array}{ll}\mathrm{A}_{11}=\mathrm{yz}, & \mathrm{A}_{12}=0 \\ \mathrm{~A}_{13}=0, & \mathrm{~A}_{21}=0 \\ \mathrm{~A}_{22}=\mathrm{xz}, & \mathrm{A}_{23}=0 \\ \mathrm{~A}_{31}=0, & \mathrm{~A}_{32}=0 \\ \mathrm{~A}_{33}=\mathrm{xy} & \end{array}$
$|A|=x y z$
$\operatorname{adj}(A)=\left[\begin{array}{ccc}y z & 0 & 0 \\ 0 & z x & 0 \\ 0 & 0 & x y\end{array}\right]$
$\therefore \mathrm{A}^{-1}=\frac{1}{\mathrm{xyz}}\left[\begin{array}{ccc}\mathrm{yz} & 0 & 0 \\ 0 & \mathrm{zx} & 0 \\ 0 & 0 & \mathrm{xy}\end{array}\right]=\left[\begin{array}{ccc}\mathrm{x}^{-1} & 0 & 0 \\ 0 & \mathrm{y}^{-1} & 0 \\ 0 & 0 & \mathrm{z}^{-1}\end{array}\right]$
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