If x, y, z are nonzero real numbers,then the | vedclass

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(A) Given the diagonal matrix $A = \begin{bmatrix} x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z \end{bmatrix}$.The determinant of $A$ is $|A| = x(yz - 0) - 0 +

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$\begin{bmatrix} x^{-1} & 0 & 0 \ 0 & y^{-1} & 0 \ 0 & 0 & z^{-1} \end{bmatrix}$

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(A) Given the diagonal matrix $A = \begin{bmatrix} x & 0 & 0 \ 0 & y & 0 \ 0 & 0 & z \end{bmatrix}$.The determinant of $A$ is $|A| = x(yz - 0) - 0 + 0 = xyz$.Since $x, y, z 
eq 0$,$|A| = xyz 
eq 0$,so $A^{-1}$ exists.The inverse of a diagonal matrix $A = 	ext{diag}(x, y, z)$ is given by $A^{-1} = 	ext{diag}(x^{-1}, y^{-1}, z^{-1})$.Calculating the adjugate matrix:$adj(A) = \begin{bmatrix} yz & 0 & 0 \ 0 & xz & 0 \ 0 & 0 & xy \end{bmatrix}$.Thus,$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{xyz} \begin{bmatrix} yz & 0 & 0 \ 0 & xz & 0 \ 0 & 0 & xy \end{bmatrix} = \begin{bmatrix} \frac{yz}{xyz} & 0 & 0 \ 0 & \frac{xz}{xyz} & 0 \ 0 & 0 & \frac{xy}{xyz} \end{bmatrix} = \begin{bmatrix} \frac{1}{x} & 0 & 0 \ 0 & \frac{1}{y} & 0 \ 0 & 0 & \frac{1}{z} \end{bmatrix} = \begin{bmatrix} x^{-1} & 0 & 0 \ 0 & y^{-1} & 0 \ 0 & 0 & z^{-1} \end{bmatrix}$.

If $x, y, z$ are nonzero real numbers,then the inverse of matrix $A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$ is

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