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

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

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

(C) $\cfrac{1}{{xyz}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&0&0\\0&y&0\\0&0&z\end{array}} \right]$

(D)$\cfrac{1}{{xyz}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$

Option a is correct

: Let A$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&0&0\\0&y&0\\0&0&z\end{array}} \right]$

$\therefore$ $|A| = xyz \ne 0,{A^{ - 1}}$ exists.

Now, cofactors of elements of A are :

${A_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}y&0\\0&z\end{array}} \right| = yz,$

${A_{12}} = {( - 1)^{1 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0\\0&z\end{array}} \right| = 0$

${A_{13}} = {( - 1)^{1 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&y\\0&0\end{array}} \right| = 0$

${A_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0\\0&z\end{array}} \right| = 0,$

${A_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&0\\0&z\end{array}} \right| = xz,$

${A_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&0\\0&0\end{array}} \right| = 0,$

${A_{31}} = {( - 1)^{3 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0\\y&0\end{array}} \right| = 0,$

${A_{32}} = {( - 1)^{3 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&0\\0&0\end{array}} \right| = 0,$

${A_{33}} = {( - 1)^{3 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&0\\0&y\end{array}} \right| = xy$

%$\therefore$ $adj\;A = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{yz}&0&0\\0&{xz}&0\\0&0&{xy}\end{array}} \right]^{\prime}} = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{yz}&0&0\\0&{xz}&0\\0&0&{xy}\end{array}} \right]$

Hence, ${A^{ - 1}} = \cfrac{1}{{xyz}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{yz}&0&0\\0&{xz}&0\\0&0&{xy}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{{x^{ - 1}}}&0&0\\0&{{y^{ - 1}}}&0\\0&0&{{z^{ - 1}}}\end{array}} \right]$