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If x, y, z are non-zero real numbers, then the inverse of matrix $$A=\left[\begin{array}{lll} {x} & {0} & {0} \\ {0} & {y} & {0} \\ {0} & {0} & {z} \end{array}\right]$$ is
• $$\frac{1}{x y z}\left[\begin{array}{lll} {x} & {0} & {0} \\ {0} & {y} & {0} \\ {0} & {0} & {z} \end{array}\right]$$
• $$\frac{1}{x y z}\left[\begin{array}{lll} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right]$$
• $$\frac{5yz}{x y z}\left[\begin{array}{lll} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right]$$
• $$\left[\begin{array}{ccc} {\mathrm{x}^{-1}} & {0} & {0} \\ {0} & {\mathrm{y}^{-1}} & {0} \\ {0} & {0} & {\mathrm{z}^{-1}} \end{array}\right]$$
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QuestionMathsClass 12

If x, y, z are non-zero real numbers, then the inverse of matrix $$A=\left[\begin{array}{lll} {x} & {0} & {0} \\ {0} & {y} & {0} \\ {0} & {0} & {z} \end{array}\right]$$ is
• $$\frac{1}{x y z}\left[\begin{array}{lll} {x} & {0} & {0} \\ {0} & {y} & {0} \\ {0} & {0} & {z} \end{array}\right]$$
• $$\frac{1}{x y z}\left[\begin{array}{lll} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right]$$
• $$\frac{5yz}{x y z}\left[\begin{array}{lll} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right]$$
• $$\left[\begin{array}{ccc} {\mathrm{x}^{-1}} & {0} & {0} \\ {0} & {\mathrm{y}^{-1}} & {0} \\ {0} & {0} & {\mathrm{z}^{-1}} \end{array}\right]$$

Here, $$A=\left[\begin{array}{lll} {x} & {0} & {0} \\ {0} & {y} & {0} \\ {0} & {0} & {z} \end{array}\right]$$
Clearly, we can see that
$$adjA$$$$=\left[\begin{array}{lll} {yz} & {0} & {0} \\ {0} & {xz} & {0} \\ {0} & {0} & {xy} \end{array}\right]~~and ~~~|A|=xyz$$
$$\therefore ~A^{-1}=\frac {adjA}{|A|}=\frac1{xyz}\left[\begin{array}{lll} {{yz}} & {0} & {0} \\ {0} & {{xz}} & {0} \\ {0} & {0} & {xy} \end{array}\right]$$
$$=\left[\begin{array}{lll} {x^{-1}} & {0} & {0} \\ {0} & {y^{-1}} & {0} \\ {0} & {0} & {z^{-1}} \end{array}\right]$$

Therefore the choice is: D