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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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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 
  • \(\frac{1}{x y z}\left[\begin{array}{lll} {x} & {0} & {0} \\ {0} & {y} & {0} \\ {0} & {0} & {z} \end{array}\right]\)
  • \(\frac{1}{x y z}\left[\begin{array}{lll} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right]\)
  • \(\frac{5yz}{x y z}\left[\begin{array}{lll} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right]\)
  • \(\left[\begin{array}{ccc} {\mathrm{x}^{-1}} & {0} & {0} \\ {0} & {\mathrm{y}^{-1}} & {0} \\ {0} & {0} & {\mathrm{z}^{-1}} \end{array}\right]\)

Answer

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
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