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 Q : 18         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

 

                    (A)\begin{bmatrix} x^-^1 &0 &0 \\ 0 &y^-^1 &0 \\ 0 & 0 & z^-^1 \end{bmatrix}                                   (B)xyz\begin{bmatrix} x^-^1 &0 &0 \\ 0 &y^-^1 &0 \\ 0 & 0 & z^-^1 \end{bmatrix}

 

                    (C)\frac{1}{xyz}\begin{bmatrix} x &0 &0 \\ 0 &y &0 \\ 0 & 0 & z \end{bmatrix}                                        (D)\frac{1}{xyz}\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 & 0 & 1 \end{bmatrix}

Answers (1)

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Given Matrix A=\begin{bmatrix} x &0 &0 \\ 0 &y &0 \\ 0 & 0 & z \end{bmatrix},

|A| = x(yz-0) =xyz

As we know, 

A^{-1} = \frac{1}{|A|}adjA

So, we  will find the adjA,

Determining its cofactor first,

A_{11} = yz  A_{12} = 0  A_{13} = 0

A_{21} = 0  A_{22} = xz  A_{23} = 0

A_{31} = 0  A_{32} = 0  A_{33} = xy

Hence A^{-1} = \frac{1}{|A|}adjA = \frac{1}{xyz}\begin{bmatrix} yz &0 &0 \\ 0& xz & 0\\ 0& 0& xy \end{bmatrix}

A^{-1} = \begin{bmatrix} \frac{1}{x} &&0 &&0 \\ 0&& \frac{1}{y} && 0\\ 0&& 0&& \frac{1}{z} \end{bmatrix}

Therefore the correct answer is (A).

 

 

Posted by

Divya Prakash Singh

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