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If x, y and z are non-zero real numbers, then the inverse of the matrix $A = \left( {\begin{array}{*{20}{c}} x&0&0 \\ 0&y&0 \\ 0&0&z \end{array}} \right)$ is A. $\left( {\begin{array}{*{20}{c}} {{x^{ - 1}}}&0&0 \\ 0&{{y^{ - 1}}}&0 \\ 0&0&{{z^{ - 1}}} \end{array}} \right)$B. $xyz\left( {\begin{array}{*{20}{c}} {{x^{ - 1}}}&0&0 \\ 0&{{y^{ - 1}}}&0 \\ 0&0&{{z^{ - 1}}} \end{array}} \right)$C. $\dfrac{1}{{xyz}}\left( {\begin{array}{*{20}{c}} x&0&0 \\ 0&y&0 \\ 0&0&z \end{array}} \right)$D. $\dfrac{1}{{xyz}}\left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right)$

Last updated date: 13th Jun 2024
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Hint: To solve this question, we have to remember that the inverse of matrix A is given by, ${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj.A$, where adj. A denotes the adjoint of matrix A and $\left| A \right|$ is the determinant of A.

We have,
$\Rightarrow A = \left( {\begin{array}{*{20}{c}} x&0&0 \\ 0&y&0 \\ 0&0&z \end{array}} \right)$
We know that, for A to be invertible, $\left| A \right| \ne 0$
So, first we will find $\left| A \right|$
$\Rightarrow x\left( {yz - 0} \right) - 0 - 0$
$\Rightarrow \left| A \right| = xyz$
We can see that $\left| A \right| \ne 0$, hence, A is invertible.
Now, we will find the adj. A.
In order to find the adj. A, we have to find the cofactor matrix of A.
We know that,
Cofactor, ${C_{ij}}$of ${a_{ij}}$ in A = ${\left[ {{a_{ij}}} \right]_{n \times n}}$ is equal to ${\left( { - 1} \right)^{i + j}}{M_{ij}}$
Where ${M_{ij}}$ is the minor.
So,
Cofactor of ${a_{11}}$ = $\left| {\begin{array}{*{20}{c}} y&0 \\ 0&z \end{array}} \right| = yz$
Cofactor of ${a_{12}}$ = $\left| {\begin{array}{*{20}{c}} 0&0 \\ 0&z \end{array}} \right| = 0$
Cofactor of ${a_{13}}$ = $\left| {\begin{array}{*{20}{c}} 0&y \\ 0&0 \end{array}} \right| = 0$
Cofactor of ${a_{21}}$ = $\left| {\begin{array}{*{20}{c}} 0&0 \\ 0&z \end{array}} \right| = 0$
Cofactor of ${a_{22}}$ = $\left| {\begin{array}{*{20}{c}} x&0 \\ 0&z \end{array}} \right| = xz$
Cofactor of ${a_{23}}$ = $\left| {\begin{array}{*{20}{c}} x&0 \\ 0&0 \end{array}} \right| = 0$
Cofactor of ${a_{31}}$ = $\left| {\begin{array}{*{20}{c}} 0&0 \\ y&0 \end{array}} \right| = 0$
Cofactor of ${a_{32}}$ = $\left| {\begin{array}{*{20}{c}} x&0 \\ 0&0 \end{array}} \right| = 0$
Cofactor of ${a_{33}}$ = $\left| {\begin{array}{*{20}{c}} x&0 \\ 0&y \end{array}} \right| = xy$
Therefore, the cofactor matrix of A is $\left( {\begin{array}{*{20}{c}} {yz}&0&0 \\ 0&{xz}&0 \\ 0&0&{xy} \end{array}} \right)$
Now, the adj. A is the transpose of the cofactor matrix of A.
Therefore,
$adj.A = \left( {\begin{array}{*{20}{c}} {yz}&0&0 \\ 0&{xz}&0 \\ 0&0&{xy} \end{array}} \right)$
We know that,
Inverse of A is given by, ${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj.A$
So,
$\Rightarrow {A^{ - 1}} = \dfrac{1}{{xyz}}\left( {\begin{array}{*{20}{c}} {yz}&0&0 \\ 0&{xz}&0 \\ 0&0&{xy} \end{array}} \right)$
Taking $\dfrac{1}{{xyz}}$ inside the matrix, we will get
$\Rightarrow {A^{ - 1}} = \left( {\begin{array}{*{20}{c}} {\dfrac{{yz}}{{xyz}}}&0&0 \\ 0&{\dfrac{{xz}}{{xyz}}}&0 \\ 0&0&{\dfrac{{xy}}{{xyz}}} \end{array}} \right)$
$\Rightarrow {A^{ - 1}} = \left( {\begin{array}{*{20}{c}} {\dfrac{1}{x}}&0&0 \\ 0&{\dfrac{1}{y}}&0 \\ 0&0&{\dfrac{1}{z}} \end{array}} \right)$
We can write this as:
$\Rightarrow {A^{ - 1}} = \left( {\begin{array}{*{20}{c}} {{x^{ - 1}}}&0&0 \\ 0&{{y^{ - 1}}}&0 \\ 0&0&{{z^{ - 1}}} \end{array}} \right)$

So, the correct answer is “Option A”.

Note: Whenever we asked such type of questions, we have to remember that a square matrix of order n is invertible if there exists a square matrix B of the same order such that $AB = {I_n} = BA$, in such a way, we can write ${A^{ - 1}} = B$ A square matrix is invertible if and only if it is non-singular. Through these things, we can easily solve the questions.