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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. $\dfrac{1}{{xyz}}\left( {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right)$

seo-qna
Last updated date: 13th Jun 2024
Total views: 402.9k
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Answer
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402.9k+ views
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.

Complete step-by-step answer:
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.