Question

Find the inverse of the following matrix​

Answer 1

$\large\underline{\sf{Solution-}}$

Given matrix is

$\rm \: A \: = \: \begin{gathered}\sf \left[\begin{array}{ccc}x&0&0\\0& y&0\\ 0&0&z\end{array}\right]\end{gathered}$

Now, to find inverse of this matrix, we use Elementary Row Transformation Method

Using Elementary Row Transformation Method, we have

$\rm \: A \: = \: I \: A$

where I is the identity matrix of order as that of A.

So, on substituting the values of A and I, we get

$\rm \: \begin{gathered}\sf \left[\begin{array}{ccc}x&0&0\\0& y&0\\ 0&0&z\end{array}\right]\end{gathered} \: = \: \begin{gathered}\sf \left[\begin{array}{ccc}1&0&0\\0& 1&0\\ 0&0&1\end{array}\right]\end{gathered}A$

$\: \: \: \: \: \: \: \: \boxed{\sf{ \:OP \: R_1 \: \to \: \frac{1}{x} \: R_1 \: }}$

$\rm \: \begin{gathered}\sf \left[\begin{array}{ccc}1&0&0\\0& y&0\\ 0&0&z\end{array}\right]\end{gathered} \: = \: \begin{gathered}\sf \left[\begin{array}{ccc} \dfrac{1}{x} &0&0\\0& 1&0\\ 0&0&1\end{array}\right]\end{gathered}A$

$\: \: \: \: \: \: \: \: \boxed{\sf{ \:OP \: R_2 \: \to \: \frac{1}{y} \: R_2\: }}$

$\rm \: \begin{gathered}\sf \left[\begin{array}{ccc}1&0&0\\0& 1&0\\ 0&0&z\end{array}\right]\end{gathered} \: = \: \begin{gathered}\sf \left[\begin{array}{ccc} \dfrac{1}{x} &0&0\\0& \dfrac{1}{y} &0\\ 0&0&1\end{array}\right]\end{gathered}A$

$\: \: \: \: \: \: \: \: \boxed{\sf{ \:OP \: R_3\: \to \: \frac{1}{z} \: R_3\: }}$

$\rm \: \begin{gathered}\sf \left[\begin{array}{ccc}1&0&0\\0& 1&0\\ 0&0&1\end{array}\right]\end{gathered} \: = \: \begin{gathered}\sf \left[\begin{array}{ccc} \dfrac{1}{x} &0&0\\0& \dfrac{1}{y} &0\\ 0&0& \dfrac{1}{z} \end{array}\right]\end{gathered}A$

which can be further rewritten as

$\rm \: \begin{gathered}\sf \left[\begin{array}{ccc}1&0&0\\0& 1&0\\ 0&0&1\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{ccc} {x}^{ - 1} &0&0\\0& {y}^{ - 1} &0\\ 0&0& {z}^{ - 1} \end{array}\right]\end{gathered}A$

We know,

$\rm \: {A \: A}^{ - 1} = I$

So, on comparing, we get

$\rm\implies \: {A}^{ - 1} \: = \: \begin{gathered}\sf \left[\begin{array}{ccc} {x}^{ - 1} &0&0\\0& {y}^{ - 1} &0\\ 0&0& {z}^{ - 1} \end{array}\right]\end{gathered}$

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Additional Information

$\rm \: A \: (adj \: A) = (adjA)A = |A| I \:$

$\rm \: {AA}^{ - 1} = {A}^{ - 1}A = I \:$

$\rm \: |adj \: A| \: = \: { |A| }^{n - 1} \:$

$\rm \: |adj(adj \: A)| \: = \: { |A| }^{(n - 1)^{2} } \:$

$\rm \: |AR| = |A| \: |R|$

$\rm \: |k \: A| = {k}^{n} |A|$

Answer 2

Given matrix,

$A\:=\: \: \color{aqua} \begin{gathered}\left[\begin{array}{ccc}x&0&0 \\ 0&y&0 \\ 0&0&z\end{array}\right]\end{gathered}$

$The \: inverse \: of \: A \: will \: be \: {A}^{ - 1}$

So,

$\color{lime}\begin{gathered}\left[\begin{array}{ccc} {x}^{ - 1} &0 &0 \\ 0 & {y}^{ - 1} &0 \\ 0 &0 & {z}^{ - 1} \end{array}\right]\end{gathered}$