Question

45.
Find inverse of the following matrices (if they exist) by elementary transformation
[2 1
[7 4]​

Answer 1

Step-by-step explanation:

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Answer 2

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

➢ Given matrix is

$\rm :\longmapsto\:\bigg[ \begin{matrix}2&1 \\ 7&4 \end{matrix} \bigg]$

Let assume that

$\rm :\longmapsto\:A = \bigg[ \begin{matrix}2&1 \\ 7&4 \end{matrix} \bigg]$

Using Elementary Row Transformation Method, we have

$\rm :\longmapsto\:A = IA$

$\rm :\longmapsto\: \bigg[ \begin{matrix}2&1 \\ 7&4 \end{matrix} \bigg] = \bigg[ \begin{matrix}1&0 \\ 0&1 \end{matrix} \bigg]A$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \purple{ \boxed{\bf{OP \: R_1 \: \to \: 4R_1}}}$

$\rm :\longmapsto\: \bigg[ \begin{matrix}8&4 \\ 7&4 \end{matrix} \bigg] = \bigg[ \begin{matrix}4&0 \\ 0&1 \end{matrix} \bigg]A$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \purple{ \boxed{\bf{OP \: R_1 \: \to \: R_1 - R_2}}}$

$\rm :\longmapsto\: \bigg[ \begin{matrix}1&0 \\ 7&4 \end{matrix} \bigg] = \bigg[ \begin{matrix}4& - 1 \\ 0&1 \end{matrix} \bigg]A$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \purple{ \boxed{\bf{OP \: R_2 \: \to \: R_2 - 7R_1}}}$

$\rm :\longmapsto\: \bigg[ \begin{matrix}1&0 \\ 0&4 \end{matrix} \bigg] = \bigg[ \begin{matrix}4& - 1 \\ - 28&8 \end{matrix} \bigg]A$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \purple{ \boxed{\bf{OP \: R_2 \: \to \: \frac{1}{4} R_2 }}}$

$\rm :\longmapsto\: \bigg[ \begin{matrix}1&0 \\ 0&1 \end{matrix} \bigg] = \bigg[ \begin{matrix}4& - 1 \\ - 7&2 \end{matrix} \bigg]A$

We know that

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

So, on comparing we get

$\red{\bf :\longmapsto\: {A}^{ - 1} = \bigg[ \begin{matrix}4& - 1 \\ - 7&2 \end{matrix} \bigg]}$

Additional Information :-

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

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

$\rm :\longmapsto\: |Aadj \: A| = { |A| }^{n}$

$\rm :\longmapsto\: | {A}^{ - 1} | = \dfrac{1}{ |A| }$

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

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