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# Find the inverse of the matrix ${\text{A = }}\left( {\begin{array}{*{20}{c}} 1&2&1 \\ 0&1&1 \\ 3&1&1 \end{array}} \right)$ by elementary transformation method?

Last updated date: 13th Jul 2024
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Hint: In this type of question where we have to find the inverse by elementary transformation method the important thing is to convert the given matrix into an identity matrix. We can use the elementary row or column transformation to convert the given matrix into an identity matrix.

Complete step-by-step solution -

It is given that we have to find the inverse of the given matrix by elementary transformation method. The given matrix is:
${\text{A = }}\left( {\begin{array}{*{20}{c}} 1&2&1 \\ 0&1&1 \\ 3&1&1 \end{array}} \right)\$ .
We know that for a matrix A, we can write:
A=AI, where
I, is the identity matrix.
Now, we will convert the matrix A on LHS to the identity matrix using elementary row or column transformation.
Putting the value of matrix A and identity matrix in above equation, we get:
$\ \left( {\begin{array}{*{20}{c}} 1&2&1 \\ 0&1&1 \\ 3&1&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right){\text{A}} \\ {\text{Applying }}{{\text{R}}_1} \to {{\text{R}}_1} - 2{{\text{R}}_2},{\text{ we get:}} \\ \left( {\begin{array}{*{20}{c}} 1&0&{ - 1} \\ 0&1&1 \\ 3&1&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&{ - 2}&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right){\text{A}}{\text{.}} \\ {\text{Applying }}{{\text{C}}_3} \to {{\text{C}}_3}{\text{ + }}{{\text{C}}_1},{\text{ we get:}} \\ \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&1 \\ 3&1&4 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&{ - 2}&1 \\ 0&1&0 \\ 0&0&1 \end{array}} \right){\text{A}}{\text{.}} \\ \ \ \\ {\text{Applying }}{{\text{C}}_3} \to {{\text{C}}_3}{\text{ - }}{{\text{C}}_2},{\text{ we get:}} \\ \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 3&1&3 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&{ - 2}&3 \\ 0&1&{ - 1} \\ 0&0&1 \end{array}} \right){\text{A}}{\text{.}} \\ {\text{Applying }}{{\text{C}}_1} \to {{\text{C}}_1}{\text{ - }}{{\text{C}}_3},{\text{ we get:}} \\ \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&1&3 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 2}&{ - 2}&3 \\ 1&1&{ - 1} \\ { - 1}&0&1 \end{array}} \right){\text{A}}{\text{.}} \\ {\text{Applying }}{{\text{R}}_3} \to {{\text{R}}_3} - {{\text{R}}_2},{\text{ we get:}} \\ \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&3 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 2}&{ - 2}&3 \\ 1&1&{ - 1} \\ { - 2}&{ - 1}&2 \end{array}} \right){\text{A}}{\text{.}} \\ {\text{Applying }}{{\text{R}}_3} \to \dfrac{{{{\text{R}}_3}}}{3},{\text{ we get:}} \\ \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 2}&{ - 2}&3 \\ 1&1&{ - 1} \\ {\dfrac{{ - 2}}{3}}&{\dfrac{{ - 1}}{3}}&{\dfrac{2}{3}} \end{array}} \right){\text{A}}{\text{.}} \\ \\ \$ (1)
Now. We know that:
${\text{A = AI}} \\ \Rightarrow {\text{I = }}{{\text{A}}^{ - 1}}{\text{A}} \\$ (2)
We have converted the matrix A into the identity matrix which is given by equation 1.
Therefore, on comparing equation 1 and 2, we get:
${\text{Inverse of given matrix A = }}{{\text{A}}^{ - 1}} = \left( {\begin{array}{*{20}{c}} { - 2}&{ - 2}&3 \\ 1&1&{ - 1} \\ {\dfrac{{ - 2}}{3}}&{\dfrac{{ - 1}}{3}}&{\dfrac{2}{3}} \end{array}} \right).$

Note: Before solving such a type of problem, you should be familiar with the basics property of the identity matrix i.e. ${\text{A = AI}}$ . On rearranging this relation, we get ${\text{I = }}{{\text{A}}^{ - 1}}{\text{A}}$ which clearly says that we have to convert the matrix A on LHS to identity matrix and the first matrix on RHS will give the inverse of the given matrix A. Also you should know how to use row and column transformation.