Question

Find the inverse of the matrix \[\left[ {\begin{array}{*{20}{c}}
2&0&{ - 1} \\
5&1&0 \\
0&1&3
\end{array}} \right]\] using elementary row or column transformation, if it exists.

Answer 1

Hint:If in a \[\left( {n \times n} \right)\] matrix A an inverse \[{A^{ - 1}}\] exists then \[A{A^{ -
1}} = {A^{ - 1}}A = I\]

In this question we need to find the inverse of the matrix using the elementary method. In the elementary transformation method, we start with a row or column and by applying various transformations on the chosen identity either on rows or the column we try to make maximum zeros of the identity. By using the concept of \[A{A^{ - 1}} = {A^{ - 1}}A = I\] we will find the inverse of the matrix.

Complete step by step solution:
Given the matrix whose inverse is to be find is \[\left[ {\begin{array}{*{20}{c}}
2&0&{ - 1} \\
5&1&0 \\
0&1&3
\end{array}} \right]\]

Let \[X = \left[ {\begin{array}{*{20}{c}}
2&0&{ - 1} \\
5&1&0 \\
0&1&3
\end{array}} \right]\]

First we will find the value of the given matrix, hence we can write
 \[\left| X \right| = \left[ {\begin{array}{*{20}{c}}
2&0&{ - 1} \\
5&1&0 \\
0&1&3

\end{array}} \right] = 2\left( {3 - 0} \right) - 0\left( {5 - 0} \right) + \left( { - 1} \right)\left( {5 - 0} \right)
= 6 - 5 = 1\]

So since the value of the matrix \[\left| X \right| \ne 0\] hence we can say the inverse of the
matrix \[{X^{ - 1}}\] exists
Now we know if inverse of a matrix exists then \[A{A^{ - 1}} = {A^{ - 1}}A = I\] where \[I\] is a singular
matrix, hence we can write
 \[{A^{ - 1}} = \dfrac{{adj\left( A \right)}}{{\left| A \right|}}\]
Now we substitute the value in the above equation, we can write
 \[{X^{ - 1}}\left[ {\begin{array}{*{20}{c}}
2&0&{ - 1} \\
5&1&0 \\
0&1&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&0&0 \\
0&1&0 \\
0&0&1
\end{array}} \right]\]
Now we will do the row transformation in the matrix, by applying \[{R_1} \to \left( {\dfrac{1}{2}}
\right){R_1}\]
 \[{X^{ - 1}}\left[ {\begin{array}{*{20}{c}}
1&0&{ - \dfrac{1}{2}} \\
5&1&0 \\
0&1&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{\dfrac{1}{2}}&0&0 \\
0&1&0 \\
0&0&1

\end{array}} \right]\]
Now apply \[{R_2} \to {R_2} + \left( { - 5} \right){R_1}\]
 \[{X^{ - 1}}\left[ {\begin{array}{*{20}{c}}
1&0&{ - \dfrac{1}{2}} \\
0&1&{\dfrac{5}{2}} \\
0&1&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{\dfrac{1}{2}}&0&0 \\
{ - \dfrac{5}{2}}&1&0 \\
0&0&1
\end{array}} \right]\]
Now apply \[{R_3} \to {R_3} + \left( { - 1} \right){R_2}\]
 \[{X^{ - 1}}\left[ {\begin{array}{*{20}{c}}
1&0&{ - \dfrac{1}{2}} \\
0&1&{\dfrac{5}{2}} \\
0&0&{\dfrac{1}{2}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{\dfrac{1}{2}}&0&0 \\
{ - \dfrac{5}{2}}&1&0 \\
{\dfrac{5}{2}}&{ - 1}&1
\end{array}} \right]\]
Now apply \[{R_3} \to \left( 2 \right){R_3}\]
 \[{X^{ - 1}}\left[ {\begin{array}{*{20}{c}}
1&0&{ - \dfrac{1}{2}} \\
0&1&{\dfrac{5}{2}} \\

0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{\dfrac{1}{2}}&0&0 \\
{ - \dfrac{5}{2}}&1&0 \\
{\dfrac{5}{2}}&{ - 1}&2
\end{array}} \right]\]
Now we apply \[{R_1} \to {R_1} + \left( {\dfrac{1}{2}} \right){R_3}\] and \[{R_2} \to {R_2} + \left( { -
\dfrac{5}{2}} \right){R_3}\] , we get
 \[{X^{ - 1}}\left[ {\begin{array}{*{20}{c}}
1&0&0 \\
0&1&0 \\
0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3&{ - 1}&1 \\
{ - 15}&6&{ - 5} \\
5&{ - 2}&2
\end{array}} \right]\]
Now since \[\left[ {\begin{array}{*{20}{c}}
1&0&0 \\
0&1&0 \\
0&0&1
\end{array}} \right] = 1\] , hence we can write this equation as
 \[{X^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
3&{ - 1}&1 \\
{ - 15}&6&{ - 5} \\
5&{ - 2}&2

\end{array}} \right]\]
Therefore the inverse of the matrix \[\left[ {\begin{array}{*{20}{c}}
2&0&{ - 1} \\
5&1&0 \\
0&1&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
3&{ - 1}&1 \\
{ - 15}&6&{ - 5} \\
5&{ - 2}&2
\end{array}} \right]\]

Note: The inverse of a matrix can also be found by using the formula \[{A^{ - 1}} = \dfrac{{adj\left( A
\right)}}{{\left| A \right|}}\] which is also known as adjoint method to find the inverse of the matrix. In this method we find the determinant value of the matrix and its joint value, so by finding their ratio we can find the inverse of the matrix.