Question

Using elementary row transformation , find the inverse of the matrix given below ,
\left( {\begin{array}{*{20}{c}} 1&3&{ - 2} \\\ { - 3}&0&{ - 1} \\\ 2&1&0 \end{array}} \right)

Answer 1

In the given question, we have been asked to find the inverse of the given matrix and we have to find it by using elementary row transformation. To solve this question, we need to take an identity matrix on one side of the “equals” sign , and the given matrix on the other side. We have to transform the given matrix as an identity matrix by doing manipulation in the row only. we have to undo the matrix operations such as addition and subtraction , that has been done to the variables.

Complete step by step solution:
Consider $A = IA$
We will get ,
\Rightarrow \left( {\begin{array}{*{20}{c}} 1&3&{ - 2} \\\ { - 3}&0&{ - 1} \\\ 2&1&0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right)A
Applying ${R_2} \leftrightarrow {R_2} + {R_1} + {R_3}$ , we will get ,
\Rightarrow \left( {\begin{array}{*{20}{c}} 1&3&{ - 2} \\\ 0&4&{ - 3} \\\ 2&1&0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 1&1&1 \\\ 0&0&1 \end{array}} \right)A
Applying ${R_3} \leftrightarrow {R_3} - 2{R_1}$ , we will get ,
\Rightarrow \left( {\begin{array}{*{20}{c}} 1&3&{ - 2} \\\ 0&4&{ - 3} \\\ 0&{ - 5}&4 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 1&1&1 \\\ { - 2}&0&1 \end{array}} \right)A
Applying ${R_2} \leftrightarrow - {R_2} - {R_3}$ , we will get ,
\Rightarrow \left( {\begin{array}{*{20}{c}} 1&3&{ - 2} \\\ 0&1&{ - 1} \\\ 0&{ - 5}&4 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 1&{ - 1}&{ - 2} \\\ { - 2}&0&1 \end{array}} \right)A
Applying ${R_3} \leftrightarrow {R_3} + 5{R_2}$ , we will get ,
\Rightarrow \left( {\begin{array}{*{20}{c}} 1&3&{ - 2} \\\ 0&1&{ - 1} \\\ 0&0&{ - 1} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 1&{ - 1}&{ - 2} \\\ 3&{ - 5}&{ - 9} \end{array}} \right)A
Applying ${R_3} \leftrightarrow - {R_3}$ , we will get ,
\Rightarrow \left( {\begin{array}{*{20}{c}} 1&3&{ - 2} \\\ 0&1&{ - 1} \\\ 0&0&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 1&{ - 1}&{ - 2} \\\ { - 3}&5&9 \end{array}} \right)A
Applying ${R_2} \leftrightarrow {R_2} + {R_3}$ , we will get ,
\Rightarrow \left( {\begin{array}{*{20}{c}} 1&3&{ - 2} \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ { - 2}&4&7 \\\ { - 3}&5&9 \end{array}} \right)A
Applying ${R_1} \leftrightarrow {R_1} + 2{R_3}$ , we will get ,
\Rightarrow \left( {\begin{array}{*{20}{c}} 1&3&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 5}&{10}&{18} \\\ { - 2}&4&7 \\\ { - 3}&5&9 \end{array}} \right)A
Applying ${R_1} \leftrightarrow {R_1} - 3{R_2}$ , we will get ,
\Rightarrow \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 3} \\\ { - 2}&4&7 \\\ { - 3}&5&9 \end{array}} \right)A
Now, can say that $I = {A^{ - 1}}A$,
\therefore {A^{ - 1}} = \left( {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 3} \\\ { - 2}&4&7 \\\ { - 3}&5&9 \end{array}} \right)
We get the required result.

Note: The important thing to recollect about any equation is that the ‘equals’ sign represents a balance. What the sign says is that what’s on the left-hand side is strictly an equal to what’s on the right-hand side. It is the type of question where only mathematical operations such as addition, subtraction, multiplication and division is used. While doing transformation you have to be very careful , one mistake will cost you the wrong answer. Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily.