Question

Use elementary row transformation, find the inverse of the matrix A = \left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 2&5&7 \\\ { - 2}&{ - 4}&{ - 5} \end{array}} \right]

Answer 1

If we have to find ${A^{ - 1}}$ using row operations, write $A = IA$ and apply a sequence of row operations on $A = IA$ till we get, $I = BA$.Then, the matrix $B$ will be the inverse of $A$. Here I = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right] is the identity matrix.

Complete step-by-step answer:
Write $A = IA$, i.e.,
\left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 2&5&7 \\\ { - 2}&{ - 4}&{ - 5} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right]A
Applying ${R_2} \to {R_2} - 2{R_1}$,
\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 0&1&1 \\\ { - 2}&{ - 4}&{ - 5} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ { - 2}&1&0 \\\ 0&0&1 \end{array}} \right]A
Applying ${R_3} \to {R_3} + 2{R_1}$,
\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 0&1&1 \\\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ { - 2}&1&0 \\\ 2&0&1 \end{array}} \right]A
Applying ${R_1} \to {R_1} - 2{R_2}$,
\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&0&1 \\\ 0&1&1 \\\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5&{ - 2}&0 \\\ { - 2}&1&0 \\\ 2&0&1 \end{array}} \right]A
Applying ${R_1} \to {R_1} - {R_3}$,
\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&1 \\\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&{ - 1} \\\ { - 2}&1&0 \\\ 2&0&1 \end{array}} \right]A
Applying ${R_2} \to {R_2} - {R_3}$,
\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&{ - 1} \\\ { - 4}&1&{ - 1} \\\ 2&0&1 \end{array}} \right]A
Since it is of the form $I = BA$, where matrix $B$ will be the inverse of $A$.
\therefore {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&{ - 1} \\\ { - 4}&1&{ - 1} \\\ 2&0&1 \end{array}} \right]

Note: A rectangular matrix does not possess an inverse matrix. It means the inverse of a matrix is defined only for a square matrix. If $B$ is the inverse of matrix $A$, then $A$ is also the inverse of matrix $B$.