Question

Obtain the inverse of the following matrix using elementary operations $A = \left[ {\begin{array}{*{20}{c}}
  0&1&2 \\
  1&2&3 \\
  3&1&1
\end{array}} \right]$

Answer 1

Hint: In this question, we need to determine the inverse of matrix A through elementary operations only. For this, we will use the property of the matrix, i.e., $A = IA$ and perform the row transformation here.

Complete step-by-step answer:
The given matrix $A = \left[ {\begin{array}{*{20}{c}}
  0&1&2 \\
  1&2&3 \\
  3&1&1
\end{array}} \right]$ is a 3x3 matrix.
The first step in determining the inverse of the square matrix is to consider $A = IA$ such that $I$ is an elementary matrix of size 3x3 having values of the cells as ${I_{3 \times 3}} = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$
So, we can write:
$\left[ {\begin{array}{*{20}{c}}
  0&1&2 \\
  1&2&3 \\
  3&1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]A - - - - (i)$
Now, we have to do elementary row transformation here so as to obtain an elementary matrix on the left-hand side of the equation (i) and thus the corresponding matrix associated with A in the right-hand side of the equation (i) will be the inverse of the matrix A as $A{A^{ - 1}} = I$.

Applying ${R_1} \leftrightarrow {R_2}$ to both sides of the equation (i), we get
$\left[ {\begin{array}{*{20}{c}}
  1&2&3 \\
  0&1&2 \\
  3&1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&1&0 \\
  1&0&0 \\
  0&0&1
\end{array}} \right]A - - - - (ii)$
Applying ${R_3} \to {R_3} - 3{R_1}$ to both sides of the equation (ii), we get$
  \left[ {\begin{array}{*{20}{c}}
  1&2&3 \\
  0&1&2 \\
  {3 - 3}&{1 - 6}&{1 - 9}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&1&0 \\
  1&0&0 \\
  0&{0 - 3}&{1 - 0}
\end{array}} \right]A \\
  \left[ {\begin{array}{*{20}{c}}
  1&2&3 \\
  0&1&2 \\
  0&{ - 5}&{ - 8}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&1&0 \\
  1&0&0 \\
  0&{ - 3}&1
\end{array}} \right]A - - - - (iii) \\
 $

Applying ${R_1} \to {R_1} - 2{R_2}$ to both sides of the equation (iii), we get$
  \left[ {\begin{array}{*{20}{c}}
  {1 - 0}&{2 - 2}&{3 - 4} \\
  0&1&2 \\
  0&{ - 5}&{ - 8}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {0 - 2}&{1 - 0}&0 \\
  1&0&0 \\
  0&{ - 3}&1
\end{array}} \right]A \\
  \left[ {\begin{array}{*{20}{c}}
  1&0&{ - 1} \\
  0&1&2 \\
  0&{ - 5}&{ - 8}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 2}&1&0 \\
  1&0&0 \\
  0&{ - 3}&1
\end{array}} \right]A - - - - (iv) \\
 $

Applying ${R_3} \to {R_3} + 5{R_2}$ to both sides of the equation (iv), we get$
  \left[ {\begin{array}{*{20}{c}}
  1&0&{ - 1} \\
  0&1&2 \\
  {0 + 0}&{ - 5 + 5}&{ - 8 + 10}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 2}&1&0 \\
  1&0&0 \\
  {0 + 5}&{ - 3 + 0}&{1 + 0}
\end{array}} \right]A \\
  \left[ {\begin{array}{*{20}{c}}
  1&0&{ - 1} \\
  0&1&2 \\
  0&0&2
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 2}&1&0 \\
  1&0&0 \\
  5&{ - 3}&1
\end{array}} \right]A - - - - (v) \\
 $
Applying ${R_3} \to \dfrac{1}{2}{R_3}$ to both sides of the equation (v), we get$\left[ {\begin{array}{*{20}{c}}
  1&0&{ - 1} \\
  0&1&2 \\
  0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 2}&1&0 \\
  1&0&0 \\
  {\dfrac{5}{2}}&{\dfrac{{ - 3}}{2}}&{\dfrac{1}{2}}
\end{array}} \right]A - - - - (vi)$

Applying ${R_1} \to {R_1} + {R_3}$ to both sides of the equation (vi), we get$
  \left[ {\begin{array}{*{20}{c}}
  {1 + 0}&{0 + 0}&{ - 1 + 1} \\
  0&1&2 \\
  0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 2 + \dfrac{5}{2}}&{1 - \dfrac{3}{2}}&{0 + \dfrac{1}{2}} \\
  1&0&0 \\
  {\dfrac{5}{2}}&{\dfrac{{ - 3}}{2}}&{\dfrac{1}{2}}
\end{array}} \right]A \\
  \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&2 \\
  0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{2}}&{ - \dfrac{1}{2}}&{\dfrac{1}{2}} \\
  1&0&0 \\
  {\dfrac{5}{2}}&{\dfrac{{ - 3}}{2}}&{\dfrac{1}{2}}
\end{array}} \right]A - - - - (vii) \\
 $

Applying ${R_2} \to {R_2} - 2{R_3}$ to both sides of the equation (vii), we get$
  \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  {0 - 0}&{1 - 0}&{2 - 2} \\
  0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{2}}&{ - \dfrac{1}{2}}&{\dfrac{1}{2}} \\
  {1 - 2\left( {\dfrac{5}{2}} \right)}&{0 - 2\left( {\dfrac{{ - 3}}{2}} \right)}&{0 - 2\left( {\dfrac{1}{2}} \right)} \\
  {\dfrac{5}{2}}&{\dfrac{{ - 3}}{2}}&{\dfrac{1}{2}}
\end{array}} \right]A \\
  \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{2}}&{ - \dfrac{1}{2}}&{\dfrac{1}{2}} \\
  { - 4}&3&{ - 1} \\
  {\dfrac{5}{2}}&{\dfrac{{ - 3}}{2}}&{\dfrac{1}{2}}
\end{array}} \right]A - - - - (viii) \\
 $

From equation (viii) we can see that the left-hand side is an elementary equation so, the matrix associated with A to the right-hand side of the equation will be the inverse of the A matrix.
Hence, ${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{2}}&{ - \dfrac{1}{2}}&{\dfrac{1}{2}} \\
  { - 4}&3&{ - 1} \\
  {\dfrac{5}{2}}&{\dfrac{{ - 3}}{2}}&{\dfrac{1}{2}}
\end{array}} \right]$.

Note: It should be noted here that the while performing elementary operations on the matrix always follow either of the row operations or the column operations but not both. Here, we have used only the row operations.