Question

Using the elementary operations, find the inverse of the following matrix
\[\left( {\begin{array}{*{20}{c}}
  { - 1}&1&2 \\
  1&2&3 \\
  3&1&1
\end{array}} \right)\]

Answer 1

Hint: Let the given matrix be A. The given matrix A can also be expressed as $A = AI$, where $I$ is the identity matrix. So the inverse of $A$ can be expressed as $I = {A^{ - 1}}A$ Apply the necessary column and row matrix. Now using the necessary row and column transformation makes the LHS matrix an identity matrix to find the ${A^{ - 1}}$ .

Complete step by step answer:
Now we can express the given matrix A as $A=AI$ , where $I$ represents the identity matrix.
So now we can express the matrix as $I = {A^{ - 1}}A$ for doing the necessary elementary transformations.
The elementary transformations should be done in such a way that it should make the LHS an identity matrix through step by step operations or procedures.
Now expressing the matrix $A$ as $A=IA$ we get
\[\left( \begin{matrix}
   -1 & 1 & 2 \\
   1 & 2 & 3 \\
   3 & 1 & 1 \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right)A\] \[\]
Applying ${{R}_{1}}\rightleftharpoons {{R}_{2}}$ , we get
\[\left( \begin{matrix}
   1 & 2 & 3 \\
   -1 & 1 & 2 \\
   3 & 1 & 1 \\
\end{matrix} \right)=\left( \begin{matrix}
   0 & 1 & 0 \\
   1 & 0 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right)A\]
Now applying ${{R}_{2}}\to {{R}_{2}}+{{R}_{1}}$ and ${{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}$ , the equation becomes
\[\left( \begin{matrix}
   1 & 2 & 3 \\
   0 & 3 & 5 \\
   0 & -5 & -8 \\
\end{matrix} \right)=\left( \begin{matrix}
   0 & 1 & 0 \\
   1 & 1 & 0 \\
   0 & -3 & 1 \\
\end{matrix} \right)A\]
Now applying ${{R}_{1}}\to {{R}_{1}}-\dfrac{2}{3}{{R}_{2}}$ , the equation becomes

\[\left( \begin{matrix}
   1 & 0 & \dfrac{-1}{3} \\
   0 & 3 & 5 \\
   0 & -5 & -8 \\
\end{matrix} \right)=\left( \begin{matrix}
   \dfrac{-2}{3} & \dfrac{1}{3} & 0 \\
   1 & 1 & 0 \\
   0 & -3 & 1 \\
\end{matrix} \right)A\]
Apply ${{R}_{2}}\to \dfrac{1}{3}{{R}_{2}}$ to get the equation as
\[\left( \begin{matrix}
   1 & 0 & \dfrac{-1}{3} \\
   0 & 1 & \dfrac{5}{3} \\
   0 & -5 & -8 \\
\end{matrix} \right)=\left( \begin{matrix}
   \dfrac{-2}{3} & \dfrac{1}{3} & 0 \\
   \dfrac{1}{3} & \dfrac{1}{3} & 0 \\
   0 & -3 & 1 \\
\end{matrix} \right)A\]
Apply ${{R}_{3}}\to {{R}_{3}}+5{{R}_{2}}$.
\[\left( \begin{matrix}
   1 & 0 & \dfrac{-1}{3} \\
   0 & 1 & \dfrac{5}{3} \\
   0 & 0 & \dfrac{1}{3} \\
\end{matrix} \right)=\left( \begin{matrix}
   \dfrac{-2}{3} & \dfrac{1}{3} & 0 \\
   \dfrac{1}{3} & \dfrac{1}{3} & 0 \\
   \dfrac{5}{3} & \dfrac{-4}{3} & 1 \\
\end{matrix} \right)A\]
Apply ${{R}_{1}}\to {{R}_{1}}+{{R}_{3}}$ and ${{R}_{2}}\to {{R}_{2}}-5{{R}_{3}}$ , the equation becomes
\[\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & \dfrac{1}{3} \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 1 & 1 \\
   -8 & 7 & -5 \\
   \dfrac{5}{3} & \dfrac{-4}{3} & 1 \\
\end{matrix} \right)A\]
Applying the final elementary transformation ${{R}_{3}}\to 3{{R}_{3}}$ to make the equation as
\[\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & -1 & 1 \\
   -8 & 7 & 5 \\
   5 & -4 & 3 \\
\end{matrix} \right)A\]
Here the LHS has become an identity matrix, so we can stop with the elementary transformation procedures.
Now taking the RHS A to the LHS, we get the ${{A}^{-1}}$ .
Hence ${{A}^{-1}}=\left( \begin{matrix}
   1 & -1 & 1 \\
   -8 & 7 & 5 \\
   5 & -4 & 3 \\
\end{matrix} \right)$
Thus the inverse of the given matrix is $\left( \begin{matrix}
   1 & -1 & 1 \\
   -8 & 7 & 5 \\
   5 & -4 & 3 \\
\end{matrix} \right)$

Note:
The elementary transformation should be done in such a way that the LHS side has to be made an identity matrix. Be careful while doing the mathematical operations in the elementary transformation as the student can usually commit a mistake in performing them. One silly mistake in the elementary transformation can make the entire solution wrong.