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# Find the inverse of matrix $\left[ \begin{matrix} 1 & 2 & 1 \\ -1 & 0 & 2 \\ 2 & 1 & -3 \\\end{matrix} \right]$ by elementary row transformation.

Last updated date: 15th Sep 2024
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Hint: In this question, we will first find the determinant of the matrix to check if its inverse exists. If the determinant is non-zero, only then will we proceed further. After that, we will use elementary row transformations to find the inverse of the matrix. We will use various operations step by step to reach our answer.

Complete step-by-step solution
We are given the matrix $A=\left[ \begin{matrix} 1 & 2 & 1 \\ -1 & 0 & 2 \\ 2 & 1 & -3 \\ \end{matrix} \right]$ . Let us find the determinant of the matrix.
\begin{align} & \left| A \right|=1\left( 0-2 \right)-2\left( 3-4 \right)+1\left( -1-0 \right) \\ & =-2+2-1 \\ & =-1\ne 0 \\ \end{align}
Hence, the inverse of the $A$ matrix exists.
Now as we know $A{{A}^{-1}}=I$, so let us put the value of $A$ and $I$ , where $I$is $3\times 3$ identity matrix.
$\left[ \begin{matrix} 1 & 2 & 1 \\ -1 & 0 & 2 \\ 2 & 1 & -3 \\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]$
Now we apply various row transformations to make $A$ as $I$ and hence find our answer.
${{R}_{1}}$ will represent row $1$.
${{R}_{2}}$ will represent row $2$.
${{R}_{3}}$ will represent row $3$.
$\left[ \begin{matrix} 1 & 2 & 1 \\ -1 & 0 & 2 \\ 2 & 1 & -3 \\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]$
Operating of ${{R}_{2}}$and ${{R}_{3}}$ simultaneously, using operations ${{R}_{2}}\to {{R}_{2}}+{{R}_{1}}$ and ${{R}_{3}}\to {{R}_{3}}-2{{R}_{1}}$, we get –
$\left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 2 & 3 \\ 0 & -3 & -5 \\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 0 & 1 \\ \end{matrix} \right]$
Operating on ${{R}_{3}}$ using operations, ${{R}_{3}}\to {{R}_{3}}+\dfrac{3}{2}{{R}_{2}}$, we get –
$\left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & -\dfrac{1}{2} \\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ -\dfrac{1}{2} & \dfrac{3}{2} & 1 \\ \end{matrix} \right]$
Again operating on ${{R}_{3}}$ using operations, ${{R}_{3}}\to \dfrac{{{R}_{3}}}{-\dfrac{1}{2}}$, we get –
$\left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 1 \\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & -3 & -2 \\ \end{matrix} \right]$
Now operating on ${{R}_{1}}$ using operations, ${{R}_{3}}\to {{R}_{1}}-{{R}_{2}}$, we get –
$\left[ \begin{matrix} 1 & 0 & -2 \\ 0 & 2 & 3 \\ 0 & 0 & 1 \\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} 0 & -1 & 0 \\ 1 & 1 & 0 \\ 1 & -3 & -2 \\ \end{matrix} \right]$
Operating on ${{R}_{1}}$ using operations, ${{R}_{3}}\to {{R}_{1}}+2{{R}_{3}}$, we get –
$\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 3 \\ 0 & 0 & 1 \\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} 2 & -7 & -4 \\ 1 & 1 & 0 \\ 1 & -3 & -2 \\ \end{matrix} \right]$
Operating on ${{R}_{2}}$ using operations, ${{R}_{2}}\to \dfrac{{{R}_{2}}}{2}$, we get –
$\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & \dfrac{3}{2} \\ 0 & 0 & 1 \\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} 2 & -7 & -4 \\ \dfrac{1}{2} & \dfrac{1}{2} & 0 \\ 1 & -3 & -2 \\ \end{matrix} \right]$
At last operating on ${{R}_{2}}$ using operations, ${{R}_{2}}\to {{R}_{2}}-\dfrac{3}{2}{{R}_{3}}$, we get –
$\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]{{A}^{-1}}=\left[ \begin{matrix} 2 & -7 & -4 \\ -1 & 5 & 3 \\ 1 & -3 & -2 \\ \end{matrix} \right]$
Hence, we have got $I{{A}^{-1}}=\left[ \begin{matrix} 2 & -7 & -4 \\ -1 & 5 & 3 \\ 1 & -3 & -2 \\ \end{matrix} \right]$
Since,$IA=AI=A$, therefore, ${{A}^{-1}}=\left[ \begin{matrix} 2 & -7 & -4 \\ -1 & 5 & 3 \\ 1 & -3 & -2 \\ \end{matrix} \right]$, which is our required answer.

Note: Students should know that while using elementary transformations, if we are using row transformation, then we have to use rows only. We cannot use column transformation in between. Also, we are not allowed to alter any row, for example, we cannot write ${{R}_{2}}\to {{R}_{1}}+{{R}_{3}}$. Also, the row which is to be transformed should come first in the equation, meaning we cannot write ${{R}_{2}}\to {{R}_{1}}-{{R}_{2}}$. We can only write ${{R}_{2}}\to {{R}_{2}}-{{R}_{1}}$.