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Question: Find the inverse of matrix \(\left[ \begin{matrix} 1 & 2 & 1 \\\ -1 & 0 & 2 \\\ 2 & 1...

Find the inverse of matrix [121 −102 21−3 ]\left[ \begin{matrix} 1 & 2 & 1 \\\ -1 & 0 & 2 \\\ 2 & 1 & -3 \\\ \end{matrix} \right] by elementary row transformation.

Explanation

Solution

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=[121 −102 21−3 ]A=\left[ \begin{matrix} 1 & 2 & 1 \\\ -1 & 0 & 2 \\\ 2 & 1 & -3 \\\ \end{matrix} \right] . Let us find the determinant of the matrix.
∣A∣=1(0−2)−2(3−4)+1(−1−0) =−2+2−1 =−1≠0 \begin{aligned} & \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{aligned}
Hence, the inverse of the AA matrix exists.
Now as we know AA−1=IA{{A}^{-1}}=I, so let us put the value of AA and II , where IIis 3×33\times 3 identity matrix.
[121 −102 21−3 ]A−1=[100 010 001 ]\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 AA as II and hence find our answer.
R1{{R}_{1}} will represent row 11.
R2{{R}_{2}} will represent row 22.
R3{{R}_{3}} will represent row 33.
[121 −102 21−3 ]A−1=[100 010 001 ]\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 R2{{R}_{2}}and R3{{R}_{3}} simultaneously, using operations R2→R2+R1{{R}_{2}}\to {{R}_{2}}+{{R}_{1}} and R3→R3−2R1{{R}_{3}}\to {{R}_{3}}-2{{R}_{1}}, we get –
[121 023 0−3−5 ]A−1=[100 010 −201 ]\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 R3{{R}_{3}} using operations, R3→R3+32R2{{R}_{3}}\to {{R}_{3}}+\dfrac{3}{2}{{R}_{2}}, we get –

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 ${{R}_{2}}\to {{R}_{2}}-{{R}_{1}}$.