Question

Find the inverse of the following matrix by using the elementary row transformation:
\[\left[ \begin{matrix}
   2 & 3 & 1 \\
   2 & 4 & 1 \\
   3 & 7 & 2 \\
\end{matrix} \right]\].

Answer 1

Hint: To find the inverse of the given matrix using the elementary row transformation, we will first find out what is elementary row transformations and what kind of transformations can we apply on any matrix. Now, we will write \[A=A{{I}_{n}}\] where \[{{I}_{n}}\] is the unit matrix and the \[{{n}^{th}}\] order. We will convert this to \[{{I}_{n}}=A{{A}^{-1}}\] and thus we will determine \[{{A}^{-1}}.\]

Complete step by step answer:
Before we solve this question, we must know what elementary row transformation is. Elementary row transformations are those operations performed on the rows of the matrices to transform it into a different form so that the calculations become simpler. We can apply three kinds of elementary row transformations which are shown below:
(i) We can interchange the rows within the matrix
(ii) We can multiply the entire row with the same non – zero number
(iii) We can add one row to another row multiplied by a non – zero number.
Now, we will find the inverse of the above matrix. Let the matrix be denoted by A. Thus, we have,
\[A=\left[ \begin{matrix}
   2 & 3 & 1 \\
   2 & 4 & 1 \\
   3 & 7 & 2 \\
\end{matrix} \right]......\left( i \right)\].
Now, if we multiply any matrix with a unit matrix, the matrix remains the same. Thus, we can say that,
\[A=A{{I}_{n}}\]
where n is the order of the matrix. Thus, we will have,
\[\Rightarrow \left[ \begin{matrix}
   2 & 3 & 1 \\
   2 & 4 & 1 \\
   3 & 7 & 2 \\
\end{matrix} \right]=A\left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]\].
Now, we will apply the elementary row transformations. In the first step, we will multiply the first row by $\dfrac{1}{2}$ i.e., $\left( {{R}_{1}}\to \dfrac{1}{2}\times {{R}_{1}} \right)$. Thus, we will get,
\[\Rightarrow \left[ \begin{matrix}
   1 & \dfrac{3}{2} & \dfrac{1}{2} \\
   2 & 4 & 1 \\
   3 & 7 & 2 \\
\end{matrix} \right]=A\left[ \begin{matrix}
   \dfrac{1}{2} & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]\].
Now, we will multiply the first row with 2 and we will subtract it from the second row i.e., $\left( {{R}_{2}}\to {{R}_{2}}-2\times {{R}_{1}} \right)$. Thus, we will get,
\[\Rightarrow \left[ \begin{matrix}
   1 & \dfrac{3}{2} & \dfrac{1}{2} \\
   0 & 1 & 0 \\
   3 & 7 & 2 \\
\end{matrix} \right]=A\left[ \begin{matrix}
   \dfrac{1}{2} & 0 & 0 \\
   -1 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]\].
Now, we will multiply the first row with 3 and we will subtract it from the third row i.e., $\left( {{R}_{3}}\to {{R}_{3}}-3\times {{R}_{1}} \right)$. Thus, we will get,
\[\Rightarrow \left[ \begin{matrix}
   1 & \dfrac{3}{2} & \dfrac{1}{2} \\
   0 & 1 & 0 \\
   0 & \dfrac{5}{2} & \dfrac{1}{2} \\
\end{matrix} \right]=A\left[ \begin{matrix}
   \dfrac{1}{2} & 0 & 0 \\
   -1 & 1 & 0 \\
   \dfrac{-3}{2} & 0 & 1 \\
\end{matrix} \right]\].
Now, we will multiply the second row with \[\dfrac{5}{2}\] and then we will subtract it from the third row i.e., $\left( {{R}_{3}}\to {{R}_{3}}-\dfrac{5}{2}\times {{R}_{2}} \right)$. Thus, we will get,
\[\Rightarrow \left[ \begin{matrix}
   1 & \dfrac{3}{2} & \dfrac{1}{2} \\
   0 & 1 & 0 \\
   0 & 0 & \dfrac{1}{2} \\
\end{matrix} \right]=A\left[ \begin{matrix}
   \dfrac{1}{2} & 0 & 0 \\
   -1 & 1 & 0 \\
   1 & \dfrac{-5}{2} & 1 \\
\end{matrix} \right]\].
Now, we will multiply the second row with \[\dfrac{3}{2}\] and we will subtract it from the first row i.e., $\left( {{R}_{1}}\to {{R}_{1}}-\dfrac{3}{2}\times {{R}_{2}} \right)$. Thus, we will get,
\[\Rightarrow \left[ \begin{matrix}
   1 & 0 & \dfrac{1}{2} \\
   0 & 1 & 0 \\
   0 & 0 & \dfrac{1}{2} \\
\end{matrix} \right]=A\left[ \begin{matrix}
   2 & \dfrac{-3}{2} & 0 \\
   -1 & 1 & 0 \\
   1 & \dfrac{-5}{2} & 1 \\
\end{matrix} \right]\].
Now, we will subtract the third row from the first row i.e., $\left( {{R}_{1}}\to {{R}_{1}}-{{R}_{3}} \right)$. Thus, we will get,
\[\Rightarrow \left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & \dfrac{1}{2} \\
\end{matrix} \right]=A\left[ \begin{matrix}
   1 & 1 & -1 \\
   -1 & 1 & 0 \\
   1 & \dfrac{-5}{2} & 1 \\
\end{matrix} \right]\].
Now, we will multiply the third row with 2 i.e., $\left( {{R}_{3}}\to 2\times {{R}_{3}} \right)$. Thus, we will get,
\[\Rightarrow \left[ \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right]=A\left[ \begin{matrix}
   1 & 1 & -1 \\
   -1 & 1 & 0 \\
   2 & -5 & 2 \\
\end{matrix} \right]\]
\[\Rightarrow I=A\left[ \begin{matrix}
   1 & 1 & -1 \\
   -1 & 1 & 0 \\
   2 & -5 & 2 \\
\end{matrix} \right].....\left( ii \right)\].
We know that, \[A{{A}^{-1}}=I.....\left( iii \right).\] So, we will get,
\[\Rightarrow {{A}^{-1}}=\left[ \begin{matrix}
   1 & 1 & -1 \\
   -1 & 1 & 0 \\
   2 & -5 & 2 \\
\end{matrix} \right]\].

Note: We have written in the solution that \[A=A{{I}_{n}}.\] In our case, the value of n = 3. The n cannot be any other value because the multiplication of A and I is possible only when they have the same order. The order of A is 3, so the value of n must also be 3.