Question

Find the inverse of following matrix using elementary row transformation:
$\left[ \begin{matrix}
   5 & 2 \\
   2 & 1 \\
\end{matrix} \right]$

Answer 1

Hint: In elementary row transformation we always try to write a given matrix in the form of $A{{A}^{-1}}=I$ by using transformation either in row or column.

Complete step-by-step answer:
The given matrix is A.
So we can write
$A=\left[ \begin{matrix}
   5 & 2 \\
   2 & 1 \\
\end{matrix} \right]$
By property of matrix we can write
$\Rightarrow A=AI$
Hence we have
$\Rightarrow \left[ \begin{matrix}
   5 & 2 \\
   2 & 1 \\
\end{matrix} \right]=A\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]$ because I is a unit matrix.
Now we will try to convert the left hand side matrix as a unit matrix by using transformation of row or columns. We will do the same changes in the right hand side unit matrix.
We can divide row 1 by 5 to make 5 as 1.
${{R}_{1}}\to \dfrac{{{R}_{1}}}{5}$
$\Rightarrow \left[ \begin{matrix}
   1 & \dfrac{2}{5} \\
   2 & 1 \\
\end{matrix} \right]=A\left[ \begin{matrix}
   \dfrac{1}{5} & 0 \\
   0 & 1 \\
\end{matrix} \right]$
Now to make 2 as zero in row 2 we can subtract row 2 by 2 times of row 1.
 ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$
$\Rightarrow \left[ \begin{matrix}
   1 & \dfrac{2}{5} \\
   2-2\times 1 & 1-\dfrac{4}{5} \\
\end{matrix} \right]=A\left[ \begin{matrix}
   \dfrac{1}{5} & 0 \\
   0-\dfrac{2}{5} & 1-2\times 0 \\
\end{matrix} \right]$
$\Rightarrow \left[ \begin{matrix}
   1 & \dfrac{2}{5} \\
   0 & \dfrac{1}{5} \\
\end{matrix} \right]=A\left[ \begin{matrix}
   \dfrac{1}{5} & 0 \\
   \dfrac{-2}{5} & 1 \\
\end{matrix} \right]$
Now we can subtract 2 times of row 2 from row 1 to make $\dfrac{2}{5}$ as zero.
${{R}_{1}}\to {{R}_{1}}-2{{R}_{2}}$
$\Rightarrow \left[ \begin{matrix}
   1-2\times 0 & \dfrac{2}{5}-2\times \dfrac{1}{5} \\
   0 & \dfrac{1}{5} \\
\end{matrix} \right]=A\left[ \begin{matrix}
   \dfrac{1}{5}-2\times \dfrac{-2}{5} & 0-2\times 1 \\
   \dfrac{-2}{5} & 1 \\
\end{matrix} \right]$
$\Rightarrow \left[ \begin{matrix}
   1 & 0 \\
   0 & \dfrac{1}{5} \\
\end{matrix} \right]=A\left[ \begin{matrix}
   1 & -2 \\
   \dfrac{-2}{5} & 1 \\
\end{matrix} \right]$
Now we can multiply row 2 by 5
${{R}_{2}}\to 5\times {{R}_{2}}$
$\Rightarrow \left[ \begin{matrix}
   1 & 0 \\
   0\times 5 & \dfrac{1}{5}\times 5 \\
\end{matrix} \right]=A\left[ \begin{matrix}
   1 & -2 \\
   \dfrac{-2}{5}\times 5 & 1\times 5 \\
\end{matrix} \right]$
$\Rightarrow \left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]=A\left[ \begin{matrix}
   1 & -2 \\
   -2 & 5 \\
\end{matrix} \right]$
Now it is in form of $I=A{{A}^{-1}}$
Hence inverse of given matrix is
${{A}^{-1}}=\left[ \begin{matrix}
   1 & -2 \\
   -2 & 5 \\
\end{matrix} \right]$

Note: To find the inverse of any matrix , matrix should be square matrix means number of rows equal to number of columns in matrix.
We can do either row transformation or column transformation at a time. We can’t do both together.