Question

Convert \[\left( \begin{matrix}
   1 & -1 \\
   2 & 3 \\
\end{matrix} \right)\]into an identity matrix by suitable row transformation?

Answer 1

Hint: We have given a\[2\times 2\]matrix i.e. the square matrix. We have been asked to convert the given matrix into an identity matrix by substituting suitable row transformations. To get the solution students need to use basic row operations in the given matrix. That is, addition of rows, subtracting the rows and multiplying a number with the row and getting to operate with the other row. In this way we are going to find an identity matrix.
Formula used:
An identity matrix is a matrix of the form of a square matrix that is having 1’s on the main diagonal and 0’s everywhere else.
The identity matrix of \[2\times 2\]matrix is given by;
\[I=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]\]

Complete step-by-step answer:
We have given the matrix,
\[\Rightarrow \left( \begin{matrix}
   1 & -1 \\
   2 & 3 \\
\end{matrix} \right)\]
Let the given matrix be ‘A’ i.e. the identity matrix
\[\Rightarrow A=\left( \begin{matrix}
   1 & -1 \\
   2 & 3 \\
\end{matrix} \right)\]
Perform: \[{{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}\]
\[\Rightarrow A=\left( \begin{matrix}
   1 & -1 \\
   2-2\left( 1 \right) & 3-2\left( -1 \right) \\
\end{matrix} \right)\]
Simplifying the terms in the above matrix,
\[\Rightarrow A=\left( \begin{matrix}
   1 & -1 \\
   0 & 5 \\
\end{matrix} \right)\]
Perform: \[{{R}_{2}}\to \dfrac{{{R}_{2}}}{5}\]
\[\Rightarrow A=\left( \begin{matrix}
   1 & -1 \\
   \dfrac{0}{5} & \dfrac{5}{5} \\
\end{matrix} \right)\]
Simplifying the terms in the above matrix,
\[\Rightarrow A=\left( \begin{matrix}
   1 & -1 \\
   0 & 1 \\
\end{matrix} \right)\]
Perform: \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}\]
\[\Rightarrow A=\left( \begin{matrix}
   1+0 & -1+1 \\
   0 & 1 \\
\end{matrix} \right)\]
Simplifying the terms in the above matrix,
\[\Rightarrow A=\left( \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right)\]
\[\therefore A=I\]
Hence, this is the required answer.

Note: We can also solve the given question in an alternative method also.
\[A{{A}^{-1}}=I\]
Where A is the given matrix, \[{{A}^{-1}}\]is the inverse matrix of the given matrix and I is the identity matrix.
Now,
\[\Rightarrow A=\left( \begin{matrix}
   1 & -1 \\
   2 & 3 \\
\end{matrix} \right)\]
As we know that,
\[A{{A}^{-1}}=I\]
Thus,
\[\Rightarrow \left( \begin{matrix}
   1 & -1 \\
   2 & 3 \\
\end{matrix} \right){{A}^{-1}}=\left( \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right)\]
Perform: \[{{R}_{1}}\to {{R}_{2}}+\dfrac{1}{3}{{R}_{2}}\]
\[\Rightarrow \left( \begin{matrix}
   \dfrac{5}{3} & 0 \\
   2 & 3 \\
\end{matrix} \right){{A}^{-1}}=\left( \begin{matrix}
   1 & \dfrac{1}{3} \\
   0 & 1 \\
\end{matrix} \right)\]
Perform: \[{{R}_{1}}\to \dfrac{3}{5}{{R}_{1}}\]
\[\Rightarrow \left( \begin{matrix}
   1 & 0 \\
   2 & 3 \\
\end{matrix} \right){{A}^{-1}}=\left( \begin{matrix}
   \dfrac{3}{5} & \dfrac{1}{5} \\
   0 & 1 \\
\end{matrix} \right)\]
Perform: \[{{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}\]
\[\Rightarrow \left( \begin{matrix}
   1 & 0 \\
   0 & 3 \\
\end{matrix} \right){{A}^{-1}}=\left( \begin{matrix}
   \dfrac{3}{5} & \dfrac{1}{5} \\
   \dfrac{-6}{5} & \dfrac{3}{5} \\
\end{matrix} \right)\]
Perform: \[{{R}_{2}}\to \dfrac{1}{3}{{R}_{2}}\]
\[\Rightarrow \left( \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right){{A}^{-1}}=\left( \begin{matrix}
   \dfrac{3}{5} & \dfrac{1}{5} \\
   \dfrac{-2}{5} & \dfrac{3}{5} \\
\end{matrix} \right)\]
\[\Rightarrow {{A}^{-1}}I={{A}^{-1}}\]
That is, multiplication of any matrix with the identity matrix results in the matrix itself. Therefore, we converted the given matrix into an identity matrix. As we converted the given matrix into an identity matrix.