Question

Using elementary transformations, find the inverse of the matrix$\left[ {\begin{array}{*{20}{l}}
  1&{ - 1} \\
  2&3
\end{array}} \right]$.

Answer 1

Hint:
For the elementary transformations, while doing operation we can either use the rows or we can use columns. Here in this question, we will use the rows while doing the operations. Also, we know that $A = AI$ there $A$ will be a matrix and we have to convert it $I$ by using the operations.

Formula used:
For any matrix $A$, the inverse will be calculated by using the formula
$A = AI$
Here, the $A$will be converted to $I$using the elementary transformations and $I$, will become ${A^{ - 1}}$ which we will call the inverse of the matrix.

Complete step by step solution:
First of all we will let the matrix be $A = \left[ {\begin{array}{*{20}{l}}
  1&{ - 1} \\
  2&3
\end{array}} \right]$
As we know that $A = AI$
So on substituting the values, we get
$ \Rightarrow \left[ {\begin{array}{*{20}{l}}
  1&{ - 1} \\
  2&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
  1&0 \\
  0&1
\end{array}} \right]A$
Now we will use the operations,
So on using the operation ${R_2} = 2{R_1} - {R_2}$
$ \Rightarrow \left[ {\begin{array}{*{20}{l}}
  1&{ - 1} \\
  0&{ - 5}
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
  1&0 \\
  2&{ - 1}
\end{array}} \right]A$
Again using the operation ${R_2} = \dfrac{{ - 1}}{5}{R_2}$
$ \Rightarrow \left[ {\begin{array}{*{20}{l}}
  1&{ - 1} \\
  0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
  1&0 \\
  {\dfrac{{ - 2}}{5}}&{\dfrac{1}{5}}
\end{array}} \right]A$
So on using the operation ${R_1} = {R_1} + {R_2}$
\[ \Rightarrow \left[ {\begin{array}{*{20}{l}}
  1&0 \\
  0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
  {\dfrac{3}{5}}&{\dfrac{1}{5}} \\
  {\dfrac{{ - 2}}{5}}&{\dfrac{1}{5}}
\end{array}} \right]A\]
Therefore, from here we can calculate the ${A^{ - 1}}$
$ \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{l}}
  {\dfrac{3}{5}}&{\dfrac{1}{5}} \\
  {\dfrac{{ - 2}}{5}}&{\dfrac{1}{5}}
\end{array}} \right]$
Therefore, $\left[ {\begin{array}{*{20}{l}}
  {\dfrac{3}{5}}&{\dfrac{1}{5}} \\
  {\dfrac{{ - 2}}{5}}&{\dfrac{1}{5}}
\end{array}} \right]$ will be the inverse of the matrix$\left[ {\begin{array}{*{20}{l}}
  1&{ - 1} \\
  2&3
\end{array}} \right]$.

Additional information:
Inverse of the matrix can be found in more than one way. We can also find it by using the determinant. In this, we have to first take the determinant of the matrix and then we have to find the adjoint of the matrix. After that on dividing the adjoint of the matrix with the determinant of the matrix, we will get the inverse of the matrix.

Note:
This type of question can have more than one way to solve it. And while solving using the elementary method we can also have used the column instead of the rows and then operate it accordingly. So we have to be careful while choosing the row or column as it might take a long time if we don’t choose well.