Question

Find the inverse of $A=\left[ \begin{aligned} & 1\text{ 0 1} \\\ & \text{0 2 3} \\\ & \text{1 2 1} \\\ \end{aligned} \right]$ by elementary column transformations.

Answer 1

Hint : The inverse of the square matrix exists, where the determinant of the matrix does not equal to zero i.e. the non-singular. It follows the property - ${{A}^{-1}}A={{A}^{-1}}A=I$ where I is the identity matrix. By using elementary column transformation, you swap (interchanged) the columns and multiply the column with non-zero constant and add a multiple of one column to another column.

Complete step-by-step answer :
To find the inverse, first find the determinant of A
Determinant is an element that determines or identifies the nature or conditions of an outcome. It is denoted by det(A), det A or $\left| A \right|$
$\Rightarrow \left| A \right|=\left| \begin{matrix} 1 & 0 & 3 \\\ 0 & 2 & 3 \\\ 1 & 2 & 1 \\\ \end{matrix} \right|$
The determinant of A is the product of the diagonal entries of the row echelon and the times of a factor $\pm 1$.
Now,
$\Rightarrow \left| A \right|=1[(2)(1)-(2)(3)]-0[(0)(1)-(1)(3)]+1[(0)(2)-(2)(1)]$
Simplify the left hand side of the equation –
Here, the middle term is equal to zero, as zero multiplied with anything becomes zero.
$\begin{aligned} \Rightarrow & \left| A \right|=1[2-6]-0+1[-2] \\\ & \left| A \right|=-4-2 \\\ \end{aligned}$
Here, both the terms are negative so according to the property, minus and minus do plus and sign of minus.
$\begin{aligned} \Rightarrow & \left| A \right|=-6 \\\ \Rightarrow & \left| A \right|\ne 0 \\\ \end{aligned}$
The determinant is not equal to zero, therefore, the inverse exists.
Now, consider ${{A}^{-1}}A=I$
Place, the values of the given matrix A in the above equation
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix} 1 & 0 & 1 \\\ 0 & 2 & 3 \\\ 1 & 2 & 1 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]$
Now, subtract column one from column third ${{C}_{3}}-{{C}_{1}}$
We get,
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 2 & 3 \\\ 1 & 2 & 0 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & -1 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]$
Now interchange column second and third,
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 3 & 2 \\\ 1 & 0 & 2 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 & 0 \\\ 0 & 0 & 1 \\\ 0 & 1 & 0 \\\ \end{matrix} \right]$
Now, subtract column third from second –
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 2 \\\ 1 & -2 & 2 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 & 0 \\\ 0 & -1 & 1 \\\ 0 & 1 & 0 \\\ \end{matrix} \right]$
Now, multiply column second with two and subtract it from column third –
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 1 & -2 & 6 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 & 2 \\\ 0 & -1 & 3 \\\ 0 & 1 & -2 \\\ \end{matrix} \right]$
Divide column third by six, $(\dfrac{{{C}_{3}}}{6})$
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 1 & -2 & 1 \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 & \dfrac{1}{3} \\\ 0 & -1 & \dfrac{1}{2} \\\ 0 & 1 & \dfrac{1}{-3} \\\ \end{matrix} \right]$
Now, use ${{C}_{1}}-{{C}_{3}}\text{ and }{{\text{C}}_{2}}+2{{C}_{3}}$ operations –
$\Rightarrow {{A}^{-1}}\left[ \begin{matrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{matrix} \right]=\left[ \begin{matrix} \dfrac{2}{3} & \dfrac{-1}{3} & \dfrac{1}{3} \\\ \dfrac{-1}{2} & 0 & \dfrac{1}{2} \\\ \dfrac{1}{3} & \dfrac{1}{3} & \dfrac{-1}{3} \\\ \end{matrix} \right]$
Now, take $\dfrac{1}{6}$ common from the right hand side of the equation –
${{A}^{-1}}=\dfrac{1}{6}\left[ \begin{matrix} 4 & -2 & 2 \\\ -3 & 0 & 3 \\\ 2 & 2 & -2 \\\ \end{matrix} \right]$
This is the required solution.

Note : Always read the given instructions twice, as the inverse of any matrix can be calculated by the row and column transformation. Also it can be calculated by the adjoint method. So, apply accordingly.