Question

If $A=\left| \begin{matrix} 8 & 27 & 125 \\\ 2 & 3 & 5 \\\ 1 & 1 & 1 \\\ \end{matrix} \right|$, then the value of ${{A}^{2}}$ is equal to

Answer 1

Hint: Reduce the given matrix by using row or column transformations to make 0’s. The best way to do this will be to apply ${{C}_{3}}\to {{C}_{3}}-{{C}_{2}}$ and then ${{C}_{2}}\to {{C}_{2}}-{{C}_{1}}$. These transformations would reduce the last row to the form of $\left( 1,0,0 \right)$ and then expand along this row to find the value of the determinant using the rule $\left| A \right|=\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right|={{a}_{31}}\left( {{a}_{12}}{{a}_{23}}-{{a}_{22}}{{a}_{13}} \right)-{{a}_{32}}\left( {{a}_{11}}{{a}_{23}}-{{a}_{21}}{{a}_{13}} \right)+{{a}_{33}}\left( {{a}_{11}}{{a}_{22}}-{{a}_{21}}{{a}_{12}} \right)$. The value of the determinant so obtained is the required value of A. Use this value of A to find ${{A}^{2}}$.

Complete step-by-step answer:

The given matrix can be transformed by using column transformations to make zeroes in the last row. For this, we first apply ${{C}_{3}}\to {{C}_{3}}-{{C}_{2}}$ and then ${{C}_{2}}\to {{C}_{2}}-{{C}_{1}}$. Applying these transformations, we obtain
$\left| A \right|=\left| \begin{matrix} 8 & 19 & 98 \\\ 2 & 1 & 2 \\\ 1 & 0 & 0 \\\ \end{matrix} \right|$ .
The determinant of this matrix can be found easily by expanding along ${{R}_{3}}$, using the formula to find determinant of a $3\times 3$ matrix which is given as
$\left| A \right|=\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right|={{a}_{31}}\left( {{a}_{12}}{{a}_{23}}-{{a}_{22}}{{a}_{13}} \right)-{{a}_{32}}\left( {{a}_{11}}{{a}_{23}}-{{a}_{21}}{{a}_{13}} \right)+{{a}_{33}}\left( {{a}_{11}}{{a}_{22}}-{{a}_{21}}{{a}_{12}} \right)$ .
In the above matrix ${{a}_{32}}=0$ and ${{a}_{33}}=0$. This reduces the above formula to
$\left| A \right|=\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right|={{a}_{31}}\left( {{a}_{12}}{{a}_{23}}-{{a}_{22}}{{a}_{13}} \right)$
Now using the values in the above determinant, we get
$\begin{aligned} & A=\left| \begin{matrix} 8 & 19 & 98 \\\ 2 & 1 & 2 \\\ 1 & 0 & 0 \\\ \end{matrix} \right|=1\left( 19\times 2-1\times 98 \right) \\\ & \Rightarrow A=38-98 \\\ & \Rightarrow A=-60 \\\ \end{aligned}$
Thus the value of the determinant of A is -60. This makes the value of ${{A}^{2}}=3600$, which is the required value.

Note: The determinant can also be found using normal row expansion method without using any transformations, but that would involve a large amount of calculations, making the question lengthier. Besides, that approach also has a higher risk of making calculation mistakes as there are a number of calculations and each may involve fairly large numbers. Hence, row and column transformations make finding the determinant value an easy task.