Question

Using properties of determinants, prove that:
$\left| \begin{matrix} {{a}^{2}}+2a & 2a+1 & 1 \\\ 2a+1 & a+2 & 1 \\\ 3 & 3 & 1 \\\ \end{matrix} \right|={{\left( a-1 \right)}^{3}}$

Answer 1

Hint: For solving this question we will perform elementary row operations in the given square matrix to get the determinant value of the matrix to prove the result.

Complete step-by-step answer:
Given:
We have a square matrix $\left[ \begin{matrix} {{a}^{2}}+2a & 2a+1 & 1 \\\ 2a+1 & a+2 & 1 \\\ 3 & 3 & 1 \\\ \end{matrix} \right]$ .
We will use the following formula of determinant to find the determinant value:
$\begin{aligned} & \left| A \right|=\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right| \\\ & \Rightarrow \left| A \right|={{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}} \right)+{{a}_{12}}\left( {{a}_{23}}{{a}_{31}}-{{a}_{21}}{{a}_{33}} \right)+{{a}_{13}}\left( {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \right) \\\ \end{aligned}$
Now, first, we will perform the elementary row operations on the given matrix to reduce it into a form in which we will have two zero elements in the same column or same row.
Elementary row operations:
1. The given matrix is $\left[ \begin{matrix} {{a}^{2}}+2a & 2a+1 & 1 \\\ 2a+1 & a+2 & 1 \\\ 3 & 3 & 1 \\\ \end{matrix} \right]$ .
2. Now, subtract the second row from the first row then, the matrix will become, $\left[ \begin{matrix} {{a}^{2}}-1 & a-1 & 0 \\\ 2a+1 & a+2 & 1 \\\ 3 & 3 & 1 \\\ \end{matrix} \right]$ .
3. Now, subtract the third row from the second row then, the matrix will become, $\begin{aligned} & \left[ \begin{matrix} {{a}^{2}}-1 & a-1 & 0 \\\ 2a-2 & a-1 & 0 \\\ 3 & 3 & 1 \\\ \end{matrix} \right] \\\ & \Rightarrow \left[ \begin{matrix} \left( a-1 \right)\left( a+1 \right) & a-1 & 0 \\\ 2\left( a-1 \right) & a-1 & 0 \\\ 3 & 3 & 1 \\\ \end{matrix} \right] \\\ \end{aligned}$
4. Now, subtract the second row from the first row then, the matrix will become,
$\begin{aligned} & \left[ \begin{matrix} \left( a-1 \right)\left( a+1 \right)-2\left( a-1 \right) & \left( a-1 \right)-\left( a-1 \right) & 0 \\\ 2\left( a-1 \right) & a-1 & 0 \\\ 3 & 3 & 1 \\\ \end{matrix} \right] \\\ & \Rightarrow \left[ \begin{matrix} \left( a-1 \right)\left( a+1-2 \right) & 0 & 0 \\\ 2\left( a-1 \right) & a-1 & 0 \\\ 3 & 3 & 1 \\\ \end{matrix} \right] \\\ & \Rightarrow \left[ \begin{matrix} \left( a-1 \right)\left( a-1 \right) & 0 & 0 \\\ 2\left( a-1 \right) & a-1 & 0 \\\ 3 & 3 & 1 \\\ \end{matrix} \right] \\\ \end{aligned}$
Now, we got a matrix $\left[ \begin{matrix} \left( a-1 \right)\left( a-1 \right) & 0 & 0 \\\ 2\left( a-1 \right) & a-1 & 0 \\\ 3 & 3 & 1 \\\ \end{matrix} \right]$ . As we know that when we perform either elementary row operations or elementary column operation the determinant value of the matrix won’t change. So, we will now calculate the determinant value of $\left[ \begin{matrix} \left( a-1 \right)\left( a-1 \right) & 0 & 0 \\\ 2\left( a-1 \right) & a-1 & 0 \\\ 3 & 3 & 1 \\\ \end{matrix} \right]$ . Then,
$\begin{aligned} & \left| \left[ \begin{matrix} \left( a-1 \right)\left( a-1 \right) & 0 & 0 \\\ 2\left( a-1 \right) & a-1 & 0 \\\ 3 & 3 & 1 \\\ \end{matrix} \right] \right| \\\ & \Rightarrow 1\times \left[ \left( a-1 \right)\left( a-1 \right)\times \left( a-1 \right) \right] \\\ & \Rightarrow {{\left( a-1 \right)}^{3}} \\\ \end{aligned}$
Thus, from the above calculation of the determinant value of the matrix, we can say that the determinant value of the given matrix is ${{\left( a-1 \right)}^{3}}$ .
$\left| \begin{matrix} {{a}^{2}}+2a & 2a+1 & 1 \\\ 2a+1 & a+2 & 1 \\\ 3 & 3 & 1 \\\ \end{matrix} \right|={{\left( a-1 \right)}^{3}}$
Hence, proved.

Note: Here, the student should not directly apply the determinant formulae for the $3\times 3$ square matrix to prove the result before applying the elementary row operations. That would be a very wrong approach as it is mentioned in the question that we have to use the properties of determinants to prove the result.