Question

Let$A$ be a square matrix of order $3\times 3$. If $\left| A \right|=4$ then find the value of $\left| 2A \right|$.

Answer 1

In this question, We are given with a square matrix $A$ of order $3\times 3$. Also we are given that $\left| A \right|=4$. The determinant of the matrix $A$ is equal to 4. Now we will use the fact that if $A$is square matrix of order $m\times m$and the determinant of the matrix $A$ is equals to $x$, then for any scalar $c$ the determinant of the matrix $cA$ is given by ${{c}^{m}}\left| A \right|$.

Complete step by step answer:
We are given a square matrix $A$ of order $3\times 3$.
Then the matrix $A$ is of the form

{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right)$$ Now we are also given that the determinant of the matrix $$A$$ is equal to 4. Hence we have $$\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| \\\ & =4......(1) \end{aligned}$$ We will now calculate the matrix $$cA$$ when $$c=2$$. That is we will multiply matrix $$A$$ with 2 and find the matrix $$2A$$ by multiplying each term of the matrix $$A$$ with 2. Then we will get $$\begin{aligned} & 2A=2\left( \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right) \\\ & =\left( \begin{matrix} 2{{a}_{11}} & 2{{a}_{12}} & 2{{a}_{13}} \\\ 2{{a}_{21}} & 2{{a}_{22}} & 2{{a}_{23}} \\\ 2{{a}_{31}} & 2{{a}_{32}} & 2{{a}_{33}} \\\ \end{matrix} \right) \end{aligned}$$ Now using the fact that for a matrix $$A$$ such that determinant of the matrix $$A$$ is equals to $$x$$, if a row is multiplied by a scalar $$\lambda $$, then the determinant of the resultant matrix becomes $$\lambda x$$. In this case we have $$2A=\left( \begin{matrix} 2{{a}_{11}} & 2{{a}_{12}} & 2{{a}_{13}} \\\ 2{{a}_{21}} & 2{{a}_{22}} & 2{{a}_{23}} \\\ 2{{a}_{31}} & 2{{a}_{32}} & 2{{a}_{33}} \\\ \end{matrix} \right)$$. That is each row of the matrix $$A$$ is multiplied by 2. Hence we will calculate the determinant of the matrix $$2A$$ using the above stated property of determinants of matrices. We will then get, $$\begin{aligned} & \left| 2A \right|=\left| \begin{matrix} 2{{a}_{11}} & 2{{a}_{12}} & 2{{a}_{13}} \\\ 2{{a}_{21}} & 2{{a}_{22}} & 2{{a}_{23}} \\\ 2{{a}_{31}} & 2{{a}_{32}} & 2{{a}_{33}} \\\ \end{matrix} \right| \\\ & =2\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ 2{{a}_{21}} & 2{{a}_{22}} & 2{{a}_{23}} \\\ 2{{a}_{31}} & 2{{a}_{32}} & 2{{a}_{33}} \\\ \end{matrix} \right| \\\ & =\left( 2\times 2 \right)\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ 2{{a}_{31}} & 2{{a}_{32}} & 2{{a}_{33}} \\\ \end{matrix} \right| \\\ & =\left( 2\times 2\times 2 \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| \\\ & =8\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right| \end{aligned}$$ Now on substituting the value in equation (1) in the above equation, we get $$\begin{aligned} & \left| 2A \right|=8\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right| \\\ & =8\left| A \right| \\\ & =8\times 4 \\\ & =32 \end{aligned}$$ **Therefore we get that the value of $$\left| 2A \right|$$ is equal to 32.** **Note:** In this problem, we can also determine the value of $$\left| 2A \right|$$ using the property of determinant of matrix that if the determinant of the matrix $$A$$ is equals to $$x$$, then for any scalar $$c$$ the determinant of the matrix $$cA$$ is given by $${{c}^{m}}\left| A \right|$$.