Answer
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Hint: 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
\[A=\left( \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)\]
Now we are also given that the determinant of the matrix \[A\] is equal to 4.
Hence we have
\[\begin{align}
& \left| A \right|=\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right| \\
& =4......(1)
\end{align}\]
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
\[\begin{align}
& 2A=2\left( \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right) \\
& =\left( \begin{matrix}
2{{a}_{11}} & 2{{a}_{12}} & 2{{a}_{13}} \\
2{{a}_{21}} & 2{{a}_{22}} & 2{{a}_{23}} \\
2{{a}_{31}} & 2{{a}_{32}} & 2{{a}_{33}} \\
\end{matrix} \right)
\end{align}\]
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,
\[\begin{align}
& \left| 2A \right|=\left| \begin{matrix}
2{{a}_{11}} & 2{{a}_{12}} & 2{{a}_{13}} \\
2{{a}_{21}} & 2{{a}_{22}} & 2{{a}_{23}} \\
2{{a}_{31}} & 2{{a}_{32}} & 2{{a}_{33}} \\
\end{matrix} \right| \\
& =2\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
2{{a}_{21}} & 2{{a}_{22}} & 2{{a}_{23}} \\
2{{a}_{31}} & 2{{a}_{32}} & 2{{a}_{33}} \\
\end{matrix} \right| \\
& =\left( 2\times 2 \right)\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
2{{a}_{31}} & 2{{a}_{32}} & 2{{a}_{33}} \\
\end{matrix} \right| \\
& =\left( 2\times 2\times 2 \right)\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right| \\
& =8\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|
\end{align}\]
Now on substituting the value in equation (1) in the above equation, we get
\[\begin{align}
& \left| 2A \right|=8\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right| \\
& =8\left| A \right| \\
& =8\times 4 \\
& =32
\end{align}\]
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 \[{{c}^{m}}\left| A \right|\].
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
\[A=\left( \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)\]
Now we are also given that the determinant of the matrix \[A\] is equal to 4.
Hence we have
\[\begin{align}
& \left| A \right|=\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right| \\
& =4......(1)
\end{align}\]
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
\[\begin{align}
& 2A=2\left( \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right) \\
& =\left( \begin{matrix}
2{{a}_{11}} & 2{{a}_{12}} & 2{{a}_{13}} \\
2{{a}_{21}} & 2{{a}_{22}} & 2{{a}_{23}} \\
2{{a}_{31}} & 2{{a}_{32}} & 2{{a}_{33}} \\
\end{matrix} \right)
\end{align}\]
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,
\[\begin{align}
& \left| 2A \right|=\left| \begin{matrix}
2{{a}_{11}} & 2{{a}_{12}} & 2{{a}_{13}} \\
2{{a}_{21}} & 2{{a}_{22}} & 2{{a}_{23}} \\
2{{a}_{31}} & 2{{a}_{32}} & 2{{a}_{33}} \\
\end{matrix} \right| \\
& =2\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
2{{a}_{21}} & 2{{a}_{22}} & 2{{a}_{23}} \\
2{{a}_{31}} & 2{{a}_{32}} & 2{{a}_{33}} \\
\end{matrix} \right| \\
& =\left( 2\times 2 \right)\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
2{{a}_{31}} & 2{{a}_{32}} & 2{{a}_{33}} \\
\end{matrix} \right| \\
& =\left( 2\times 2\times 2 \right)\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right| \\
& =8\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right|
\end{align}\]
Now on substituting the value in equation (1) in the above equation, we get
\[\begin{align}
& \left| 2A \right|=8\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right| \\
& =8\left| A \right| \\
& =8\times 4 \\
& =32
\end{align}\]
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 \[{{c}^{m}}\left| A \right|\].
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