Question

If A is any square matrix of order $3\times 3$ then $\left| 3A \right|$ is equal to:
(a) $3\left| A \right|$
(b) $\dfrac{1}{3}\left| A \right|$
(c) $27\left| A \right|$
(d) $9\left| A \right|$

Answer 1

Hint:Let us assume any $3\times 3$ square matrix then multiply the matrix by 3 and after multiplication take the determinant of the matrix 3A and then compare with the results of the options given in the question.

Complete step-by-step answer:
Let us assume a $3\times 3$ matrix:
$A=\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)$
Multiplying 3 on both the sides of the above equation we get,
$\begin{align}
  & 3A=3\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right) \\
 & \Rightarrow 3A=\left( \begin{matrix}
   3{{a}_{11}} & 3{{a}_{12}} & 3{{a}_{13}} \\
   3{{a}_{21}} & 3{{a}_{22}} & 3{{a}_{23}} \\
   3{{a}_{31}} & 3{{a}_{32}} & 3{{a}_{33}} \\
\end{matrix} \right) \\
\end{align}$
Now, taking determinant on both the sides we get,
$\left| 3A \right|=\left| \begin{align}
  & 3{{a}_{11}}\text{ 3}{{a}_{12}}\text{ 3}{{a}_{13}} \\
 & 3{{a}_{21}}\text{ 3}{{a}_{22}}\text{ 3}{{a}_{23}} \\
 & 3{{a}_{31}}\text{ 3}{{a}_{32}}\text{ 3}{{a}_{33}} \\
\end{align} \right|$
There is a property of the determinant that if a row or a column has a number in common then we can take the common number outside from the determinant. So in the above determinant, as you can see that 3 is common in the first row so taking 3 outside from the determinant will make the determinant look as follows:
$\left| 3A \right|=3\left| \begin{align}
  & {{a}_{11}}\text{ }{{a}_{12}}\text{ }{{a}_{13}} \\
 & 3{{a}_{21}}\text{ 3}{{a}_{22}}\text{ 3}{{a}_{23}} \\
 & 3{{a}_{31}}\text{ 3}{{a}_{32}}\text{ 3}{{a}_{33}} \\
\end{align} \right|$
Similarly, we can take 3 outside from the second and third row so then the determinant will look like:
$\left| 3A \right|={{3}^{3}}\left| \begin{align}
  & {{a}_{11}}\text{ }{{a}_{12}}\text{ }{{a}_{13}} \\
 & {{a}_{21}}\text{ }{{a}_{22}}\text{ }{{a}_{23}} \\
 & {{a}_{31}}\text{ }{{a}_{32}}\text{ }{{a}_{33}} \\
\end{align} \right|$
$\Rightarrow \left| 3A \right|=27\left| \begin{align}
  & {{a}_{11}}\text{ }{{a}_{12}}\text{ }{{a}_{13}} \\
 & {{a}_{21}}\text{ }{{a}_{22}}\text{ }{{a}_{23}} \\
 & {{a}_{31}}\text{ }{{a}_{32}}\text{ }{{a}_{33}} \\
\end{align} \right|$……… Eq. (1)
We can write the matrix A in the determinant form as follows:
$\left| A \right|=\left| \begin{align}
  & {{a}_{11}}\text{ }{{a}_{12}}\text{ }{{a}_{13}} \\
 & {{a}_{21}}\text{ }{{a}_{22}}\text{ }{{a}_{23}} \\
 & {{a}_{31}}\text{ }{{a}_{32}}\text{ }{{a}_{33}} \\
\end{align} \right|$
Now, we can write the eq. (1) as:
$\left| 3A \right|=27\left| A \right|$
From the above discussion, we have found that $\left| 3A \right|$ is equal to $27\left| A \right|$.
Now, comparing the answer of $\left| 3A \right|$ that we have got from above with the options given in the question we have found that the correct option is (c).
Hence, the correct option is (c).

Note: If instead of using the property of the determinant you will find the value of $\left| 3A \right|$ by expanding the determinant along the first row and then comparing this answer with all the options given in the question is a time consuming process. We are showing the method below but it’s better to use the property of the determinant and then proceed with the question.
We are showing from the step when we have taken the determinant on both the sides.
$\left| 3A \right|=\left| \begin{align}
  & 3{{a}_{11}}\text{ 3}{{a}_{12}}\text{ 3}{{a}_{13}} \\
 & 3{{a}_{21}}\text{ 3}{{a}_{22}}\text{ 3}{{a}_{23}} \\
 & 3{{a}_{31}}\text{ 3}{{a}_{32}}\text{ 3}{{a}_{33}} \\
\end{align} \right|$
Expanding the determinant along the first row will give:
$\begin{align}
  & \left| 3A \right|=3{{a}_{11}}\left( 9{{a}_{22}}{{a}_{33}}-9{{a}_{23}}{{a}_{32}} \right)-3{{a}_{12}}\left( 9{{a}_{21}}{{a}_{33}}-9{{a}_{31}}{{a}_{23}} \right)+3{{a}_{13}}\left( 9{{a}_{21}}{{a}_{32}}-9{{a}_{22}}{{a}_{31}} \right) \\
 & \Rightarrow \left| 3A \right|=27\left[ {{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}} \right)-{{a}_{12}}\left( {{a}_{21}}{{a}_{33}}-{{a}_{31}}{{a}_{23}} \right)+{{a}_{13}}\left( {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \right) \right] \\
\end{align}$
From the above expression, you can see that the expression in the square bracket is the expansion of the first row of $\left| A \right|$ so we can write the expression written in the square bracket as $\left| A \right|$.
$\left| 3A \right|=27\left| A \right|$
Hence, compare this answer with the options given in the question.