Question

If we have a matrix $\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|=5$, then find the value of $\left| \begin{matrix} 3a & 3b \\\ 3c & 3d \\\ \end{matrix} \right|$.

Answer 1

This is a simple question of determinant which can be easily solved by using the property of determinant. The property is multiplication of determinant by a constant. When we multiply any determinant by a constant then every element of a particular row or column of the determinant gets multiplied by that constant.

Complete step-by-step answer :
It is given in the question that $\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|=5..................(1)$.
We know that when we multiply any determinant by constant, then every element of a particular row or column of the determinant gets multiplied by that constant.
Let D = $\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} \\\ {{a}_{21}} & {{a}_{22}} \\\ \end{matrix} \right|$be a determinant of order 2.
Let k be a constant, and we multiply k with determinant $\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} \\\ {{a}_{21}} & {{a}_{22}} \\\ \end{matrix} \right|$, we will get:
$\therefore k\times D=k\times \left| \begin{matrix} {{a}_{11}} & {{a}_{12}} \\\ {{a}_{21}} & {{a}_{22}} \\\ \end{matrix} \right|$
$\Rightarrow kD=\left| \begin{matrix} k{{a}_{11}} & k{{a}_{12}} \\\ {{a}_{21}} & {{a}_{22}} \\\ \end{matrix} \right|$
Now, when we multiply again the above determinant by k then, we will get:
$\Rightarrow k\times k\times D=\left| \begin{matrix} k{{a}_{11}} & k{{a}_{12}} \\\ k{{a}_{21}} & k{{a}_{22}} \\\ \end{matrix} \right|$
We have to find the value of the determinant $\left| \begin{matrix} 3a & 3b \\\ 3c & 3d \\\ \end{matrix} \right|$.
Here, it can be seen that 3 is common to all element row 1 and row 2 both of the above determinant. So, when we take out 3 common from row 1, we will get:
So, $\left| \begin{matrix} 3a & 3b \\\ 3c & 3d \\\ \end{matrix} \right|$ can be rewritten as:
$\Rightarrow 3\times \left| \begin{matrix} a & b \\\ 3c & 3d \\\ \end{matrix} \right|$
Now, we will take out 3 common again from row 2, then we will get:
$\Rightarrow 3\times 3\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|$
From, equation (1), we know that $\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|=5$
$\therefore 3\times 3\times \left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|=3\times 3\times 5$
So, $\left| \begin{matrix} 3a & 3b \\\ 3c & 3d \\\ \end{matrix} \right|=45$
Hence, 45 is our required answer.

Note : There is also an alternate method of solving the above question. Instead of taking common 3 from $\left| \begin{matrix} 3a & 3b \\\ 3c & 3d \\\ \end{matrix} \right|$, when we directly multiply $\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|=5$by 3 two times both sides, we will get:
$\Rightarrow 3\times 3\times \left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|=3\times 3\times 5$
And, as I have explained before that when we multiply any determinant by constant, then every element of a particular row or column of the determinant gets multiplied by that constant. So,
$\Rightarrow 3\times 3\times \left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|=3\times \left| \begin{matrix} 3a & 3b \\\ c & d \\\ \end{matrix} \right|=\left| \begin{matrix} 3a & 3b \\\ 3c & 3d \\\ \end{matrix} \right|$
Therefore, $\left| \begin{matrix} 3a & 3b \\\ 3c & 3d \\\ \end{matrix} \right|=45$.