Question

If $A$ is a $3 \times 3$ matrix and {\text{det}}\left( {3A} \right) = k\left\\{ {\det \left( A \right)} \right\\}, then the value of $k =$
A. Nine
B. Six
C. One
D. Twenty-seven

Answer 1

In this question, we will use one of the properties of the determinants i.e., $\det \left( {qA} \right) = {q^n}\det \left( A \right)$ where $n$ is the order of the square matrix. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer :
Given $A$ is a matrix of order $3 \times 3$. So, it is a square matrix.
We know that $\det \left( {qA} \right) = {q^n}\det \left( A \right)$, where $n$ is the order of the matrix $A$.
So, here the order of the matrix $A$ is 3 i.e., $n = 3$.
Now, consider the value of $\det \left( {3A} \right)$ by using the formula

\Rightarrow \det \left( {3A} \right) = {3^n}\left\\{ {\det \left( A \right)} \right\\} \\\ \Rightarrow \det \left( {3A} \right) = {3^3}\left\\{ {\det \left( A \right)} \right\\}{\text{ }}\left[ {\because n = 3} \right] \\\ \therefore \det \left( {3A} \right) = 27\left\\{ {\det \left( A \right)} \right\\} \\\

But given that {\text{det}}\left( {3A} \right) = k\left\\{ {\det \left( A \right)} \right\\}
By comparing the above two values, we have $k = 27$.
Thus, the correct option is D. Twenty-seven

Note : To check whether the formula we used correct or not, let us consider an example with A = {\left[ {\begin{array}{*{20}{c}} 1&2 \\\ 3&4 \end{array}} \right]_{2 \times 2}}and consider the value of $\det \left( {2A} \right)$.
First let us find 2A = 2 \times \left[ {\begin{array}{*{20}{c}} 1&2 \\\ 3&4 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2&4 \\\ 6&8 \end{array}} \right]
Now \det \left( {2A} \right) = \left| {\begin{array}{*{20}{c}} 2&4 \\\ 6&8 \end{array}} \right| = \left\\{ {\left( 2 \right)\left( 8 \right) - \left( 4 \right)\left( 6 \right)} \right\\} = \left( {16 - 24} \right) = - 8
And \det \left( A \right) = \left| {\begin{array}{*{20}{c}} 1&2 \\\ 3&4 \end{array}} \right| = \left\\{ {\left( 1 \right)\left( 4 \right) - \left( 2 \right)\left( 3 \right)} \right\\} = 4 - 6 = - 2
We have used the formula $\det \left( {qA} \right) = {q^n}\det \left( A \right)$. Let us check for this by substituting the value of $n = 2$ and $q = 2$.

$$\Rightarrow \det \left( {2A} \right) = {2^2}\det \left( A \right) \\\ \Rightarrow \det \left( {2A} \right) = 4\left( { - 2} \right) \\\ \therefore \det \left( {2A} \right) = - 8 \\\$$

Since, both the values are equal, the formula we have used is correct.