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

Hint: 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.