Question

If \[A\] is a matrix of order 3, then \[\det \left( {kA} \right)\] is
A.\[{k^3}\det \left( A \right)\]
B.\[{k^2}\det \left( A \right)\]
C.\[k\det \left( A \right)\]
D.\[\det \left( A \right)\]

Answer 1

Hint: Here we need to find the value of determinant when a matrix of the given order is multiplied by a constant term. We will first assume the matrix with variable elements and then we will find its determinant. We will multiply the given matrix by a constant term and then we will again find its determinant. From there, we will get our required answer.

Complete step-by-step answer:
Let the matrix \[A\] of order 3 be \[\left[ {\begin{array}{*{20}{c}}a&b&c\\d&e&f\\g&h&i\end{array}} \right]\]
\[ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}a&b&c\\d&e&f\\g&h&i\end{array}} \right]\]
Now, we will find the determinant of the matrix \[A\] using the rule.
\[\det \left( A \right) = a\left( {e \times i - f \times h} \right) - b\left( {d \times i - f \times g} \right) + c\left( {d \times h - e \times g} \right)\] ………… \[\left( 1 \right)\]
Now, we will multiply the given matrix i.e. matrix \[A\] by the constant \[k\] . We know that, when you multiply a matrix by any constant, then all of its elements get multiplied by that constant term.
Therefore,
On multiplying the matrix by a constant, we get
\[ \Rightarrow k \times A = k \times \left[ {\begin{array}{*{20}{c}}a&b&c\\d&e&f\\g&h&i\end{array}} \right]\]
\[ \Rightarrow k \times A = \left[ {\begin{array}{*{20}{c}}{k \times a}&{k \times b}&{k \times c}\\{k \times d}&{k \times e}&{k \times f}\\{k \times g}&{k \times h}&{k \times i}\end{array}} \right]\]
Now, we will find the determinant of this matrix i.e. we will find \[\det \left( {kA} \right)\].
\[\begin{array}{l} \Rightarrow \det \left( {kA} \right) = K \times a\left( {k \times e \times k \times i - k \times f \times k \times h} \right) - k \times b\left( {k \times d \times k \times i - k \times f \times k \times g} \right)\\ + k \times c\left( {k \times d \times k \times h - k \times e \times k \times g} \right)\end{array}\]
Now, we will take common terms out of the bracket.
\[ \Rightarrow \det \left( {kA} \right) = {k^3} \times a\left( {e \times i - f \times h} \right) - {k^3} \times b\left( {d \times i - f \times g} \right) + {k^3} \times c\left( {d \times h - e \times g} \right)\]
On taking the term \[{k^3}\] common, we get
\[ \Rightarrow \det \left( {kA} \right) = {k^3} \times \left( {a\left( {e \times i - f \times h} \right) - b\left( {d \times i - f \times g} \right) + c\left( {d \times h - e \times g} \right)} \right)\]
We know from equation 1 that
\[\det \left( A \right) = a\left( {e \times i - f \times h} \right) - b\left( {d \times i - f \times g} \right) + c\left( {d \times h - e \times g} \right)\]
Now, we will substitute this value here.
\[ \Rightarrow \det \left( {kA} \right) = {k^3} \times \det \left( A \right)\]
Therefore, the value of \[\det \left( {kA} \right)\] is equal to \[{k^3}\det \left( A \right)\].
Hence, the correct option is option A.

Note: In order to solve this question, we need to keep in mind some basic properties of determinants. If all elements of a row (or column) of a determinant are multiplied by some constant number, then the value of the new determinant will be constant times the value of the given determinant. If we multiply a constant term \[k\] to a \[n \times n\] matrix \[A\] , then the value of the determinant will change by a factor \[{k^n}\] .