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Question: If \[A\] is a matrix of order 3, then \[\det \left( {kA} \right)\] is A.\[{k^3}\det \left( A \righ...

If AA is a matrix of order 3, then det⁡(kA)\det \left( {kA} \right) is
A.k3det⁡(A){k^3}\det \left( A \right)
B.k2det⁡(A){k^2}\det \left( A \right)
C.kdet⁡(A)k\det \left( A \right)
D.det⁡(A)\det \left( A \right)

Explanation

Solution

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 AA 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 AA using the rule.
det⁡(A)=a(e×i−f×h)−b(d×i−f×g)+c(d×h−e×g)\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) ………… (1)\left( 1 \right)
Now, we will multiply the given matrix i.e. matrix AA by the constant kk . 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⁡(kA)\det \left( {kA} \right).
⇒det⁡(kA)=K×a(k×e×k×i−k×f×k×h)−k×b(k×d×k×i−k×f×k×g) +k×c(k×d×k×h−k×e×k×g)\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.
⇒det⁡(kA)=k3×a(e×i−f×h)−k3×b(d×i−f×g)+k3×c(d×h−e×g)\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 k3{k^3} common, we get
⇒det⁡(kA)=k3×(a(e×i−f×h)−b(d×i−f×g)+c(d×h−e×g))\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⁡(A)=a(e×i−f×h)−b(d×i−f×g)+c(d×h−e×g)\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.
⇒det⁡(kA)=k3×det⁡(A)\Rightarrow \det \left( {kA} \right) = {k^3} \times \det \left( A \right)
Therefore, the value of det⁡(kA)\det \left( {kA} \right) is equal to k3det⁡(A){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 kk to a n×nn \times n matrix AA , then the value of the determinant will change by a factor kn{k^n} .