Class CBSE Class 12 Mathematics Matrices and Determinants Q #1399
KNOWLEDGE BASED
REMEMBER
2 Marks 2025 AISSCE(Board Exam) VSA
Let A and B be two square matrices of order 3 such that $\det(A) = 3$ and $\det(B) = -4$. Find the value of $\det(-6AB)$.
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Detailed Solution

Step 1: Recall determinant properties

We need to use the following properties of determinants:\n\n1. $\det(kA) = k^n \det(A)$, where $A$ is an $n \times n$ matrix.\n2. $\det(AB) = \det(A) \det(B)$.

Step 2: Apply the properties to the given expression

We are given $\det(A) = 3$ and $\det(B) = -4$. We want to find $\det(-6AB)$.\n\nUsing the properties, we have:\n$$\det(-6AB) = \det(-6I \cdot AB) = \det(-6I) \det(A) \det(B)$$\nwhere $I$ is the identity matrix of order 3.\n\nSince $-6I$ is a $3 \times 3$ matrix, $\det(-6I) = (-6)^3 = -216$.\n\nTherefore,\n$$\det(-6AB) = (-6)^3 \det(A) \det(B) = (-216) \cdot (3) \cdot (-4)$$\n$$\det(-6AB) = -216 \cdot (-12) = 2592$$

Final Answer: 2592

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because it requires recalling and applying the properties of determinants.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply a specific procedure (using determinant properties) to arrive at the solution.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly tests the knowledge of determinant properties as covered in the textbook.

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