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UPSC NDA I 2019 Mathematics Solved Paper

Question Numbers: 27-28

Direction: Consider the following for the next 02 (two) items:

Let A and B be (3 × 3) matrices with det A = 4 and det B = 3

Question : 27 of 120

Marks: +1, -0

What is det (2AB) equal to?

Solution:

Concept:

For square matrices of order

n

\det(kAB) = k^n \cdot \det(A) \cdot \det(B)

where

k

is a scalar.

Explanation:

Given

A

and

B

are

3 \times 3

matrices with

\det A = 4

and

\det B = 3

We compute

\det(2AB)

using the property:

\det(2AB) = 2^3 \cdot \det(A) \cdot \det(B)

Thus,

\det(2AB) = 8 \times 4 \times 3 = 96

Answer:

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