Algebra

Matrices & Determinants

Medium

applied

Let $A$ and $B$ be square matrices of order 3, then det $[(A - A^T) + (B - B^T)]$ is equal to

-1

✅ Correct Option: 1

We need to find det $[(A - A^T) + (B - B^T)]$ where $A$ and $B$ are $3 \times 3$ matrices.

For any matrix $M = A - A^T$ :

$M^T = (A - A^T)^T$

$= A^T - (A^T)^T$

$= A^T - A$

$= -(A - A^T)$

$= -M$

Therefore, $A - A^T$ is skew-symmetric. Similarly, $B - B^T$ is also skew-symmetric.

Let $S = (A - A^T) + (B - B^T)$

$S^T = [(A - A^T) + (B - B^T)]^T$

$= (A - A^T)^T + (B - B^T)^T$

$= -(A - A^T) - (B - B^T)$

$= -[(A - A^T) + (B - B^T)]$

$= -S$

The sum of two skew-symmetric matrices is also skew-symmetric.

The determinant of a skew-symmetric matrix of odd order is always zero.

Since $A$ and $B$ are $3 \times 3$ matrices (order 3 is odd), the matrix $S$ is also $3 \times 3$ .

Therefore, det $[(A - A^T) + (B - B^T)] = 0$

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(A) $|A^{-1}| = |A|$

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Choose the correct answer from the options given below:

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Let AAA and BBB be square matrices of order 3, then det [(A−AT)+(B−BT)][(A - A^T) + (B - B^T)][(A−AT)+(B−BT)] is equal to

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Let $A$ and $B$ be square matrices of order 3, then det $[(A - A^T) + (B - B^T)]$ is equal to