Let A and B are square matrices of order 3 such that \(A + B =\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}\). If A is a symmetric matrix and B is a skew-symmetric matrix, then the value of \(|B|\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
B is a skew-symmetric matrix of odd order (order 3), so \(B^T = -B\) and
\(|B| = |B^T| = |-B| = (-1)^3|B| = -|B|\)
\(2|B| = 0,\quad |B| = 0\)
As a check, B is the skew-symmetric part of \(M = A + B\). Since \(M^T = A - B\):
\( \displaystyle B = \frac{M - M^T}{2} = \begin{bmatrix}0&-\frac{1}{2}&-\frac{5}{2}\\ \frac{1}{2}&0&3\\ \frac{5}{2}&-3&0\end{bmatrix} \)
Its determinant is 0, as expected.
Note: It seems that there is a mistake in the original NTA question. It gives A + B and says only that A is symmetric, but says nothing about B, so \(|B|\) cannot be found: for example, A = O gives \(|B| = 0\), while A = 2I gives \(|B| = -2\). The answer given by NTA (0) is correct only if B is skew-symmetric. For the benefit of students and to ensure that the question is correct and has a single valid answer, “and B is a skew-symmetric matrix” has been added to the question.