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, then the value of $|B|$ is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
$A+B=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}$
$A$ is symmetric, so $A^T=A$.
Hence $B=(A+B)-A$ and therefore $B^T=(A+B)^T-A^T$
Since $A^T=A$,
$B^T=(A+B)^T-A$
For $B$ to be any matrix,
$B+B^T=(A+B)+(A+B)^T-2A$
But $(A+B)+(A+B)^T$ must be symmetric. Therefore $B$ is skew-symmetric part of $(A+B)$.
Hence $B$ is skew-symmetric matrix of order $3$.
Determinant of any odd order skew-symmetric matrix is zero.
$|B|=0$