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

We're given:

$A + B = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$

where $A$ is a symmetric matrix (meaning $A^T = A$, so elements mirror across the main diagonal).

We need to find $|B|$ (the determinant of matrix $B$).

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Any matrix can be split into a symmetric part and a skew-symmetric part.

Let $M = A + B$

Taking transpose of both sides:

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

$A^T + B^T = M^T$

Since $A^T = A$:

$A + B^T = M^T$

We also have $A + B = M$

Subtracting the first from the second:

$B - B^T = M - M^T$

This means $B$ is skew-symmetric (where $B^T = -B$).

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For the skew-symmetric part:

$B = \dfrac{M - M^T}{2}$

First, find $M^T$:

$M^T = \begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}$

Now calculate $M - M^T$:

$M - M^T = \begin{bmatrix} 2-2 & -2-(-1) & -4-1 \\ -1-(-2) & 3-3 & 4-(-2) \\ 1-(-4) & -2-4 & -3-(-3) \end{bmatrix} = \begin{bmatrix} 0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0 \end{bmatrix}$

Therefore:

$B = \dfrac{1}{2}\begin{bmatrix} 0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0 \end{bmatrix}$

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Key property: For any skew-symmetric matrix of odd order (like $3 \times 3$), the determinant is always $0$.

Here's why: For skew-symmetric matrix $B$, we have $B^T = -B$

Taking determinant of both sides:

$|B^T| = |-B|$

$|B| = (-1)^3|B|$

$|B| = -|B|$

This means:

$2|B| = 0$

$|B| = 0$

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$|B| = 0$