If A = begin{bmatrix} -1 & x & -3 2 & 4 & z | vedclass

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(A) For matrix $A$ to be symmetric,$A = A^T$. Comparing elements,we get $x = 2$,$y = -3$,and $z = 5$. Thus,$A = \begin{bmatrix} -1 & 2 & -3 \ 2 & 4 &

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$xyz + pqr$

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(A) For matrix $A$ to be symmetric,$A = A^T$. Comparing elements,we get $x = 2$,$y = -3$,and $z = 5$. Thus,$A = \begin{bmatrix} -1 & 2 & -3 \ 2 & 4 & 5 \ -3 & 5 & -6 \end{bmatrix}$. The determinant $|A| = -1(-24 - 25) - 2(-12 + 15) - 3(10 + 12) = -1(-49) - 2(3) - 3(22) = 49 - 6 - 66 = -23$. For matrix $B$ to be skew-symmetric,$B^T = -B$. This implies diagonal elements must be $0$,so $s = 0$. Also,$p = -2$,$q = 3$,and $r = 4$. Thus,$B = \begin{bmatrix} 0 & 2 & 3 \ -2 & 0 & -4 \ -3 & 4 & 0 \end{bmatrix}$. The determinant $|B| = 0 - 2(0 - 12) + 3(-8 - 0) = 24 - 24 = 0$. Since $|B| = 0$,$|AB| = |A| 	imes |B| = -23 	imes 0 = 0$. Therefore,$|A| + |B| - |AB| = -23 + 0 - 0 = -23$. Checking the options,none of the provided expressions evaluate to $-23$ based on the variables $x=2, y=-3, z=5, p=-2, q=3, r=4$. However,if we evaluate the expression $xyz + pqr = (2)(-3)(5) + (-2)(3)(4) = -30 - 24 = -54$. Given the standard nature of such problems,there may be a typo in the question options.

If $A = \begin{bmatrix} -1 & x & -3 \\ 2 & 4 & z \\ y & 5 & -6 \end{bmatrix}$ is a symmetric matrix and $B = \begin{bmatrix} 0 & 2 & q \\ p & 0 & -4 \\ -3 & r & s \end{bmatrix}$ is a skew-symmetric matrix,then $|A| + |B| - |AB| = $

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