If A is a square matrix of order 3 with |A| = | vedclass

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(C) Let $A$ be a square matrix of order $n = 3$.Consider the matrix $B = A - A^T$.The transpose of $B$ is $B^T = (A - A^T)^T = A^T - (A^T)^T = A^T - A

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(C) Let $A$ be a square matrix of order $n = 3$.Consider the matrix $B = A - A^T$.The transpose of $B$ is $B^T = (A - A^T)^T = A^T - (A^T)^T = A^T - A = -(A - A^T) = -B$.Since $B^T = -B$,$B$ is a skew-symmetric matrix.For any skew-symmetric matrix $B$ of odd order $n$,the determinant $|B| = 0$.Since the order of $A$ is $3$,the order of $(A - A^T)$ is $3$,which is odd.Therefore,$|A - A^T| = 0$.Similarly,$|A^T - A| = 0$.Now,$|(A - A^T)^5| = |A - A^T|^5 = 0^5 = 0$.And $|(A^T - A)^3| = |A^T - A|^3 = 0^3 = 0$.Thus,$|(A - A^T)^5| + |(A^T - A)^3| = 0 + 0 = 0$.

If $A$ is a square matrix of order $3$ with $|A| = 2$,then the value of $|(A - A^T)^5| + |(A^T - A)^3|$ is-

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