If A is a square matrix of 3 times 3 order,an | vedclass

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(A) Let $B = A - A^T$. Then $B^T = (A - A^T)^T = A^T - (A^T)^T = A^T - A = -(A - A^T) = -B$. Since $B^T = -B$,$B$ is a skew-symmetric matrix. The dete

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(A) Let $B = A - A^T$. Then $B^T = (A - A^T)^T = A^T - (A^T)^T = A^T - A = -(A - A^T) = -B$. Since $B^T = -B$,$B$ is a skew-symmetric matrix. The determinant of a skew-symmetric matrix of odd order is always $0$. Since $A$ is a $3 	imes 3$ matrix,$B = A - A^T$ is also a $3 	imes 3$ matrix,which is of odd order. Therefore,$|A - A^T| = 0$. Similarly,$|A^T - A| = |-(A - A^T)| = (-1)^3 |A - A^T| = -1 	imes 0 = 0$. Now,$|(A - A^T)^6| = |A - A^T|^6 = 0^6 = 0$. And $|(A^T - A)^7| = |A^T - A|^7 = 0^7 = 0$. Thus,$|(A - A^T)^6| + |(A^T - A)^7| = 0 + 0 = 0$.

If $A$ is a square matrix of $3 \times 3$ order,and $|A| = 2$,then $|(A-A^T)^6| + |(A^T-A)^7|$ is equal to (where $A^T$ denotes the transpose of matrix $A$).

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