If A^T denotes the transpose of the matrix A | vedclass

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(C) Given $A = \begin{bmatrix} 0 & 0 & a \ 0 & b & c \ d & e & f \end{bmatrix}$. The determinant $|A| = 0(bf - ce) - 0(0 - cd) + a(0 - bd) = -abd$.S

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$2^3$

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(C) Given $A = \begin{bmatrix} 0 & 0 & a \ 0 & b & c \ d & e & f \end{bmatrix}$. The determinant $|A| = 0(bf - ce) - 0(0 - cd) + a(0 - bd) = -abd$.Since $abd 
eq 0$,the matrix is invertible.The cofactor matrix is calculated as:$C_{11} = bf - ce, C_{12} = -(-cd) = cd, C_{13} = -bd$$C_{21} = -(-ae) = ae, C_{22} = -ad, C_{23} = 0$$C_{31} = -ab, C_{32} = 0, C_{33} = 0$Thus,$	ext{adj}(A) = \begin{bmatrix} bf-ce & ae & -ab \ cd & -ad & 0 \ -bd & 0 & 0 \end{bmatrix}$.$A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{-abd} \begin{bmatrix} bf-ce & ae & -ab \ cd & -ad & 0 \ -bd & 0 & 0 \end{bmatrix}$.Given $A^{-1} = A^T$,where $A^T = \begin{bmatrix} 0 & 0 & d \ 0 & b & e \ a & c & f \end{bmatrix}$.Equating $A^{-1} = A^T$ leads to the condition $A A^T = I$ (since $A^{-1} = A^T \implies A A^T = I$).Calculating $A A^T = \begin{bmatrix} 0 & 0 & a \ 0 & b & c \ d & e & f \end{bmatrix} \begin{bmatrix} 0 & 0 & d \ 0 & b & e \ a & c & f \end{bmatrix} = \begin{bmatrix} a^2 & ac & af \ ac & b^2+c^2 & be+cf \ ad & ae+cf & d^2+e^2+f^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$.From $a^2 = 1$,$a = \pm 1$.From $ac = 0$,$c = 0$.From $af = 0$,$f = 0$.From $b^2+c^2 = 1$,$b^2 = 1 \implies b = \pm 1$.From $be+cf = 0$,$be = 0 \implies e = 0$.From $d^2+e^2+f^2 = 1$,$d^2 = 1 \implies d = \pm 1$.Thus,$a, b, d \in \{1, -1\}$ and $c=e=f=0$.The number of such matrices is $2 	imes 2 	imes 2 = 2^3 = 8$.

If $A^T$ denotes the transpose of the matrix $A = \begin{bmatrix} 0 & 0 & a \\ 0 & b & c \\ d & e & f \end{bmatrix}$,where $a, b, c, d, e$ and $f$ are integers such that $abd \neq 0$,then the number of such matrices for which $A^{-1} = A^T$ is

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