There is a 3×3 matrix with diagonal elements 0.The rest of the elements can be 1 or -1.Find the number of invertible matrices

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{A}=\left[\begin{array}{ccc}0& \mathrm{a}& \mathrm{b}\\ \mathrm{f}& 0& \mathrm{c}\\ \mathrm{e}& \mathrm{d}& 0\end{array}\right] \\ \mathrm{Now} \mathrm{for} \mathrm{all} \mathrm{of} \mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d},\mathrm{e},\mathrm{f} \mathrm{we} \mathrm{have} 2 \mathrm{options} \left\{1,-1\right\} \\ \mathrm{Hence} \\ \mathrm{Total} \mathrm{matrices}=2^6=64 \\ \mathrm{For} \mathrm{invertible} \mathrm{matrix} , \mathrm{determinant} \mathrm{must} \mathrm{not} \mathrm{be} 0. \\ \left|\mathrm{A}\right|=0-\mathrm{a}\left(0-\mathrm{ec}\right)+\mathrm{b}\left(\mathrm{fd}-0\right) \\ =\mathrm{aec}+\mathrm{bfd} \\ \mathrm{Both} \mathrm{aec} \mathrm{and} \mathrm{bfd} \mathrm{can} \mathrm{either} \mathrm{be} 1 \mathrm{or} -1 \\ \mathrm{For} \mathrm{determinant} \mathrm{to} \mathrm{be} 0, \mathrm{they} \mathrm{should} \mathrm{be} \mathrm{opposite} \mathrm{sign} \\ \mathrm{When} \mathrm{aec}=1 \\ \mathrm{a}=\mathrm{e}=\mathrm{c}=1 \\ \mathrm{or} \\ \mathrm{a}=\mathrm{e}=-1 \mathrm{and} \mathrm{c}=1 \\ \mathrm{a}=\mathrm{c}=-1 \mathrm{and} \mathrm{e}=1 \\ \mathrm{c}=\mathrm{e}=-1 \mathrm{and} \mathrm{a}=1 \\ \mathrm{Hence} 4 \mathrm{possibilities} \\ \mathrm{When} \mathrm{bfd}=-1 \\ \mathrm{b}=\mathrm{f}=\mathrm{d}=-1 \\ \mathrm{or} \\ \mathrm{b}=\mathrm{f}=1 \mathrm{and} \mathrm{d}=-1 \\ \mathrm{b}=\mathrm{d}=1 \mathrm{and} \mathrm{f}=-1 \\ \mathrm{f}=\mathrm{d}=1 \mathrm{and} \mathrm{b}=-1 \\ \mathrm{Hence} \mathrm{we} \mathrm{have} 4 \mathrm{possibilties} \\ \mathrm{Hence} \\ \mathrm{When} \mathrm{aec}=1 \mathrm{and} \mathrm{bfd}=-1 \\ \mathrm{we} \mathrm{have} 4\times 4=16 \mathrm{cases} \\ \mathrm{Similarly} \\ \mathrm{When} \mathrm{aec}=-1 \mathrm{and} \mathrm{bfd}=1 \\ \mathrm{we} \mathrm{have} 4\times 4=16 \mathrm{cases} \\ \mathrm{Above} \mathrm{are} \mathrm{cases} \mathrm{when} \mathrm{determinant} \mathrm{is} 0. \\ \mathrm{Hence} \mathrm{number} \mathrm{of} \mathrm{invertible} \mathrm{matrices}=64-\left(16+16\right)=32 \left[\mathrm{Answer}\right] \end{gathered}$$

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