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Let $A = \left( {\begin{array}{*{20}{c}}0&0&{ - 1}\\0&{ - 1}&0\\{ - 1}&0&0\end{array}} \right)$, the only correct statement about the matrix $A$ is
${A^2} = I$
$A = ( - 1)\,I,$ where I is a unit matrix
${A^{ - 1}}$ does not exist
$A$ is a zero matrix
Solution
(a) Let $A = \left( {\begin{array}{*{20}{c}}0&0&{ – 1}\\0&{ – 1}&0\\{ – 1}&0&0\end{array}} \right)$
Check by options.
$(i)$ ${A^2} = \left( {\begin{array}{*{20}{c}}0&0&{ – 1}\\0&{ – 1}&0\\{ – 1}&0&0\end{array}} \right)\,\,\left( {\begin{array}{*{20}{c}}0&0&{ – 1}\\0&{ – 1}&0\\{ – 1}&0&0\end{array}} \right)$
${A^2} = \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right) = I$
$(ii)$ $( – 1)\,I = \left( {\begin{array}{*{20}{c}}{ – 1}&0&0\\0&{ – 1}&0\\0&0&{ – 1}\end{array}} \right) \ne A$.
$(iii)$ $|A| = 1 \ne 0 \Rightarrow {A^{ – 1}}$ exists.
$(iv)$ Clearly $A$, is not a zero matrix.
