Question
Let \[A = \left( \begin{array}{l}
\,\,0\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\, - 1\\
\,\,0\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,0\\
- 1\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,0
\end{array} \right).\] The only correct statement about the matrix $$A$$ is
A.
$${A^2} = I$$
B.
$$A = \left( { - 1} \right)I,$$ where $$I$$ is a unit matrix
C.
$${A^{ - 1}}$$ does not exist
D.
$$A$$ is a zero matrix
Answer :
$${A^2} = I$$
Solution :
\[A = \left[ \begin{array}{l}
\,\,0\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\, - 1\\
\,\,0\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,0\\
- 1\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,0
\end{array} \right]\]
$${\text{clearly }}A \ne 0.\,{\text{Also }}\left| A \right| = - 1 \ne 0$$
\[\therefore \,\,{A^{ - 1}}\,{\rm{exists,}}\,\,{\rm{further }}\left( { - 1} \right)I = \left[ \begin{array}{l}
- 1\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,0\\
\,\,\,0\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,0\\
\,\,\,0\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\, - 1
\end{array} \right] \ne A\]
\[{\rm{Also }}\,\,{A^2} = \left[ \begin{array}{l}
\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\, - 1\\
\,\,0\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,0\\
- 1\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,0
\end{array} \right]\left[ \begin{array}{l}
\,\,0\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\, - 1\\
\,\,0\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,0\\
- 1\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,0
\end{array} \right]\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ \begin{array}{l}
\,1\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,0\\
0\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,0\\
0\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,1
\end{array} \right] = I\]