Question

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. ${A^2} = I$
B. $A = ( - 1)I$ , where $I$ is a unit matrix
C. ${A^{ - 1}}$ does not exist
D. $A$ is a zero matrix

Answer 1

Hint: The matrix is referred to as singular if the determinant is zero, in which case there would not be an inverse. Inverses are only present in non-singular matrices. Further, check if all of the main diagonal elements are $1$ and the remaining elements are $0$ in the given matrix. The matrix is an identity matrix if the criterion is met.

Formula Used: If the determinant of the matrix $A$ i.e., $\left| A \right| \ne 0$ , then the inverse of matrix $A$ exists or if $\left| A \right| = 0$ , then the inverse does not exist.

Complete step-by-step solution:
We have been provided with the matrix $A = \left[ {\begin{array}{*{20}{c}}
  0&0&{ - 1} \\
  0&{ - 1}&0 \\
  { - 1}&0&0
\end{array}} \right]$
If we multiply the given matrix $A$ with itself, then we will have
$A \cdot A = \left[ {\begin{array}{*{20}{c}}
  0&0&{ - 1} \\
  0&{ - 1}&0 \\
  { - 1}&0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  0&0&{ - 1} \\
  0&{ - 1}&0 \\
  { - 1}&0&0
\end{array}} \right]$
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$
$ \Rightarrow {A^2} = I$
Suppose the identity matrix is multiplied by $( - 1)$ , we get
$( - 1)I = \left[ {\begin{array}{*{20}{c}}
  { - 1}&0&0 \\
  0&{ - 1}&0 \\
  0&0&{ - 1}
\end{array}} \right]$
But matrix $A = \left[ {\begin{array}{*{20}{c}}
  0&0&{ - 1} \\
  0&{ - 1}&0 \\
  { - 1}&0&0
\end{array}} \right]$
$ \Rightarrow A \ne ( - 1)I$
Now, we determine the determinant of the matrix $A$
$\left| A \right| = \left| {\begin{array}{*{20}{c}}
  0&0&{ - 1} \\
  0&{ - 1}&0 \\
  { - 1}&0&0
\end{array}} \right|$
$\left| A \right| = 0(0 - 0) - 0(0 + 0) + ( - 1)(0 - 1)$
$ \Rightarrow \left| A \right| = 1$
So, we have $\left| A \right| = 1 \ne 0$ , which implies that ${A^{ - 1}}$ of the matrix exists.
And as we can see clearly, the given matrix $A$ does not have all of its elements as zero.
$ \Rightarrow A$ is not a zero matrix.
Hence, the correct option would be (A) as ${A^2} = I$ .

Note: Another way of finding inverse of matrix $A = \left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]$ . Use the formula ${A^{ - 1}} = \dfrac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}}
  d&{ - b} \\
  { - c}&a
\end{array}} \right]$ , where $(ad - bc)$ is the determinant and $(ad - bc) \ne 0$ . If this is the condition, then the inverse of the matrix will exist. Otherwise, if $(ad - bc) = 0$ then the inverse would not exist.