Question

solve this;; mods/stars. ✴️✴️✨​

Answer 1

$\large\underline{\sf{Solution-}}$

➢ Given matrix is

$\rm :\longmapsto\:A \: = \: \begin{gathered}\sf \left[\begin{array}{ccc}0&0& - 1\\0& - 1&0\\ - 1&0&0\end{array}\right]\end{gathered}$

For part (a)

➢ To check whether inverse of matrix A exist or not, we have to check |A|.

So, Consider,

$\rm :\longmapsto\: |A|$

$\rm \: = \: \: 0 - 0 - 1(0 - 1)$

$\rm \: = \: \: 1$

Since,

$\rm :\longmapsto\: |A| \: \ne \: 0$

Hence,

$\rm :\longmapsto\: { |A| }^{ - 1} \: exist.$

For part (b) ,

We know,

➢ The value of identity matrix of order 3 × 3 is

$\rm :\longmapsto\:I = \begin{gathered}\sf \left[\begin{array}{ccc}1&0&0\\0& 1&0\\ 0&0&1\end{array}\right]\end{gathered}$

Now, given matrix is

$\rm :\longmapsto\:A \: = \: \begin{gathered}\sf \left[\begin{array}{ccc}0&0& - 1\\0& - 1&0\\ - 1&0&0\end{array}\right]\end{gathered}$

can be rewritten as

$\rm :\longmapsto\:A \: = - \: \begin{gathered}\sf \left[\begin{array}{ccc}0&0&1\\0&1&0\\ 1&0&0\end{array}\right]\end{gathered}$

$\bf\implies \:A \: \ne \: - \: I$

For part (c)

We know,

➢ Unit matrix of order 3 × 3 is given by

$\rm :\longmapsto\:I = \begin{gathered}\sf \left[\begin{array}{ccc}1&0&0\\0& 1&0\\ 0&0&1\end{array}\right]\end{gathered}$

and

Given matrix is

$\rm :\longmapsto\:A \: = \: \begin{gathered}\sf \left[\begin{array}{ccc}0&0& - 1\\0& - 1&0\\ - 1&0&0\end{array}\right]\end{gathered}$

So,

$\bf\implies \:A \: \ne \: \: I$

For part (d)

Given matrix is

$\rm :\longmapsto\:A \: = \: \begin{gathered}\sf \left[\begin{array}{ccc}0&0& - 1\\0& - 1&0\\ - 1&0&0\end{array}\right]\end{gathered}$

Now,

Consider,

$\rm :\longmapsto\: {A}^{2}$

$\rm \: = \: \: A \times A$

$\rm \: = \: \: \begin{gathered}\sf \left[\begin{array}{ccc}0&0& - 1\\0& - 1&0\\ - 1&0&0\end{array}\right]\end{gathered} \times \begin{gathered}\sf \left[\begin{array}{ccc}0&0& - 1\\0& - 1&0\\ - 1&0&0\end{array}\right]\end{gathered}$

$\rm \: = \: \: \begin{gathered}\sf \left[\begin{array}{ccc}0 + 0 + 1&0 + 0 + 0& 0 + 0 + 0\\0 + 0 + 0&0 + 1 + 0&0 + 0 + 0\\ 0 + 0 + 0&0 + 0 + 0&0 + 0 + 1\end{array}\right]\end{gathered}$

$\rm \: = \: \: \begin{gathered}\sf \left[\begin{array}{ccc}1&0&0\\0& 1&0\\ 0&0&1\end{array}\right]\end{gathered}$

$\rm \: = \: \: I$

So,

$\bf\implies \: {A}^{2} = I$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf{ \: Option \: (d) \: is \: correct}}$