Mathematics - Determinant Question with Solution | TestHub

Let $A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right]$.Then, $|A|=-1$ If $\mathrm{A}$ is a prefect square of matrix $A_1$. Then, $A_1^2=A$ $\Rightarrow {\left|A_1\right|}^2=|A|\Rightarrow {\left|A_1\right|}^2=-1\Rightarrow A_1$ cannot be a real matrix So, option correct. Let $R=\left[\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right]$ be the sqare of a $3\times 3$ matrix $B_1$, Then, $B=B_1^2\Rightarrow {\left|B_1\right|}^2=|B|\Rightarrow {\left|B_1\right|}^2=\left|\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right|=-1$ So, B cannot be the perfect square of a real matrix. In option, the given matrix is the identity matrix $I_3$ such that $I_3=I_3^2$ $\begin{array}{rl}& \text{ Conisder the matrix }\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right]\text{ given in option. Clearly, }\\ & \left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& -1& 0\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& -1& 0\end{array}\right]\end{array}$ Thus, the matrix given in option is the perfect square of a real matrix.