Question

If $I$ denotes the $3\times 3$ identity matrix and $P$ is a matrix obtained by rearranging the columns of $I$. Then,

• A. there are six distinct choices for $P$ and $det\left(P\right)=1$
• B. there are six distinct choices for $P$ and $det\left(P\right)=\pm 1$
• C. there are more than one choices for $P$ and some of them are not invertible
• D. there are more choices for $P$ and $P^{-1}=I$ in each choice

Answer 1

The correct option is B

there are six distinct choices for $P$ and $det\left(P\right)=\pm 1$

Explanation for correct option:

Find the relation:

We know,

$$\begin{gathered} I=\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right| \\ \Rightarrow \left|I\right|=1 \end{gathered}$$

$P$ is a matrix after rearranging the columns of $I$.

This can be done in $6$ different combinations.

$\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|or\left|\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right|or\left|\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right|or\left|\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right|or\left|\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right|or\left|\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right|$

$P=-\left|I_1\right|=-1·.·\left|I_1\right|=1$
If we interchange columns then we will get a $\left(-\right)$ sign.
Here there are $6$ different choices for $P$.

The determinant of $P$ will have $+1or–1.$

$P=\pm 1$

Hence, the correct option is B.