Question

Let $I$ denote the $3\times 3$ identity matrix and $P$ be 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 than one choices for $P$ and $P^{-1}=I$ in each choice

Answer 1

Answer: (B)

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

Given, $I=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
Then, $det\left(I\right)=1$
If we take $I$ as
$A_1=\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$
Then, $det\left(I_1\right)=-1$
Similarly, there are four other possibilities,
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right],\left[\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right]$
$\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right],\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]$
who will give a determinant either $-1$ or $1$.
Hence, there are six distinct choices for $P$ and $det\left(P\right)=\pm 1$.