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(B) The identity matrix $I$ is given by $I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$.Rearranging the columns of $I$ results

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there are six distinct choices for $P$ and $\operatorname{det}(P)=\pm 1$

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(B) The identity matrix $I$ is given by $I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$.Rearranging the columns of $I$ results in permutation matrices.For a $3 	imes 3$ matrix,the number of ways to rearrange the $3$ columns is given by $3! = 3 	imes 2 	imes 1 = 6$.Thus,there are $6$ distinct choices for $P$.Since $P$ is a permutation matrix,its determinant is equal to the sign of the permutation,which is either $1$ or $-1$.Therefore,$\operatorname{det}(P) = \pm 1$.Hence,there are six distinct choices for $P$ and $\operatorname{det}(P) = \pm 1$.

Let $I$ denote the $3 \times 3$ identity matrix and $P$ be a matrix obtained by rearranging the columns of $I$. Then,

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