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Let $I$ denote the $3 \times 3$ identity matrix and $P$ be a matrix obtained by rearranging the columns of
$I$. Then,
Options:
$I$. Then,
Solution:
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Verified Answer
The correct answer is:
there are six distinct choices for $P$ and $\operatorname{det}(P)=\pm 1$
Given, $I=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
Then, $\quad \operatorname{det}(I)=1$
If we take $I$ as
$$
A_{1}=\left[\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right]
$$
Then, $\quad \operatorname{det}\left(I_{1}\right)=-1$
Similarly, there are four other possibilities,
$$
\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array}\right] \cdot\left[\begin{array}{lll}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{array}\right]
$$
$$
\left[\begin{array}{lll}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{array}\right] \cdot\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]
$$
who will give a determinant either -1 or 1 . Hence, there are six distinct choices for $P$ and
$\operatorname{det}(P)=\pm 1$
Then, $\quad \operatorname{det}(I)=1$
If we take $I$ as
$$
A_{1}=\left[\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right]
$$
Then, $\quad \operatorname{det}\left(I_{1}\right)=-1$
Similarly, there are four other possibilities,
$$
\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array}\right] \cdot\left[\begin{array}{lll}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{array}\right]
$$
$$
\left[\begin{array}{lll}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{array}\right] \cdot\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]
$$
who will give a determinant either -1 or 1 . Hence, there are six distinct choices for $P$ and
$\operatorname{det}(P)=\pm 1$
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