Which of the following matrix has rank 3?
Which of the following matrix has rank 3?
- $\left[\begin{array}{ccc}10 & 11 & 12 \\ 11 & 12 & 13 \\ 12 & 13 & 14\end{array}\right]$
- $\left[\begin{array}{ccc}0 & -51 & 101 \\ 51 & 0 & -581 \\ -101 & 581 & 0\end{array}\right]$
- $\left[\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & 5 \\ -2 & 7 & 0\end{array}\right]$
- $\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9\end{array}\right]$
Solution
Option (a) Let $A=\left[\begin{array}{lll}10 & 11 & 12 \\ 11 & 12 & 13 \\ 12 & 13 & 14\end{array}\right]$
Applying $R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3$
$
\left[\begin{array}{ccc}
-1 & -1 & -1 \\
-1 & -1 & -1 \\
12 & 13 & 14
\end{array}\right]
$
Applying $R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3+12 R_1$,
$
\left[\begin{array}{ccc}
-1 & -1 & -1 \\
0 & 0 & 0 \\
0 & 1 & 2
\end{array}\right]
$
Applying $R_2 \leftrightarrow R_3$
$\left[\begin{array}{ccc}-1 & -1 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0\end{array}\right]$ this is echelon form
$\therefore$ Rank of matrix $A=2$
Option (b) $B=\left[\begin{array}{ccc}0 & -51 & 101 \\ 51 & 0 & -581 \\ -101 & 581 & 0\end{array}\right]$
$\because B^T=-B, \therefore B$ is skew-symmetric matrix.
$\because$ Rank of skew-symmetric matrix is always an even number rank of $B \neq 3$.
Option (c) $C=\left[\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & 5 \\ -2 & 7 & 0\end{array}\right]$
Applying $R_2 \leftrightarrow R_1$,
$
\begin{aligned}
& {\left[\begin{array}{ccc}
-1 & 0 & 5 \\
0 & 1 & 2 \\
-2 & 7 & 0
\end{array}\right]} \\
& R_1 \rightarrow R_1-R_3 \Rightarrow\left[\begin{array}{ccc}
1 & -7 & 5 \\
0 & 1 & 2 \\
-2 & 7 & 0
\end{array}\right]
\end{aligned}
$
Applying $R_3 \rightarrow R_3+2 R_1,\left[\begin{array}{ccc}1 & -7 & 5 \\ 0 & 1 & 2 \\ 0 & -7 & 10\end{array}\right]$
Applying $R_3 \rightarrow R_3+7 R_2, R_1 \rightarrow R_1+7 R_2$
$\left[\begin{array}{ccc}1 & 0 & 19 \\ 0 & 1 & 2 \\ 0 & 0 & 24\end{array}\right]$
Applying $R_3 \rightarrow \frac{1}{24} R_3,\left[\begin{array}{ccc}1 & 0 & 19 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right]$
Applying $R_1 \rightarrow R_1-19 R_3, R_2 \rightarrow R_2-2 R_3$
$\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\therefore$ Rank of matrix $C=3$
Applying $R_2 \rightarrow R_2-2 R_1, R_3 \rightarrow R_3-3 R_1$
$
\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]
$
$\therefore$ Rank of matrix $D=1$
Hence, option (c) is the correct answer
Asked in: AP EAMCET 2021 (24 Aug Shift 1)
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