AP EAPCET 24-Aug-2021 Shift 1 Paper

Option (a) Let $A=\begin{array}{ccc}10&11&12\\11&12&13\\12&13&14\end{array}$ Applying $R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}$ $\begin{bmatrix} -1 & -1 & -1 \\ -1 & -1 & -1 \\ 12 & 13 & 14 \end{bmatrix}$ Applying $R_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow R_{3}+12 R_{1}$, $\begin{bmatrix} -1 & -1 & -1 \\ 0 & 0 & 0 \\ 0 & 1 & 2 \end{bmatrix}$ Applying $R_{2} \leftrightarrow R_{3}$ $\begin{bmatrix} -1 & -1 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix}$ this is echelon form ∴ Rank of matrix $A=2$ Option(b) $B=\begin{bmatrix} 0 & -51 & 101 \\ 51 & 0 & -581 \\ -101 & 581 & 0 \end{bmatrix}$ ∵ $B^{T}=-B, \therefore B$ is skew-symmetric matrix. ∵ Rank of skew-symmetric matrix is always an even number rank of $B \neq 3$ Option (c) $C=\begin{bmatrix} 0 & 1 & 2 \\ -1 & 0 & 5 \\ -2 & 7 & 0 \end{bmatrix}$ Applying $R_{2} \leftrightarrow R_{1}$. $\begin{bmatrix} -1 & 0 & 5 \\ 0 & 1 & 2 \\ -2 & 7 & 0 \end{bmatrix}$ $R_{1} \rightarrow R_{1}-R_{3}$ $\Rightarrow \begin{bmatrix} 1 & -7 & 5 \\ 0 & 1 & 2 \\ -2 & 7 & 0 \end{bmatrix}$ Applying $R_{3} \rightarrow R_{3}+2 R_{1} \begin{bmatrix} 1 & -7 & 5 \\ 0 & 1 & 2 \\ 0 & -7 & 10 \end{bmatrix}$ Applying $R_{3} \rightarrow R_{3}+7 R_{2}, R_{1} \rightarrow R_{1}+7 R_{2}$ $\begin{bmatrix} 1 & 0 & 19 \\ 0 & 1 & 2 \\ 0 & 0 & 24 \end{bmatrix}$ Applying $R_{3} \rightarrow \frac{1}{24} R_{3} \begin{bmatrix} 1 & 0 & 19 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}$ Applying $R_{1} \rightarrow R_{1}-19 R_{3}, R_{2} \rightarrow R_{2}-2 R_{3}$ $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ ∴ Rank of matrix $C=3$ Option (d) $D=\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix}$ Applying $R_{2} \rightarrow R_{2}-2 R_{1}, R_{3} \rightarrow R_{3}-3 R_{1}$ $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ ∴ Rank of matrix $D=1$ Hence, option (c) is the correct answer.