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Reduce the matrix A to normal form and hence find its rank where $ \left[ \begin{array}{cccc}1&-1& 3& 6\\1& 3&-3 &-4\\5& 3 & 3 & 11\end{array}\right] $
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A=$ \left[ \begin{array}{cccc} 1&-1& 3& 6 \\ 1& 3&-3 &-4 \\ 5& 3 & 3 & 11 \end{array}\right] \\ \; \\ \; \\ \; \\ R_2 \rightarrow R_2-R_1 \; , \; R_3 \rightarrow R_3 -5R_1 \\ \; \\ \; \\ \therefore A= \left[ \begin{array}{cccc} 1&-1& 3& 6 \\ 0& 4&-6 &-10 \\ 0& 8 & -12 & -19 \end{array}\right] \\ \; \\ \; \\ \; \\ C_3 \rightarrow C_3+C_2 \; \\ \; \\ \; \\ \therefore A= \left[ \begin{array}{cccc} 1&-1& 2& 6 \\ 0& 4&-2 &-10 \\ 0& 8 & -4 & -19 \end{array}\right] \\ \; \\ \; \\ \; \\ R_3 \rightarrow R_3-2R_2 \; \\ \; \\ \; \\ \therefore A= \left[ \begin{array}{cccc} 1&-1& 2& 6 \\ 0& 4&-2 &-10 \\ 0& 0 & 0 & 1 \end{array}\right] \\ \; \\ \; \\ \; \\ R_1 \rightarrow R_1-6R_3 \; , \; R_2 \rightarrow R_2-10R_3 \\ \; \\ \; \\ \therefore A= \left[ \begin{array}{cccc} 1&-1& 2& 0 \\ 0& 4&-2 &0 \\ 0& 0 & 0 & 1 \end{array}\right] \\ \; \\ \; \\ \; \\ R_1 \rightarrow R_1+R_2 \; , \; C_3 \rightarrow C_3 \times 2 \\ \; \\ \; \\ \therefore A= \left[ \begin{array}{cccc} 1& 3& 0& 0 \\ 0& 4&-4 &0 \\ 0& 0 & 0 & 1 \end{array}\right] \\ \; \\ \; \\ \; \\ C_2 \rightarrow C_2-3C_1 \; \\ \; \\ \; \\ \therefore A= \left[ \begin{array}{cccc} 1& 0& 0& 0 \\ 0& 4&-4 &0 \\ 0& 0 & 0 & 1 \end{array}\right] \\ \; \\ \; \\ \; \\ C_3 \rightarrow C_3+C_2 \; , \; C_3 \longleftrightarrow C_4 \\ \; \\ \; \\ \therefore A= \left[ \begin{array}{cccc} 1& 0& 0& 0 \\ 0& 4& 0 &0 \\ 0& 0 & 1 & 0 \end{array}\right] \\ \; \\ \; \\ \; \\ R_2 \rightarrow R_2 \times \frac{1}{4} \\ \; \\ \; \\ \therefore A= \left[ \begin{array}{cccc} 1& 0& 0& 0 \\ 0& 1& 0 &0 \\ 0& 0 & 1 & 0 \end{array}\right] \\ \; \\ \; \\ \; \\ Hence \; Rank \; = \; 3 $

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