Question

Exercise1. Write the following matrices into row echelon form.

 

(a) $A=\left(\begin{array}{cccc}1& 1& 3& 2\\ -2& 2& 1& 0\\ 0& 4& 7& 4\end{array}\right)$ 

(c) $C=\left(\begin{array}{cccccc}1& 2& 1& 4& 5& 7\\ 2& 1& 0& 1& 2& 1\\ 3& 3& 1& 5& 7& 8\end{array}\right)$ 

(b) $B=\left(\begin{array}{ccc}2& 3& 4\\ 3& 0& 4\\ 1& 3& 1\end{array}\right)$ 

(d) $D=\left(\begin{array}{cccc}1& 1& 1& 1\\ 1& -1& 1& -1\\ -1& -1& 1& 1\\ -1& 1& \lambda & \lambda \end{array}\right)$

Answer 1

Solution : Write the following matrices into row echelon form.

(a). $A=\left(\begin{array}{cccc}1& 1& 3& 2\\ -2& 2& 1& 0\\ 0& 4& 7& 4\end{array}\right)R_2\to R_2+2R_1$$\sim \left(\begin{array}{cccc}1& 1& 3& 2\\ 0& 4& 7& 4\\ 0& 4& 7& 4\end{array}\right)R_3\to R_3-R_2$ 

$\sim \left(\begin{array}{cccc}1& 1& 3& 2\\ 0& 4& 7& 4\\ 0& 0& 0& 0\end{array}\right)R_2\to R_2\times \frac{1}{4}$ $\sim \left(\begin{array}{cccc}1& 1& 3& 2\\ 0& 1& 7/4& 1\\ 0& 0& 0& 0\end{array}\right)$ 

Therefore $rank\left(A\right)=2,null\left(A\right)=2■$

 (b) $B=\left(\begin{array}{ccc}2& 3& 4\\ 3& 0& 4\\ 1& 3& 1\end{array}\right)R_1\leftrightarrow R_3\left(\begin{array}{ccc}1& 3& 1\\ 3& 0& 4\\ 2& 3& 4\end{array}\right)\begin{array}{c}{R}_2\to R_2-3R_1\\ R_3\to R_3-2R_1\end{array}$ 

$\sim \left(\begin{array}{ccc}1& 3& 1\\ 0& -9& 1\\ 0& -3& 2\end{array}\right)\sim \left(\begin{array}{ccc}1& 3& 1\\ 0& 1& -\frac{1}{9}\\ 0& -3& 2\end{array}\right)\sim \left(\begin{array}{ccc}1& 3& 1\\ 0& 1& -\frac{1}{9}\\ 0& 0& \frac{1}{3}\end{array}\right)\sim \left(\begin{array}{ccc}1& 3& 1\\ 0& 1& -\frac{1}{9}\\ 0& 0& 1\end{array}\right)$

Therefore $rank\left(B\right)=3,null\left(B\right)=0■$ 

(c) $C=\left(\begin{array}{cccccc}1& 2& 1& 4& 5& 7\\ 2& 1& 0& 1& 2& 1\\ 3& 3& 1& 5& 7& 8\end{array}\right)\begin{array}{c}{R}_2\to R_2-2R_1\\ R_3\to R_3-3R_1\end{array}$ 

$\sim \left(\begin{array}{cccccc}1& 2& 1& 4& 5& 7\\ 0& -3& -2& -7& -8& -13\\ 0& -3& -2& -7& -8& -13\end{array}\right)\begin{array}{c}{R}_2\to R_2-\frac{1}{3}{R}_2\\ R_3\to R_3-R_2\end{array}$ 

$\sim \left(\begin{array}{cccccc}1& 2& 1& 4& 5& 7\\ 0& 1& \frac{2}{3}& \frac{7}{3}& \frac{8}{3}& \frac{13}{3}\\ 0& 0& 0& 0& 0& 0\end{array}\right)$ Therefore. $rank\left(C\right)=2,null\left(C\right)=4■$ 

(d) $D=\left(\begin{array}{cccc}1& 1& 1& 1\\ 1& -1& 1& -1\\ -1& -1& 1& 1\\ -1& 1& \lambda & \lambda \end{array}\right)\begin{array}{c}{R}_2\to R_2-R_1\\ R_3\to R_3+R_1\\ R_4\to R_4+R_1\end{array}$ 

$\sim \left(\begin{array}{cccc}1& 1& 1& 1\\ 0& -2& 0& -2\\ 0& 0& 2& 2\\ 0& 2& \lambda +1& {\lambda }_{+1}\end{array}\right)\begin{array}{c}{R}_2\to -\frac{1}{2}{R}_2\\ R_3\to R_3\times \frac{1}{2}\\ R_4\to R_4+R_2\end{array}$ 

$\sim \left(\begin{array}{cccc}1& 1& 1& 1\\ 0& 1& 0& 1\\ 0& 0& 1& 1\\ 0& 0& \lambda +1& \lambda -1\end{array}\right)R_4\to R_4-(\lambda +1)R_3$ $\sim \left(\begin{array}{cccc}1& 1& 1& 1\\ 0& 1& 0& 1\\ 0& 0& 1& 1\\ 0& 0& 0& 1\end{array}\right)$ 

Therefore $\mathrm{rank}\left(D\right)=4,null\left(D\right)=0■$