91Ó°ÊÓ

We have $A=\left[\begin{array}{ccccc}1& 2& 3& 2& 1\\ 3& 6& 9& 6& 3\\ 1& 2& 4& 1& 2\\ 2& 4& 9& 1& 2\end{array}\right]$

$$\begin{gathered} ⇒A=\left[\begin{array}{ccccc}1& 2& 3& 2& 1\\ 0& 0& 0& 0& 0\\ 0& 0& 1& -1& 1\\ 0& 0& 3& -3& 0\end{array}\right]\left(\begin{array}{c}{R}_2→R_2-3R_1\\ R_3→R_3-R_1\\ R_4→R_4-2R_1\end{array}\right) \\ ⇒A=\left[\begin{array}{ccccc}1& 2& 3& 2& 1\\ 0& 0& 1& -1& 1\\ 0& 0& 0& 0& -3\\ 0& 0& 0& 0& 0\end{array}\right]\left(\begin{array}{c}{R}_2↔R_4\\ R_3↔R_2\end{array}\right) \\ ⇒A=\left[\begin{array}{ccccc}1& 2& 0& 5& -2\\ 0& 0& 1& -1& 1\\ 0& 0& 0& 0& -3\\ 0& 0& 0& 0& 0\end{array}\right]\left(R_1→R_1-3R_2\right) \end{gathered}$$

$$\begin{gathered} ⇒A=\left[\begin{array}{ccccc}1& 2& 0& 5& 0\\ 0& 0& 1& -1& 1\\ 0& 0& 0& 0& -3\\ 0& 0& 0& 0& 0\end{array}\right]\left(C_5→C_5+C_2\right) \\ ⇒A=\left[\begin{array}{ccccc}1& 2& 0& 5& 0\\ 0& 0& 1& -1& 0\\ 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0\end{array}\right]\left(\begin{array}{c}{C}_5→C_5-C_5\\ C_5→\frac{1}{3}{C}_5\end{array}\right) \end{gathered}$$

Thus the reduced row-echelon form of A is $\left[\begin{array}{ccccc}1& 2& 0& 5& 0\\ 0& 0& 1& -1& 0\\ 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0\end{array}\right]$.

Let${\stackrel{→}{v}}_1=\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right],{\stackrel{→}{v}}_2=\left[\begin{array}{c}2\\ 0\\ 0\\ 0\end{array}\right],{\stackrel{→}{v}}_3=\left[\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right],{\stackrel{→}{v}}_4=\left[\begin{array}{c}5\\ -1\\ 0\\ 0\end{array}\right],{\stackrel{→}{v}}_5=\left[\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right]$

$\mathrm{Here}\mathrm{we}\mathrm{have}{\stackrel{→}{v}}_1=2{\stackrel{→}{v}}_{1}and{\stackrel{→}{v}}_4=5{\stackrel{→}{v}}_1-{\stackrel{→}{v}}_3$

$⇒{\stackrel{→}{v}}_2\mathrm{and}{\stackrel{→}{v}}_4$areredundant vectors ${\stackrel{→}{v}}_1,{\stackrel{→}{v}}_3\mathrm{and}{\stackrel{→}{v}}_5$and are non-redundant vectors.

The non-redundant column vectors of A form the basis of the image of A.

Since column vectors ${\stackrel{→}{v}}_1\mathrm{and}{\stackrel{→}{v}}_5$are non-redundant vectors of matrix A, thus the basis of the image A = $\left\{\left[\begin{array}{c}1\\ 3\\ 1\\ 2\end{array}\right],\left[\begin{array}{c}3\\ 9\\ 4\\ 9\end{array}\right],\left[\begin{array}{c}1\\ 3\\ 2\\ 2\end{array}\right]\right\}$.

Since the vectors ${\stackrel{→}{v}}_2\mathrm{and}{\stackrel{→}{v}}_4$are redundant vectors such that

$$\begin{gathered} {\stackrel{→}{v}}_2=2{\stackrel{→}{v}}_1\mathrm{and}{\stackrel{→}{v}}_4=5{\stackrel{→}{v}}_1-{\stackrel{→}{v}}_3 \\ ⇒2{\stackrel{→}{v}}_1-{\stackrel{→}{v}}_2=0\mathrm{and}5{\stackrel{→}{v}}_1-{\stackrel{→}{v}}_3-{\stackrel{→}{v}}_4=0 \end{gathered}$$

Thus the vectorsin the kernel of A are ${\stackrel{→}{w}}_1=\left[\begin{array}{c}2\\ -1\\ 0\\ 0\\ 0\end{array}\right],{\stackrel{→}{w}}_2=\left[\begin{array}{c}5\\ 0\\ -1\\ -1\\ 0\end{array}\right]$.

The reduced row-echelon form of the given matrix A is $\left[\begin{array}{ccccc}1& 2& 0& 5& 0\\ 0& 0& 1& -1& 0\\ 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0\end{array}\right]$.

The basis of the image of A = $\left\{\left[\begin{array}{c}1\\ 3\\ 1\\ 2\end{array}\right],\left[\begin{array}{c}3\\ 9\\ 4\\ 9\end{array}\right],\left[\begin{array}{c}1\\ 3\\ 2\\ 2\end{array}\right]\right\}$.

The basis of the kernel of A = $\left\{\left[\begin{array}{c}2\\ -1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}5\\ 0\\ -1\\ -1\\ 0\end{array}\right]\right\}$.