Question

Problem 2 A matrix A is given by 2 3 0 1 7 2 1 13 16 3 -5 -3 8 22 -1 -1 -11 -18 Find a basis for N(A) (the null space of A). Find a basis for RaneA) = C(A) (the range, or column space of A)

Answer 1

2 3 0 1 0 13 7 2 1 16 3 -5 -3 8 22 -1 -11-18, -1 8 00

R R

7 2 1 13 16 2 3 0 1 0 -5-3 8 22 -1-11-18 -1 00

2 Ri R2 R2

3 Ri R3 R3

R4 R4 R1

2 1 13 16 17 19 0 24 0 1 106 10 58 0

RA

2 1 13 16 58 6 64 110 0 106 0 19 0

41 R2 R3 R3 58

$R_4\:\leftarrow \:R_4-\frac{17}{58}\cdot \:R_2$

7 2 1 13 16 6 64 110 0 11 0 0 0 0 29 29 29 29 하13

R3 RA R4117

7 2 1 13 16 6 64 110 0 0 29 0 29 0 29 0 0 0

29 R3 R3 117

$=\begin{pmatrix}7&2&1&13&16\\ 0&\frac{58}{7}&-\frac{6}{7}&-\frac{64}{7}&-\frac{110}{7}\\ 0&0&1&1&-1\\ 0&0&0&0&0\end{pmatrix}$

$R_2\:\leftarrow \:R_2+\frac{6}{7}\cdot \:R_3$

$R_1\:\leftarrow \:R_1-1\cdot \:R_3$

$=\begin{pmatrix}7&2&0&12&17\\ 0&\frac{58}{7}&0&-\frac{58}{7}&-\frac{116}{7}\\ 0&0&1&1&-1\\ 0&0&0&0&0\end{pmatrix}$

$R_2\:\leftarrow \frac{7}{58}\cdot \:R_2$

R1 R1 2 R2

$=\begin{pmatrix}7&0&0&14&21\\ 0&1&0&-1&-2\\ 0&0&1&1&-1\\ 0&0&0&0&0\end{pmatrix}$

$R_1\:\leftarrow \frac{1}{7}\cdot \:R_1$

$=\begin{pmatrix}{\color{Red} 1}&0&0&2&3\\ 0&{\color{Red} 1}&0&-1&-2\\ 0&0&{\color{Red} 1}&1&-1\\ 0&0&0&0&0\end{pmatrix}$

there are 3 pivot entry at first second and third column so basis of column space is

${\color{Red} Col(A): \begin{pmatrix}2\\ 7\\ 3\\ -1\end{pmatrix},\:\begin{pmatrix}3\\ 2\\ -5\\ 8\end{pmatrix},\:\begin{pmatrix}0\\ 1\\ -3\\ -1\end{pmatrix}}$

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reduced system is

/1 0 0 2 0 1 0--2 0 01 0 0 0 0 3 y 1 1 0

$x+2w+3v=0..............x=-2w-3v$

$y-w-2v=0......................y=w+2v$

$z+w-v=0....................z=-w+v$

$w=w$.....................free

$v=v$.....................free

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general solution is

basis of null space is

${\color{Red} N(A):\begin{pmatrix}-2\\ 1\\ -1\\ 1\\ 0\end{pmatrix},\:\begin{pmatrix}-3\\ 2\\ 1\\ 0\\ 1\end{pmatrix}}$