Find the spanning set of NullspaceA Please show all work Le

Find the spanning set of Nullspace(A). Please show all work :)

Let A = [-1 6 3 1 2 9 1 5 -1 -15 -4 -6 -3 -3 2 -4] Find a spanning set of nullspace(A)

Solution

A =    - 1          6            3             1

            2          9            1             5

            -1         -15        -4           -6

            -3         -3           2           -4

we multiply it with vector X to get a null matrix

          - 1          6            3             1            x1                   0

            2          9            1             5       *    x2          =      0

            -1         -15        -4           -6             x3                   0

            -3         -3           2           -4             x4                   0

so we multiply these matrix

-x1 + 6x2 +3x3 +x4 = 0

2x1 + 9x2 + x3 + 5x4 =0

-x1 -15x2 -4x3 -6x4 =0

-3x1 -3x2 +2x3 -4x4 =0

now we make a Augemented matrix

-1    6     3    1       0

2     9     1     5       0

-1   -15 -4    -6      0

-3     -3   2     -4      0

now we make this into reduced row echelon form

R4 -> R4 -3R1

-1    6     3    1       0

2     9     1     5       0

-1   -15 -4    -6      0

0     -21   -7   -7      0

R2 -> R2 +2*R1

-1    6     3    1       0

0    21     7     7       0

-1   -15 -4    -6      0

0     -21   -7   -7      0

R4 ->R4 +R2

-1    6     3    1       0

0    21     7     7       0

-1   -15 -4    -6      0

0      0     0    0      0

R3 ->R3 -R1

-1    6     3    1       0

0    21     7     7       0

0   -21   -7    -7      0

0      0     0    0      0

R3 ->R3+R2

-1    6     3      1       0

0    21     7    7       0

0   0      0   0      0

0      0     0    0      0

R1 ->R1 - (2/7) *R2

-1    0     1    -1       0

0    21     7    7       0

0   0      0   0      0

0      0     0    0      0

now we get

-x1 +x3 -x4 = 0   => x1 = x3 -x4

21x2 +7x3 +7x4 =0 => x2 = -x3/3 -x4/3

therefore

x1            1             -1

x2   = x3 (-1/3 )   + x4 (-1/3)

x3           1               0

x4           0               1

we can write null space as

N(A) = span {[1 ]         [-1]}

                     -1/3        -1/3

                       1              0

                        0             1