Get vector $\vec{v}$ and on matrix $A$

linear algebramatricesvectors

I have a surjective matrix such that

$$A * V = A * [1,1,1]$$

$$A = \begin{bmatrix}1&2&3\\2&1&1\end{bmatrix}$$

I need to find V that is of shape (3,).

This is what I did so far

$$\begin{bmatrix}1&2&3\\2&1&1\\0&0&1\end{bmatrix}.\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}=\begin{bmatrix}6\\4\\?\end{bmatrix}$$

I used a value of 2 for the ?

So

$$\begin{bmatrix}1&2&3\\2&1&1\\0&0&1\end{bmatrix}.\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}=\begin{bmatrix}6\\4\\2\end{bmatrix}$$

Solving this got me

$$v_1 + 2v_2 + 3v_3 = 6$$
$$2v_1 + v_2 + v_3 = 4$$
$$v_3 = 2$$

which implies

$$v_1 = -2, v_2 = 6, v_3 = 2$$

But apparently, this is wrong. I am not sure what I am doing wrong. Any help is appreciated. Thanks in advance

Best Answer

[1 2 3][v1] = [1 2 3][1] = [6]
[2 2 1][v2]   [2 1 1][1]   [4]
       [v3]          [1]

v1+2v2+3v3=6 (1)
2v1+v2+v3=4  (2)

From (2): v3=4-v2-2v1

Plug into (1): v1+2v2+12-3v2-6v1=6
v2=-5v1+6

Plug back in (2): v3=4+5v1-6-2v1
v3=-2+3v1

One possible answer is:
Set v1=2
v2=-4
v3=4

All possible answers of the form:
[0 ] + [1 ]
[6 ]   [-5]x
[-2]   [3 ]
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