Linear algebra: $Ax=0$, there are only two free variables and it confuses me a lot

Question

Describe all solutions of Ax = 0 in parametric vector form, where A is row equivalent to the matrix:

$$\begin{bmatrix}1 & 2 & 0 & 0 & -7 & 3\\0 & 0 & 1 & 0 & -3 & 3\\ 0&0&0&1&3&26\\0&0&0&0&0&1\end{bmatrix}$$

My answer would be: $$X= x_2\begin{bmatrix}-2\\1\\0\\0\\0\\0\end{bmatrix}+ x_5\begin{bmatrix}7\\0\\3\\-3\\1\\0\end{bmatrix}$$

Is that correct? if not, can you explain why?

You might find this helpful for sorting things out when the pivots don’t line up nicely. — amd, Feb 21 '18 at 06:03
Looks fine to me. It doesn’t really matter what you call the two free variables. — amd, Feb 21 '18 at 06:06
haha yeah its just that so far there were always three of them and so I got confused here — mrnobody, Feb 21 '18 at 06:07

score 3 · Answer 1 · answered Feb 21 '18 at 05:05

3

Yup, they are correct.

The nullity is $2$ since there are $4$ pivot columns and those two vectors are indeed form a basis.

answered Feb 21 '18 at 05:05

Siong Thye Goh

1 Answers1