0
$\begingroup$

I have the following matrix $$ A= \begin{bmatrix} 1 & 3 & 0 & 1 & 0\\ 1 & 2 & 1 &-2 & 1\\ -1 & 1 &-1 & 3 & 2\\ \end{bmatrix} $$ I have calculated the column space and I've got $\text{ Col}(A) = \big\{(1,1,-1)^T,(3,2,1)^T,(0,1,-1)^T\big\}$

Now the question is if the vector $\vec b = (1,1,1)^T$ is a part of the column space of $A$?
And what is the solution set for $A\vec x = \vec b$?

Tried to do Gauss-elimination with the values of the vector on the right side. Is that the correct first step? If not, can someone explain how it is done?

$\endgroup$
2
  • $\begingroup$ if the column space is $3$-dimensional, then, yes, it contains any element of $\mathbb R^3$, including $(1,1,1)$ $\endgroup$ Commented Nov 15, 2020 at 16:23
  • $\begingroup$ Ok. So, what does the solution set become? $\endgroup$
    – MathLord
    Commented Nov 15, 2020 at 16:47

1 Answer 1

1
$\begingroup$

$b\in C(A)$ if and only if you can write it as a linear combination of vectors that are in $C(A)$.

In that case, you can set up a system of equations $\alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3=(1,1,1)$ ($\alpha_i\in\mathbb{R}, v_i \in C(A), i\in[|1,3|]$).

Another approuch would be that $C(A)$ you got are linearly indpendent vectors (it should be if you got it right), and then the answer is yes, because it spans $\mathbb{R}^3$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .