Suppose that A is a 3 x 3 matrix and that the solution to $A\vec{x} = \left[\begin{array}{ccc}8\\10\\5\end{array}\right]$ is

$\vec{x} = \left[\begin{array}{ccc}4\\-1\\0\end{array}\right] + s\left[\begin{array}{ccc}2\\3\\-1\end{array}\right], s \in \mathbb R$

The problem:
Is this enough information to find the solution to $A \vec{x} = \vec{0}$?
If yes, what is the solution?
If no, explain why not.

The solution:

1) Yes. Solution to $A \vec{x} = \vec{0}$ is $\vec{x} = s\left[\begin{array}{ccc}2\\3\\-1\end{array}\right], s \in \mathbb R$

2) If the solution to $A\vec{x} = \left[\begin{array}{ccc}8\\10\\5\end{array}\right]$ is $\vec{x} = \left[\begin{array}{ccc}4\\-1\\0\end{array}\right] + s\left[\begin{array}{ccc}2\\3\\-1\end{array}\right]$, then

3) $\left[\begin{array}{c|c}&8\\A&10\\&5\end{array}\right] \xrightarrow[\text{}]{\text{rref}} \left[\begin{array}{ccc|c}1&0&-2&4\\0&1&3&-1\\0&0&0&0\end{array}\right]$ then

4) $\left[\begin{array}{c|c}&0\\A&0\\&0\end{array}\right] \xrightarrow[\text{}]{\text{rref}} \left[\begin{array}{ccc|c}1&0&-2&0\\0&1&3&0\\0&0&0&0\end{array}\right]$

My question:
How am I supposed to start this question?
(the solution is given, but I do not know how to achieve even the first step)

  • $\begingroup$ Are you sure that in step 3 and 4 the entries $-2$ and $3$ are given to be opposite signs? $\endgroup$
    – imranfat
    Oct 3, 2017 at 19:16
  • $\begingroup$ @imranfat It is, according to the given solution $\endgroup$
    – arsyyuhjb1
    Oct 3, 2017 at 19:25
  • $\begingroup$ Then there is an issue. From RREF I read in the first line $a-2c=4$, second line $b+3c=-1$ and $c$ = free. This gives $a=4+2c$, $b=-1-3c$ and $c$ free. This means that the vector as given for $s$ is incorrect. That should be $<-2,3,-1>$ for $c=-1$. It doesn't invalidate Ziad's solution though. $\endgroup$
    – imranfat
    Oct 3, 2017 at 21:52

2 Answers 2


If the solution of $A\vec{x} = b$ is of the form $\vec x=\vec x_1+s\vec x_2$, then

$$A\vec x_1+sA\vec x_2=b.$$

But for this relation to be true for any $s$, we must have

$$A\vec x_2=0$$ and $x=s\vec x_2$ is a solution of $A\vec x=0$.

  • $\begingroup$ I understand the beginning of your solution, but how did you get "... and $x=s\vec{x}_{2}$ is a solution of $A\vec{x} = 0$"? $\endgroup$
    – arsyyuhjb1
    Oct 3, 2017 at 20:18
  • $\begingroup$ @arsyyuhjb: hem, if $A\vec x_2=0$, then $A(s\vec x_2)=0$. $\endgroup$
    – user65203
    Oct 4, 2017 at 7:00
  • 1
    $\begingroup$ @YvesDaoust That's a neat approach $\endgroup$
    – imranfat
    Oct 4, 2017 at 18:08

This actually does not require much work, just a bit of thought. Remember that the matrix transformation is linear meaning if we multiplied $A$ by the sum of 2 vectors $v+ w$ then we get $Av + Aw$ more concretely $$A(v+w) = A(v) + A(w)$$

Now we are told that the solution to your equation involves 2 vectors, lets call them $v$ and $w$. Now the solution is $x = v+ sw$ where $s$ can be any real number. But why can $s$ be any real number, simple because $Aw = 0$. Let me show you. Let $s$ be substiuted by 2 arbitrary numbers $1$ and $2$ $$Ax = A(v+ sw) = Av + sAw = Av + 1(Aw) = Av + 2(Aw)= \begin{pmatrix}8 \\ 10 \\5 \end{pmatrix}$$ $$Av + Aw = Av + 2Aw$$ $$Aw = 0$$

As a result, we can see why it does not matter what $s$ is. $$Ax = A(v+ sw) = Av + sAw = Av + s(0) = \begin{pmatrix}8 \\ 10 \\5 \end{pmatrix}$$ So it didnt matter what $s$ was as $Aw = 0$. So what is the solution to $Ax = 0$ then?

  • $\begingroup$ Let me get this right. So you are letting $v = [8, 10, 5]$ and $w = [2, 3, -1]$? Also, I don't get why you can say $Aw = 0$ (I know it is first step in solution), but why is it 0? $\endgroup$
    – arsyyuhjb1
    Oct 3, 2017 at 19:37
  • $\begingroup$ Yes you are right about $v,w$ as for the second part of the question I will edit my answer. $\endgroup$ Oct 4, 2017 at 17:41

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