The exercise I have to solve is the following:
"Suppose $A$ is a $5\times 4$ matrix with columns $a_1, a_2, a_3, a_4$, and $a_1 + a_4 = a_2 + a_3$.
Which of the following vectors is (certainly) a solution of the equation $Ax = 0$ ?
$$\begin{split}
\mathbf u &= [\begin{matrix}1& -1& -1& 1\end{matrix}]\\
\mathbf v &= [\begin{matrix}1& 1& 1& 1\end{matrix}]\\
\mathbf w &= [\begin{matrix}1& -1& 0& 0\end{matrix}]\\
\end{split}$$
(all of these are $4\times 1$ matrices)
I know the correct answer is matrix $\mathbf u$ however I can't understand why. In my solution I wrote matrix $A$ as follows while taking into consideration the given relation between its columns:
$$A=\begin{bmatrix}
1 &-1 &-1& 1\\
0 & 0 & 0 &0\\
0 & 0 & 0 &0\\
0 & 0 & 0 &0\\
\end{bmatrix}$$
Therefore the solution here considering only column $a_1$ is a pivot column is the following ($s, t, u$ arbitrary scalars):
$$[\begin{matrix}1& 1& 0& 0\end{matrix}]b
+ [\begin{matrix}1 &0 &1& 0\end{matrix}]c
+ [\begin{matrix}-1& 0 &0 &1\end{matrix}d
$$
(again all of these are $4\times1$ matrices)
And then I found that only vector from above can be written in this form is vector $\mathbf v$. So what am I missing? Thanks in advance!
Best Answer
HINT
Notice that $x = (x_{1},x_{2},x_{3},x_{4})\in\mathbb{R}^{4}$ is a solution of $Ax = 0$ iff \begin{align*} x_{1}a_{1} + x_{2}a_{2} + x_{3}a_{3} + x_{4}a_{4} = 0 & \Longleftrightarrow x_{1}(a_{2} + a_{3} - a_{4}) + x_{2}a_{2} + x_{3}a_{3} + x_{4}a_{4} = 0\\\\ & \Longleftrightarrow (x_{1} + x_{2})a_{1} + (x_{1} + x_{3})a_{3} + (x_{4} - x_{1})a_{4} = 0 \end{align*}
Based on such relation, can you realize why $u = (1,-1,-1,1)^{T}$ is certainly a solution?