Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I just took an exam and theres a question that's driving me crazy because I can't seem to figure it out. It said that I was to find a matrix $A$ such that $A\vec x=\vec 0$ and it has the solution $\vec x=\left[\begin{matrix}1\\2\\-3\end{matrix}\right]$ and $A$ isn't all $0$'s. I'm completely stumped and I know I will find out how to do it eventually but it's bugging me. I really want to know. All I could really figure out was that this is what I wanted as my end product of the matrix after it gets row reduced: $$\left[\begin{matrix}1&0&0&1\\ 0&1&0&2\\ 0&0&1&-3\end{matrix}\right]$$ Where the last row is the augmented column.

share|cite|improve this question
Why must your matrix be square? An $A$ which has the form you give here would compute a solution to the system of equations $x_1=1$, $x_2=2$, $x_3=-3$ which is very different from the system $1\cdot x_1+2\cdot x_2-3\cdot x_3=0$. – Barbara Osofsky Feb 20 '13 at 4:47
up vote 3 down vote accepted

your $x \in null(A)$

one choice of $A$:

$$A = \begin{pmatrix} 1&1&1\\ 1&1&1\\ 1&1&1 \end{pmatrix}$$

share|cite|improve this answer

You want to choose $A$ so that all its rows are orthogonal to $\vec{x}$. Assuming $A$ is $3\times 3$, then for each row, you can arbitrarily choose the first two elements and then compute the third so that that row is orthogonal to $\vec{x}$. For example, let $$ A = \left[\begin{array}{ccc}1 & 2 & ? \\ 0 & 1 & ? \\ 1 & 0 & ? \end{array}\right]. $$ Consider each row of $A$ in turn and compute the ? so that the row is orthogonal to $\vec{x}$. In this case, we have $$ A = \left[\begin{array}{ccc}1 & 2 & 5/3 \\ 0 & 1 & 2/3 \\ 1 & 0 & 1/3 \end{array}\right]. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.