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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.

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  • $\begingroup$ 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 $1\cdot x_1+2\cdot x_2-3\cdot x_3=0$. $\endgroup$ – Barbara Osofsky Feb 20 '13 at 4:47
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your $x \in null(A)$

one choice of $A$:

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

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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 $$ A = \left[\begin{array}{ccc}1 & 2 & 5/3 \\ 0 & 1 & 2/3 \\ 1 & 0 & 1/3 \end{array}\right]. $$

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