In the last 2 lectures of linear algebra we have talked about linear mappings and other stuff, but I missed actually the last one and I am quite in bad situation.

What matrix transforms $\left(\begin{matrix} 1 \\ 0\end{matrix}\right)$ into $\left(\begin{matrix} 2 \\ 6\end{matrix}\right)$ and tranforms $\left(\begin{matrix} 0 \\ 1\end{matrix}\right)$ into $\left(\begin{matrix} 4 \\ 8\end{matrix}\right)$?

I think I understood what I need to find: a matrix that multiplies our initial matrix formed by our initial vectors $$\left(\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}\right)$$

and the resulting matrix is: $$\left(\begin{matrix} 2 & 6 \\ 4 & 8\end{matrix}\right)$$

Am I right?

Is there a way to automate this process?

  • $\begingroup$ Pretend such a matrix $A$ exists and multiply it by each of your vectors. Then solve for the entries of $A$. $\endgroup$
    – Jon
    May 4, 2015 at 20:50

1 Answer 1


No, the question is to find $a,b,c,d$ so that $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}2\\6\end{pmatrix} $$


$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}4\\8\end{pmatrix} $$

You can find this by performing the multiplication explicitly in the first equation, obtaining two equations in $a,b,c,d$ (one from the upper component of the result, one from the lower) and then similarly in the second equation. Then you have four equations in $a,b,c,d$ which you can solve to find the matrix you seek.

When you do find that matrix, you will look at it and say “Oh, is that all?”


You must log in to answer this question.