What matrix transforms $(1, 0)$ into $(2, 6)$ and tranforms $(0, 1)$ into $(4, 8)$?

Question

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?

• Jon: Pretend such a matrix $A$ exists and multiply it by each of your vectors. Then solve for the entries of $A$.

Answer 1

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

and

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

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?"