# Find the standard matrix of the linear transformation

Suppose there is a linear transformation $T:\mathbb{R^2} \rightarrow \mathbb{R^2}$ such that $$T\left( \begin{array}{ccc}2 \\ 1 \end{array} \right)=\left( \begin{array}{ccc}1\\ 4 \end{array} \right)$$ and $$T\left( \begin{array}{ccc}3 \\ 2 \end{array} \right)=\left( \begin{array}{ccc}3 \\ 3 \end{array} \right)$$

What is the standard matrix of $T$?

So I thought I could just solve the system:

$$\left( \begin{array}{ccc} 2 & 3 &|& 1 & 3\\ 1 & 2 &|& 4 & 3\\ \end{array} \right)$$

to obtain:

$$\left( \begin{array}{ccc} 1 & 0 &|& -10 &-3\\ 0 & 1 &|& 7 & 3\\ \end{array} \right)$$

and that the standard matrix would just be:

$$\left( \begin{array}{ccc} -10 &-3\\ 7 & 3\\ \end{array} \right)$$

However, this doesn't seem to be the case.

Let $A$ be the standard matrix of $T$. What we have is $$A \pmatrix{2&3\\1&2} = \pmatrix{1&3\\4&3}$$ So, in fact, the answer to this question will be $$A = \pmatrix{1&3\\4&3}\pmatrix{2&3\\1&2}^{-1}$$ The result you found was $$\pmatrix{-10 &-3\\ 7 & 3\\} = \pmatrix{2&3\\1&2}^{-1}\pmatrix{1&3\\4&3}$$ which is a different matrix altogether.