Find matrix of linear transformation A linear transformation T:R2→R2 is given by T(i)=i+j T(j)=2i−j

Step 1
If ${\displaystyle T\left(i\right)=(1,1)}$ and ${\displaystyle T\left(j\right)=(2,-1)}$ and
${\displaystyle e_1=i-j=(1,-1)}$ and ${\displaystyle e_2=3i+j=(3,1)}$, then
${\displaystyle T\left(e_1\right)=T(i-j)}$
${\displaystyle =T\left(i\right)-T\left(j\right)}$
${\displaystyle =(1,2)}$
${\displaystyle =a_1{e}_1+b_1{e}_2}$ (say)
and
${\displaystyle T\left(e_2\right)=T(3i+j)}$
${\displaystyle =3(1,1)+(2,-1)}$
${\displaystyle =(5,2)}$
${\displaystyle =a_2{e}_1+b_2{e}_2}$ (say)
Thus the matrix of T w.r.t. the new basis ${\displaystyle \{e_1,e_2\}}$ is
${\displaystyle \left(\begin{array}{cc}{a}_1& a_2\\ b_1& b_2\end{array}\right)}$
and you need to find the values of ${\displaystyle a_1,a_2,b_1}$ and ${\displaystyle b_2}$. The above systems of equations reduces to
${\displaystyle a_1+2b_1=-1}$
${\displaystyle -a_1+b_1=2}$
and
${\displaystyle a_2+3b_2=5}$
${\displaystyle -a_2+b_2=2}$
Solve these equations to obtain
${\displaystyle a_1=-\frac{7}{4}}$
${\displaystyle b_1=\frac{1}{4}}$
${\displaystyle a_2=-\frac{1}{4}}$
and
${\displaystyle b_2=\frac{7}{4}}$

Step 1
The matrix of the transformation of basis is:
${\displaystyle P=\left[\begin{array}{cc}1& 3\\ -1& 1\end{array}\right]}$
and
${\displaystyle P^{-1}=\frac{1}{4}\left[\begin{array}{cc}1& -3\\ 1& 1\end{array}\right]}$
Your transformed matrix is:
${\displaystyle P^{-1}TP=\frac{1}{4}\left[\begin{array}{cc}1& -3\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 1& -1\end{array}\right]\left[\begin{array}{cc}1& 3\\ -1& 1\end{array}\right]}$
${\displaystyle =\frac{1}{4}\left[\begin{array}{cc}-7& -1\\ 1& 7\end{array}\right]}$
Step 2
The figure illustrate how operate the given transformation in the two basis.
Indices ${\displaystyle \{i,j\}}$ refer to the canonical basis, indices ${\displaystyle \{1,2\}}$ to the new basis ${\displaystyle e_1,e_2}$.
In the figure we have the vector ${\displaystyle v=v_{i,j}={[1,2]}^T}$ (in canonical basis), that is transformed in ${\displaystyle v^{\prime }=v_{i,j}^{\prime }={[5,-1]}^T}$
In the new basis we have:
${\displaystyle v_{1,2}=P^{-1}{v}_{i,j}=\frac{1}{4}\left[\begin{array}{c}-5\\ 3\end{array}\right]}$
and, for the tranformed vector:
${\displaystyle v_{1,2}^{\prime }=P^{-1}{v}_{i,j}=\frac{1}{4}\left[\begin{array}{c}2\\ 1\end{array}\right]}$
${\displaystyle =P^{-1}{T}_{i.j}{v}_{i,j}}$
${\displaystyle =P^{-1}{T}_{i,j}P{v}_{1,2}}$
${\displaystyle =T_{1,2}{v}_{1,2}}$
ao we have:
${\displaystyle T_{1,2}=P^{-1}{T}_{i,j}P=\frac{1}{4}\left[\begin{array}{cc}-7& -1\\ 1& 7\end{array}\right]}$
I hope this can be helpful.

Step 1 $e_1+e_2=4i$ $i=\frac{e_1+e_2}{4}=P(i|j{)}^T$ $e_2-3e_1=4j$ $j=\frac{e_2-3e_1}{4}=P(i|j{)}^T$ Now compute the Matrix $P^{-1}$ and your T in the new Basis is given by $P^{-1}TP$