Vtgyjfy

I know that the answer is $left[begin{matrix} 2 & -1 \ 1 & 1 end{matrix}right]$, but how to get the answer?

Let $mathcal{B} = { mathbf{b}_1 , mathbf{b}_2 }$ be the basis for $mathbb{R}^2$ with $mathbf{b}_1 = left [ begin{matrix} 1 \ 1 end{matrix} right ]$, $mathbf{b}_2 = left [ begin{matrix} 0 \ 1 end{matrix} right ]$. Furthermore, let $T: mathbb{R}^2 to mathbb{R}^2$ be a linear transformation. The matrix representation of $T$ with respect to $mathcal{B}$ is $[T]_mathcal{B} = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]$.
What is the standard matrix of $T$?

Original problem:

You have
$$bbox{T = left [ begin{matrix} t_{11} & t_{12} \ t_{21} & t_{22} end{matrix} right ]}, quad bbox{mathcal{B} = left [ begin{matrix} 1 & 0 \ 1 & 1 end{matrix} right ]}$$
The change of basis is
$$bbox{[T]_mathcal{B} = mathcal{B}^{-1} T mathcal{B} = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]}$$
First step is to calculate $mathcal{B}^{-1}$. A 2×2 matrix is easiest to invert (if possible) via its adjugate matrix. Simply put,
$$bbox{mathbf{M} = left [ begin{matrix} m_{11} & m_{12} \ m_{21} & m_{22} end{matrix} right ]} quad iff quad bbox{mathbf{M}^{-1} = frac{1}{m_{11} m_{22} - m_{12} m_{21}} left [ begin{matrix} m_{22} & -m_{12} \ -m_{21} & m_{11} end{matrix} right ]}$$
Applying this to $mathcal{B}$, we get
$$bbox{mathcal{B}^{-1} = left [ begin{matrix} 1 & 0 \ -1 & 1 end{matrix} right ]}$$
Thus, you need to solve
$$bbox{ left [ begin{matrix} 1 & 0 \ -1 & 1 end{matrix} right ] left [ begin{matrix} t_{11} & t_{12} \ t_{21} & t_{22} end{matrix} right ] left [ begin{matrix} 1 & 0 \ 1 & 1 end{matrix} right ] = left [ begin{matrix} 1 & -1 \ 1 & 2 end{matrix} right ]}$$
for $t_{11}$, $t_{12}$, $t_{21}$, and $t_{22}$.