# Standard matrix of a transformation, matrix representation [closed]

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:

## closed as off-topic by Dietrich Burde, Namaste, max_zorn, metamorphy, Ali CaglayanJan 5 at 19:31

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• Hint – John Doe Jan 5 at 13:50
• You'll get a better response if you write the question here rather than linking to an image, and also explain what you have tried and which step you got stuck on. – littleO Jan 5 at 13:50
• @JohnDoe the hint helped, thanks :) – Antoni Malecki Jan 5 at 14:49
• @AntoniMalecki great! :) – John Doe Jan 5 at 14:51

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 $$t_{22}$$.