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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: The Problem

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

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  • $\begingroup$ Hint $\endgroup$ – John Doe Jan 5 at 13:50
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    $\begingroup$ 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. $\endgroup$ – littleO Jan 5 at 13:50
  • $\begingroup$ @JohnDoe the hint helped, thanks :) $\endgroup$ – Antoni Malecki Jan 5 at 14:49
  • $\begingroup$ @AntoniMalecki great! :) $\endgroup$ – John Doe Jan 5 at 14:51
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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 $t_{22}$.

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