(1 point) Let $f: R^{2} \rightarrow R^{2}$ be the linear transformation defined by
\[
f(x)=\left[\begin{array}{cc}
2 & -4 \\
-5 & 0
\end{array}\right] \mathbf{x}
\]
Let
\[
\begin{array}{l}
\mathcal{B}=\{\langle-1,1\rangle,\langle 2,-1\rangle\} \\
\mathcal{C}=\{\langle 1,-1\rangle,\langle 1,-2\rangle\}
\end{array}
\]
be two different bases for $R^{2}$.
a. Find the matrix $[f]_{\mathcal{B}}^{\mathcal{B}}$ for $f$ relative to the basis $\mathcal{B}$.
\[
[f]_{\mathcal{B}}^{\mathcal{B}}=[
\]
b. Find the matrix $[f]_{\mathcal{C}}^{\mathcal{C}}$ for $f$ relative to the basis $\mathcal{C}$.
c. Find the transition matrix $[I]_{\mathcal{C}}^{\mathcal{B}}$ from $\mathcal{C}$ to $\mathcal{B}$
\[
\left[I_{\mathcal{C}}^{\mathcal{B}}=\left[\begin{array}{lll}
\square & \\
\square & & \\
\square
\end{array}\right]\right.
\]
d. Find the transition matrix $[I]_{\mathcal{B}}^{\mathcal{C}}$ from $\mathcal{B}$ to $\mathcal{C}$. (Note: $[I]_{\mathcal{B}}^{\mathcal{C}}=\left([I]_{\mathcal{C}}^{\mathcal{B}}\right)^{-1}$,)
\[
[I]_{\mathcal{B}}^{\mathcal{C}}=\left[\begin{array}{lll}
\square & & \\
\square & & \\
\square & & \\
& &
\end{array}\right]
\]
e. On paper, check that $[I]_{\mathcal{B}}^{\mathcal{C}}[f]_{\mathcal{B}}^{\mathcal{B}}[I]_{\mathcal{C}}^{\mathcal{B}}=[f]_{\mathcal{C}}^{\mathcal{C}}$
Answer
\(\boxed{[f]_{\mathcal{B}}^{\mathcal{B}} = \left[\begin{array}{cc} 4 & -12 \\ -1 & -2 \end{array}\right]}\)
Step 1 ：First, we apply the transformation \(f\) to each vector in the basis \(\mathcal{B}\).
Step 2 ：We then express the result as a linear combination of the basis vectors in \(\mathcal{B}\).
Step 3 ：The coefficients of these linear combinations will form the columns of the matrix representation.
Step 4 ：By performing these steps, we find that the matrix \([f]_{\mathcal{B}}^{\mathcal{B}}\) for \(f\) relative to the basis \(\mathcal{B}\) is \(\left[\begin{array}{cc} 4 & -12 \\ -1 & -2 \end{array}\right]\).
Step 5 ：\(\boxed{[f]_{\mathcal{B}}^{\mathcal{B}} = \left[\begin{array}{cc} 4 & -12 \\ -1 & -2 \end{array}\right]}\)
