Let’s break this problem into parts as outlined in the question.

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Step-by-step guide to solving this:

a. Find the transition matrix $T_C^B$:

The transition matrix from basis $C$ to basis $B$, denoted as $T_C^B$, transforms coordinates in $C$-basis to $B$-basis. To compute it:

1. Write each vector in $C$ as a linear combination of the basis vectors in $B$.
2. The coefficients of these linear combinations will form the columns of $T_C^B$.

Given:

• $B = \left\{ \begin{bmatrix} -3 & -2 \\ 0 & 3 \end{bmatrix}, \begin{bmatrix} -1 & 1 \\ 0 & -3 \end{bmatrix}, \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} \right\}$
• $C = \left\{ \begin{bmatrix} 2 & -2 \\ 0 & -3 \end{bmatrix}, \begin{bmatrix} 0 & 2 \\ 0 & -2 \end{bmatrix}, \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \right\}$

To find the transition matrix:

• Express each vector in $C$ as a linear combination of the vectors in $B$. Let $\vec{c}_i = a_1\vec{b}_1 + a_2\vec{b}_2 + a_3\vec{b}_3$, where $a_1, a_2, a_3$ are scalars.
• Solve the corresponding system of equations for each vector $\vec{c}_i$ (i.e., $c_1, c_2, c_3$).

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b. Find the coordinates of $M$ in the ordered basis $B$:

The relation between coordinates in different bases is given by: [ [M]_B = T_C^B [M]_C. ]

Given:

• $[M]_C = \begin{bmatrix} 1 \\ -2 \\ -1 \end{bmatrix}$
• Compute $[M]_B$ by multiplying $T_C^B$ (from part a) with $[M]_C$.

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c. Find $M$:

The actual matrix $M$ can be recovered using: $M = [M]_B \cdot B,$ where $B$ is the matrix of basis vectors for $B$.

• Express $M$ as a linear combination of the basis vectors in $B$, using the coordinates $[M]_B$ obtained in part (b).

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Let’s compute this together.

Would you like me to handle the actual computation for part (a) first, or do you need help with a specific section?