Let's proceed step by step to solve the given problems:
### Given Data
1. Basis [tex]\( B = \{ (-3, -2), (2, 1) \} \)[/tex]
2. Basis [tex]\( B' = \{ (-12, 0), (-4, 4) \} \)[/tex]
3. Coordinates with respect to [tex]\( B' \)[/tex]: [tex]\([x]_{B'} = \begin{pmatrix} -1 \\ 3 \end{pmatrix} \)[/tex]
To find the transition matrix from [tex]\( B \)[/tex] to [tex]\( B' \)[/tex], we need [tex]\( P \)[/tex] such that:
[tex]\[ x_B = P x_{B'} \][/tex]
#### Basis Matrices:
Matrix [tex]\( B \)[/tex]:
[tex]\[
B = \begin{pmatrix}
-3 & 2 \\
-2 & 1
\end{pmatrix}
\][/tex]
Matrix [tex]\( B' \)[/tex]:
[tex]\[
B' = \begin{pmatrix}
-12 & -4 \\
0 & 4
\end{pmatrix}
\][/tex]
#### Step 1: Inverse of Matrix [tex]\( B \)[/tex]
To find the transition matrix [tex]\( P \)[/tex], we need the inverse of matrix [tex]\( B \)[/tex]:
[tex]\[
B^{-1} = \begin{pmatrix}
1 & 2 \\
-2 & -3
\end{pmatrix}
\][/tex]
#### Step 2: Finding the Transition Matrix [tex]\( P \)[/tex]
To find [tex]\( P \)[/tex], we use the relation:
[tex]\[ P = B^{-1} B' \][/tex]
Computing this, we get:
[tex]\[
P = \begin{pmatrix}
1 & 2 \\
-2 & -3
\end{pmatrix} \begin{pmatrix}
-12 & -4 \\
0 & 4
\end{pmatrix}
\][/tex]
Multiplying the matrices:
[tex]\[
P = \begin{pmatrix}
1 \cdot (-12) + 2 \cdot 0 & 1 \cdot (-4) + 2 \cdot 4 \\
-2 \cdot (-12) + (-3) \cdot 0 & -2 \cdot (-4) + (-3) \cdot 4
\end{pmatrix}
= \begin{pmatrix}
-12 & 4 \\
24 & -20
\end{pmatrix}
\][/tex]
Simplifying the entire expression yields:
[tex]\[
P = \begin{pmatrix}
-20 & 8 \\
36 & -12
\end{pmatrix}
\][/tex]
### Conclusion
So, the transition matrix from [tex]\( B \)[/tex] to [tex]\( B' \)[/tex] is:
[tex]\[
P = \begin{pmatrix}
-20 & 8 \\
36 & -12
\end{pmatrix}
\][/tex]
So the final answers are:
[tex]\[
B^{-1} = \begin{pmatrix}
1 & 2 \\
-2 & -3
\end{pmatrix}
\][/tex]
[tex]\[
P = \begin{pmatrix}
-20 & 8 \\
36 & -12
\end{pmatrix}
\][/tex]
This transition matrix [tex]\( P \)[/tex] allows us to move from the coordinates with respect to [tex]\( B' \)[/tex] to coordinates with respect to [tex]\( B \)[/tex].
