To find the transition matrix from basis B_2 to basis B_1 , we need to express each vector of B_2 as a linear combination of the vectors in B_1 .
### Step-by-Step Solution
#### Given:
- B_1 = \{ \mathbf{v}_1 = (1, 1, 2), \mathbf{v}_2 = (2, 1, 2), \mathbf{v}_3 = (1, -2, -1) \}
- B_2 = \{ \mathbf{w}_1 = (2, 2, 2), \mathbf{w}_2 = (1, 3, -1), \mathbf{w}_3 = (0, -2, 1) \}
We want to find the transition matrix P such that:
[\mathbf{w}]_{B_1} = P [\mathbf{w}]_{B_2}
This involves expressing each vector in B_2 in terms of the vectors in B_1 .
#### 1. Express \mathbf{w}_1 in terms of B_1:
\mathbf{w}_1 = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3
(2, 2, 2) = c_1(1, 1, 2) + c_2(2, 1, 2) + c_3(1, -2, -1)
This gives us the system of equations:
2 = c_1 + 2c_2 + c_3
2 = c_1 + c_2 - 2c_3
2 = 2c_1 + 2c_2 - c_3
#### 2. Solve the system for c_1, c_2, and c_3:
Solving this system, we obtain:
c_1 = -\frac{2}{5}, c_2 = \frac{9}{5}, c_3 = \frac{2}{5}
So:
\mathbf{w}_1 = -\frac{2}{5}\mathbf{v}_1 + \frac{9}{5}\mathbf{v}_2 + \frac{2}{5}\mathbf{v}_3
#### 3. Express \mathbf{w}_2 in terms of B_1:
\mathbf{w}_2 = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3
(1, 3, -1) = c_1(1, 1, 2) + c_2(2, 1, 2) + c_3(1, -2, -1)
This gives us the system of equations:
1 = c_1 + 2c_2 + c_3
3 = c_1 + c_2 - 2c_3
-1 = 2c_1 + 2c_2 - c_3
Solving this system, we obtain:
c_1 = 1, c_2 = 1, c_3 = 0
So:
\mathbf{w}_2 = 1\mathbf{v}_1 + 1\mathbf{v}_2 + 0\mathbf{v}_3
#### 4. Express \mathbf{w}_3 in terms of B_1:
\mathbf{w}_3 = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3
(0, -2, 1) = c_1(1, 1, 2) + c_2(2, 1, 2) + c_3(1, -2, -1)
This gives us the system of equations:
0 = c_1 + 2c_2 + c_3
-2 = c_1 + c_2 - 2c_3
1 = 2c_1 + 2c_2 - c_3
Solving this system, we obtain:
c_1 = 0, c_2 = -1, c_3 = 1
So:
\mathbf{w}_3 = 0\mathbf{v}_1 - 1\mathbf{v}_2 + 1\mathbf{v}_3
#### Construct the Transition Matrix P :
P = \begin{pmatrix} -\frac{2}{5} & 1 & 0 \ \frac{9}{5} & 1 & -1 \ \frac{2}{5} & 0 & 1 \end{pmatrix}
Thus, the transition matrix from B_2 to B_1 is:
P = \begin{pmatrix} -\frac{2}{5} & 1 & 0 \ \frac{9}{5} & 1 & -1 \ \frac{2}{5} & 0 & 1 \end{pmatrix}