Question

# BASE CHANGE Given the Bases, obtain the transition matrix from B2 to B1 B1= { $112$ ,$212$ ,$1−2−1$ } B2= { $222$ ,$13−1$ ,$0−21$ }

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## Answer to a math question BASE CHANGE Given the Bases, obtain the transition matrix from B2 to B1 B1= { $112$ ,$212$ ,$1−2−1$ } B2= { $222$ ,$13−1$ ,$0−21$ }

Corbin
4.6
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}

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