Question: Find the transition matrix from $B$ to $\mathbf{B'}$ for the following bases in $\mathbb{R}^3$. $B=\{(1,1,0),(1,0,1),(0,1,1)\}$ and $\mathbf{B'}=\{(1,0,1),(2,1,1),(1,2,1)\}$.
1. **State the problem:** We want to find the transition matrix from basis $B$ to basis $\mathbf{B'}$ in $\mathbb{R}^3$. This matrix converts coordinates of vectors from the $B$-basis representation to the $\mathbf{B'}$-basis representation.
2. **Recall the formula:** The transition matrix from $B$ to $\mathbf{B'}$, denoted $P_{B \to \mathbf{B'}}$, satisfies:
$$[v]_{\mathbf{B'}} = P_{B \to \mathbf{B'}} [v]_B$$
for any vector $v$.
3. **Key idea:** To find $P_{B \to \mathbf{B'}}$, express each vector of $B$ in terms of the basis $\mathbf{B'}$. The columns of $P_{B \to \mathbf{B'}}$ are the coordinates of the vectors of $B$ expressed in $\mathbf{B'}$.
4. **Given bases:**
$$B = \{b_1=(1,1,0), b_2=(1,0,1), b_3=(0,1,1)\}$$
$$\mathbf{B'} = \{b'_1=(1,0,1), b'_2=(2,1,1), b'_3=(1,2,1)\}$$
5. **Express $b_1$ in terms of $\mathbf{B'}$: find scalars $x,y,z$ such that**
$$b_1 = x b'_1 + y b'_2 + z b'_3$$
which means
$$ (1,1,0) = x(1,0,1) + y(2,1,1) + z(1,2,1) $$
6. **Set up the system of equations:**
$$\begin{cases}
1 = x + 2y + z \\
1 = 0x + y + 2z \\
0 = x + y + z
\end{cases}$$
7. **Solve the system:**
From the third equation:
$$0 = x + y + z \implies x = -y - z$$
Substitute into the first:
$$1 = (-y - z) + 2y + z = y$$
So,
$$y = 1$$
Substitute $y=1$ into the second:
$$1 = 1 + 2z \implies 2z = 0 \implies z = 0$$
Then,
$$x = -1 - 0 = -1$$
8. **Coordinates of $b_1$ in $\mathbf{B'}$ are:**
$$(-1, 1, 0)$$
9. **Repeat for $b_2 = (1,0,1)$:**
$$ (1,0,1) = x b'_1 + y b'_2 + z b'_3 = (x + 2y + z, y + 2z, x + y + z) $$
System:
$$\begin{cases}
1 = x + 2y + z \\
0 = y + 2z \\
1 = x + y + z
\end{cases}$$
From the second:
$$0 = y + 2z \implies y = -2z$$
Substitute into the first:
$$1 = x + 2(-2z) + z = x - 4z + z = x - 3z$$
Substitute into the third:
$$1 = x + (-2z) + z = x - z$$
From the third:
$$x = 1 + z$$
Substitute into the first:
$$1 = (1 + z) - 3z = 1 - 2z \implies -2z = 0 \implies z = 0$$
Then,
$$y = -2(0) = 0$$
$$x = 1 + 0 = 1$$
10. **Coordinates of $b_2$ in $\mathbf{B'}$ are:**
$$(1, 0, 0)$$
11. **Repeat for $b_3 = (0,1,1)$:**
$$ (0,1,1) = x b'_1 + y b'_2 + z b'_3 = (x + 2y + z, y + 2z, x + y + z) $$
System:
$$\begin{cases}
0 = x + 2y + z \\
1 = y + 2z \\
1 = x + y + z
\end{cases}$$
From the first and third:
Subtract first from third:
$$(1) - (0) = (x + y + z) - (x + 2y + z) = y - 2y = -y = 1 \implies y = -1$$
From second:
$$1 = -1 + 2z \implies 2z = 2 \implies z = 1$$
From first:
$$0 = x + 2(-1) + 1 = x - 2 + 1 = x - 1 \implies x = 1$$
12. **Coordinates of $b_3$ in $\mathbf{B'}$ are:**
$$(1, -1, 1)$$
13. **Form the transition matrix $P_{B \to \mathbf{B'}}$ with these coordinates as columns:**
$$P_{B \to \mathbf{B'}} = \begin{bmatrix} -1 & 1 & 1 \\ 1 & 0 & -1 \\ 0 & 0 & 1 \end{bmatrix}$$
**Final answer:**
$$\boxed{P_{B \to \mathbf{B'}} = \begin{bmatrix} -1 & 1 & 1 \\ 1 & 0 & -1 \\ 0 & 0 & 1 \end{bmatrix}}$$