Transition Matrix 4Dd75D

1. **State the problem:** We want to find the transition matrix from basis $B$ to basis $\mathbf{B}'$ in $\mathbb{R}^3$. 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. **Method:** To find $P_{B \to \mathbf{B}'}$, express each vector of $B$ in terms of $\mathbf{B}'$ coordinates. 4. **Write bases explicitly:** $$B = \{b_1=(1,1,0), b_2=(1,0,1), b_3=(0,1,1)\}$$ $$\mathbf{B}' = \{b'_1=(1,0,1), b'_2=(2,1,1), b'_3=(1,2,1)\}$$ 5. **Express $b_1$ in $\mathbf{B}'$ coordinates:** 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 system of equations:** From coordinates: - $x + 2y + z = 1$ (first coordinate) - $0x + y + 2z = 1$ (second coordinate) - $x + y + z = 0$ (third coordinate) 7. **Solve system for $b_1$:** From second equation: $$ y + 2z = 1 $$ From third equation: $$ x + y + z = 0 \implies x = -y - z $$ Substitute $x$ into first equation: $$ (-y - z) + 2y + z = 1 \implies (-y - z) + 2y + z = 1 \implies y = 1 $$ From $y + 2z = 1$: $$ 1 + 2z = 1 \implies 2z = 0 \implies z = 0 $$ Then $$ x = -y - z = -1 - 0 = -1 $$ So, $$ b_1 = -1 b'_1 + 1 b'_2 + 0 b'_3 $$ 8. **Express $b_2$ in $\mathbf{B}'$ coordinates:** Find $x,y,z$ such that $$ (1,0,1) = x(1,0,1) + y(2,1,1) + z(1,2,1) $$ Equations: - $x + 2y + z = 1$ - $0x + y + 2z = 0$ - $x + y + z = 1$ From second: $$ y + 2z = 0 $$ From third: $$ x + y + z = 1 \implies x = 1 - y - z $$ Substitute $x$ into first: $$ (1 - y - z) + 2y + z = 1 \implies 1 - y - z + 2y + z = 1 \implies 1 + y = 1 \implies y = 0 $$ From $y + 2z = 0$: $$ 0 + 2z = 0 \implies z = 0 $$ Then $$ x = 1 - 0 - 0 = 1 $$ So, $$ b_2 = 1 b'_1 + 0 b'_2 + 0 b'_3 $$ 9. **Express $b_3$ in $\mathbf{B}'$ coordinates:** Find $x,y,z$ such that $$ (0,1,1) = x(1,0,1) + y(2,1,1) + z(1,2,1) $$ Equations: - $x + 2y + z = 0$ - $0x + y + 2z = 1$ - $x + y + z = 1$ From third: $$ x = 1 - y - z $$ Substitute into first: $$ (1 - y - z) + 2y + z = 0 \implies 1 - y - z + 2y + z = 0 \implies 1 + y = 0 \implies y = -1 $$ From second: $$ y + 2z = 1 \implies -1 + 2z = 1 \implies 2z = 2 \implies z = 1 $$ Then $$ x = 1 - (-1) - 1 = 1 + 1 - 1 = 1 $$ So, $$ b_3 = 1 b'_1 - 1 b'_2 + 1 b'_3 $$ 10. **Form the transition matrix:** Columns are coordinates of $b_1, b_2, b_3$ in $\mathbf{B}'$: $$ P_{B \to \mathbf{B}'} = \begin{bmatrix} -1 & 1 & 1 \\ 1 & 0 & -1 \\ 0 & 0 & 1 \end{bmatrix} $$ **Final answer:** $$\boxed{P_{B \to \mathbf{B}'} = \begin{bmatrix} -1 & 1 & 1 \\ 1 & 0 & -1 \\ 0 & 0 & 1 \end{bmatrix}}$$