its) Let \( v=(-5,8,-5) \in \mathbf{R}^{\circ} \). Let \( B \) and \( C \) be bases of \( \mathbf{R}^{\circ} \) \( B=\{(2,1,1),(2,-1,1),(1,2,1)\}, C=\{(3,1,-5),(1,1,-3),(-1,0,2)\} \) Find the coordinates \( [v]_{B} \) of vector \( v \) with respect to \( B \). Find the transition matrix \( P \) from \( B \) to \( C \). Using \( [v]_{B} \) and \( P \) find \( [v]_{C} \).

To find the coordinates of the vector \( v = (-5, 8, -5) \) with respect to the basis \( B \), we need to express \( v \) as a linear combination of the vectors in \( B \). The basis \( B \) consists of the vectors: \[ b_1 = (2, 1, 1), \quad b_2 = (2, -1, 1), \quad b_3 = (1, 2, 1) \] We want to find scalars \( x_1, x_2, x_3 \) such that: \[ v = x_1 b_1 + x_2 b_2 + x_3 b_3 \] This leads to the following system of equations: \[ x_1(2, 1, 1) + x_2(2, -1, 1) + x_3(1, 2, 1) = (-5, 8, -5) \] This can be written as a matrix equation: \[ \begin{pmatrix} 2 & 2 & 1 \\ 1 & -1 & 2 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} -5 \\ 8 \\ -5 \end{pmatrix} \] We can solve this system using row reduction. The augmented matrix is: \[ \begin{pmatrix} 2 & 2 & 1 & | & -5 \\ 1 & -1 & 2 & | & 8 \\ 1 & 1 & 1 & | & -5 \end{pmatrix} \] Performing row operations to reduce this matrix: 1. Divide the first row by 2: \[ \begin{pmatrix} 1 & 1 & 0.5 & | & -2.5 \\ 1 & -1 & 2 & | & 8 \\ 1 & 1 & 1 & | & -5 \end{pmatrix} \] 2. Subtract the first row from the second and third rows: \[ \begin{pmatrix} 1 & 1 & 0.5 & | & -2.5 \\ 0 & -2 & 1.5 & | & 10.5 \\ 0 & 0 & 0.5 & | & -2.5 \end{pmatrix} \] 3. Divide the second row by -2: \[ \begin{pmatrix} 1 & 1 & 0.5 & | & -2.5 \\ 0 & 1 & -0.75 & | & -5.25 \\ 0 & 0 & 0.5 & | & -2.5 \end{pmatrix} \] 4. Divide the third row by 0.5: \[ \begin{pmatrix} 1 & 1 & 0.5 & | & -2.5 \\ 0 & 1 & -0.75 & | & -5.25 \\ 0 & 0 & 1 & | & -5 \end{pmatrix} \] 5. Back substitute to eliminate the third variable: \[ \begin{pmatrix} 1 & 1 & 0 & | & -0.5 \\ 0 & 1 & 0 & | & -3 \\ 0 & 0 & 1 & | & -5 \end{pmatrix} \] From the second row, we have \( x_2 = -3 \). Substituting \( x_2 \) into the first row gives: \[ x_1 - 3 = -0.5 \implies x_1 = 2.5 \] And from the third row, we have \( x_3 = -5 \). Thus, the coordinates of \( v \) with respect to the basis \( B \) are: \[ [v]_{B} = \begin{pmatrix} 2.5 \\ -3 \\ -5 \end{pmatrix} \] Next, we need to find the transition matrix \( P \) from basis \( B \) to basis \( C \). The columns of \( P \) are the coordinates of the basis vectors of \( B \) expressed in terms of the basis \( C \). To find \( P \), we first express each vector in \( B \) in terms of the basis \( C \): \[ C = \{(3, 1, -5), (1, 1, -3), (-1, 0, 2)\} \] We need to find the inverse of the matrix formed by the basis vectors of \( C \): \[ C = \begin{pmatrix} 3 & 1 & -1 \\ 1 & 1 & 0 \\ -5 & -3 & 2 \end{pmatrix} \] Calculating the inverse of \( C \): 1. Find the determinant of \( C \): \[ \text{det}(C) = 3(1 \cdot 2 - 0 \cdot -3) - 1(1 \cdot 2 - 0 \cdot -5) - 1(-5 \cdot 0 - -3 \cdot 1) = 6 - 2 + 3 = 7 \] 2. The inverse \( C^{-1} \) can be calculated using the adjugate method or row reduction. For simplicity, we can use a calculator or software to find: \[ C^{-1} = \frac{1}{7} \begin{pmatrix} 2 & 1 & 1 \\ -5 & -3 & 3 \\ 1 & 0 & 1 \end{pmatrix} \] Now, we can find the coordinates of the basis vectors of \( B \) in terms of \( C \): \[ P = C^{-1} B \] Calculating \( P \): \[ P = \frac{1}{7} \begin{pmatrix} 2 & 1 & 1 \\ -5 & -3 & 3 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 2 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & 1 \end{pmatrix} \] Calculating the product: 1. First column: \[ \begin{pmatrix} 2 \cdot 2 + 1 \cdot 2 + 1 \cdot 1 \\ -5 \cdot 2 - 3 \cdot 2 + 3 \cdot 1 \\ 1 \cdot 2 + 0 \cdot 2 + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} 4 + 2 + 1 \\ -10 - 6 + 3 \\ 2 + 0 + 1 \end{pmatrix} = \begin{pmatrix} 7 \\ -13 \\ 3 \end{pmatrix} \] 2. Second column: \[ \begin{pmatrix} 2 \cdot 2 + 1 \cdot -1 + 1 \cdot 2 \\ -5 \cdot 2 - 3 \cdot -1 + 3 \cdot 2 \\ 1 \cdot 2 + 0 \cdot -1 + 1 \cdot 2 \end{pmatrix} = \begin{pmatrix} 4 - 1 + 2 \\ -10 + 3 + 6 \\ 2 + 0 + 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -1 \\ 4 \end{pmatrix} \] 3. Third column: \[ \begin{pmatrix} 2 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 \\ -5 \cdot 1 - 3 \cdot 1 + 3 \cdot 1 \\ 1 \cdot 1 + 0 \cdot 1 + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} 2 + 1 + 1 \\ -5 - 3 + 3 \\ 1 + 0 + 1 \end{pmatrix} = \begin{pmatrix} 4 \\ -5 \\ 2 \end{pmatrix} \] Thus, the transition matrix \( P \) is: \[ P = \frac{1}{7} \begin{pmatrix} 7 & 5 & 4 \\ -13 & -1 & -5 \\ 3 & 4 & 2 \end{pmatrix} = \begin{pmatrix} 1 & \frac{5}{7} & \frac{4}{7} \\ - \frac{13}{7} & - \frac{1}{7} & - \frac{5}{7} \\ \frac{3}{7} & \frac{4}{7} & \frac{2}{7} \end{pmatrix} \] Finally, to find \( [v]_{C} \), we use: \[ [v]_{C} = P