Let's solve the problem step by step: --- **Step 1: Find \( [\mathbf{u}]_{\mathcal{B}} \) (the coordinate vector of \(\mathbf{u}\) relative to basis \(\mathcal{B}\)).** Given: - \(\mathbf{u} = \begin{bmatrix} 3 \\ -4 \end{bmatrix}\) - \(\mathcal{B} = \left\{ \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right\}\) We want to find \( c_1, c_2 \) such that: \[ \mathbf{u} = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} \] So, \[ \begin{bmatrix} 3 \\ -4 \end{bmatrix} = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2c_1 + c_2 \\ c_1 + c_2 \end{bmatrix} \] This gives the system: \[ 2c_1 + c_2 = 3 \\ c_1 + c_2 = -4 \] Subtract the second equation from the first: \[ (2c_1 + c_2) - (c_1 + c_2) = 3 - (-4) \implies c_1 = 7 \] Substitute \(c_1 = 7\) into \(c_1 + c_2 = -4\): \[ 7 + c_2 = -4 \implies c_2 = -11 \] **Explanation:** We expressed \(\mathbf{u}\) as a linear combination of the basis vectors in \(\mathcal{B}\), and solved for the coefficients. So, \[ [\mathbf{u}]_{\mathcal{B}} = \begin{bmatrix} 7 \\ -11 \end{bmatrix} \] --- **Step 2: Find \([T]_{\mathcal{B}}^{\mathcal{C}}(P_{\mathcal{B}}(\mathbf{u}))\) using the Fundamental Theorem of Matrix Representations.** The theorem says: \[ [T]_{\mathcal{B}}^{\mathcal{C}}([\mathbf{u}]_{\mathcal{B}}) = [T(\mathbf{u})]_{\mathcal{C}} \] So, we need \(T(\mathbf{u})\) and then express it in the basis \(\mathcal{C}\). First, calculate \(T(\mathbf{u})\): \[ T\left(\begin{bmatrix} 3 \\ -4 \end{bmatrix}\right) = \begin{bmatrix} -3(3) - (-4) \\ 3(3) - 2(-4) \end{bmatrix} = \begin{bmatrix} -9 + 4 \\ 9 + 8 \end{bmatrix} = \begin{bmatrix} -5 \\ 17 \end{bmatrix} \] Now, express \(\begin{bmatrix} -5 \\ 17 \end{bmatrix}\) as a linear combination of \(\mathcal{C} = \left\{ \begin{bmatrix} 5 \\ 2 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \end{bmatrix} \right\}\): Let \(a, b\) such that \[ \begin{bmatrix} -5 \\ 17 \end{bmatrix} = a \begin{bmatrix} 5 \\ 2 \end{bmatrix} + b \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 5a + 2b \\ 2a + b \end{bmatrix} \] So, \[ 5a + 2b = -5 \\ 2a + b = 17 \] From the second equation: \(b = 17 - 2a\) Plug into the first equation: \[ 5a + 2(17 - 2a) = -5 \\ 5a + 34 - 4a = -5 \\ a = -39 \\ b = 17 - 2(-39) = 17 + 78 = 95 \] **Explanation:** We calculated \(T(\mathbf{u})\), then wrote it as a linear combination of the basis vectors in \(\mathcal{C}\) to get the coordinates. So, \[ [T]_{\mathcal{B}}^{\mathcal{C}}(P_{\mathcal{B}}(\mathbf{u})) = \begin{bmatrix} -39 \\ 95 \end{bmatrix} \] --- **Final Answer:** \[ \boxed{\begin{bmatrix} -39 \\ 95 \end{bmatrix}} \]