Let's solve the problem step by step. Given the system: \[ \begin{align*} 3x_1 + 5x_2 &= 3 \\ 3x_1 + 4x_2 &= 4 \\ 2x_1 + 3x_2 &= 5 \\ \end{align*} \] Let’s define: - \( A = \begin{bmatrix} 3 & 5 \\ 3 & 4 \\ 2 & 3 \end{bmatrix} \) - \( \mathbf{b} = \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix} \) --- ## (a) Find the projection \( \mathbf{b}_p \) of \( \mathbf{b} \) onto the column space of \( A \) The projection of \( \mathbf{b} \) onto the column space of \( A \) is: \[ \mathbf{b}_p = A \hat{\mathbf{x}} \quad \text{where} \quad \hat{\mathbf{x}} = (A^T A)^{-1} A^T \mathbf{b} \] Let's compute step by step. --- ### **Step 1: Compute \( A^T A \)** \[ A^T = \begin{bmatrix} 3 & 3 & 2 \\ 5 & 4 & 3 \end{bmatrix} \] \[ A^T A = \begin{bmatrix} 3 & 3 & 2 \\ 5 & 4 & 3 \end{bmatrix} \begin{bmatrix} 3 & 5 \\ 3 & 4 \\ 2 & 3 \end{bmatrix} \] Let’s compute each entry: - (1,1): \( 3*3 + 3*3 + 2*2 = 9 + 9 + 4 = 22 \) - (1,2): \( 3*5 + 3*4 + 2*3 = 15 + 12 + 6 = 33 \) - (2,1): (same as (1,2)): 33 - (2,2): \( 5*5 + 4*4 + 3*3 = 25 + 16 + 9 = 50 \) \[ A^T A = \begin{bmatrix} 22 & 33 \\ 33 & 50 \end{bmatrix} \] --- ### **Step 2: Compute \( A^T \mathbf{b} \)** \[ A^T \mathbf{b} = \begin{bmatrix} 3 & 3 & 2 \\ 5 & 4 & 3 \end{bmatrix} \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix} \] - First entry: \( 3*3 + 3*4 + 2*5 = 9 + 12 + 10 = 31 \) - Second entry: \( 5*3 + 4*4 + 3*5 = 15 + 16 + 15 = 46 \) \[ A^T \mathbf{b} = \begin{bmatrix} 31 \\ 46 \end{bmatrix} \] --- ### **Step 3: Solve \( (A^T A) \hat{\mathbf{x}} = A^T \mathbf{b} \)** \[ \begin{bmatrix} 22 & 33 \\ 33 & 50 \end{bmatrix} \begin{bmatrix} \hat{x}_1 \\ \hat{x}_2 \end{bmatrix} = \begin{bmatrix} 31 \\ 46 \end{bmatrix} \] Let’s use Cramer’s rule or matrix inversion. First, find the determinant: \[ \det = 22*50 - 33*33 = 1100 - 1089 = 11 \] Inverse: \[ (A^T A)^{-1} = \frac{1}{11} \begin{bmatrix} 50 & -33 \\ -33 & 22 \end{bmatrix} \] So, \[ \hat{\mathbf{x}} = (A^T A)^{-1} A^T \mathbf{b} = \frac{1}{11} \begin{bmatrix} 50 & -33 \\ -33 & 22 \end{bmatrix} \begin{bmatrix} 31 \\ 46 \end{bmatrix} \] Compute: First entry: \[ 50*31 + (-33)*46 = 1550 - 1518 = 32 \] Second entry: \[ -33*31 + 22*46 = -1023 + 1012 = -11 \] So: \[ \hat{\mathbf{x}} = \frac{1}{11} \begin{bmatrix} 32 \\ -11 \end{bmatrix} = \begin{bmatrix} \frac{32}{11} \\ -1 \end{bmatrix} \] --- ### **Step 4: Compute \( \mathbf{b}_p = A \hat{\mathbf{x}} \)** \[ A \hat{\mathbf{x}} = \begin{bmatrix} 3 & 5 \\ 3 & 4 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} \frac{32}{11} \\ -1 \end{bmatrix} \] First row: \( 3*\frac{32}{11} + 5*(-1) = \frac{96}{11} - 5 = \frac{96 - 55}{11} = \frac{41}{11} \) Second row: \( 3*\frac{32}{11} + 4*(-1) = \frac{96}{11} - 4 = \frac{96 - 44}{11} = \frac{52}{11} \) Third row: \( 2*\frac{32}{11} + 3*(-1) = \frac{64}{11} - 3 = \frac{64 - 33}{11} = \frac{31}{11} \) So, \[ \boxed{ \mathbf{b}_p = \begin{bmatrix} \frac{41}{11} \\ \frac{52}{11} \\ \frac{31}{11} \end{bmatrix} } \] --- ## (b) Compute the solution to the projected system \( A \mathbf{x} = \mathbf{b}_p \) By construction, \( \hat{\mathbf{x}} \) from above is the (least squares) solution to \( A\mathbf{x} = \mathbf{b}_p \): \[ \boxed{ \mathbf{x} = \begin{bmatrix} \frac{32}{11} \\ -1 \end{bmatrix} } \] --- ## (c) Find the projection of \( \mathbf{b} \) onto the **left null-space** of \( A \) The left null-space of \( A \) is the nullspace of \( A^T \), i.e., all \( \mathbf{y} \) such that \( A^T\mathbf{y} = 0 \). Let’s find a basis for the left null-space. Let \( \mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} \), then: \[ A^T \mathbf{y} = 0 \\ \begin{bmatrix} 3 & 3 & 2 \\ 5 & 4 & 3 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = \begin{bmatrix} 3y_1 + 3y_2 + 2y_3 \\ 5y_1 + 4y_2 + 3y_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \] So, 1. \( 3y_1 + 3y_2 + 2y_3 = 0 \) 2. \( 5y_1 + 4y_2 + 3y_3 = 0 \) Let’s solve for \( y_1 \), \( y_2 \), \( y_3 \): Let’s express \( y_2 \) and \( y_1 \) in terms of \( y_3 \): From (1): \( 3y_1 + 3y_2 + 2y_3 = 0 \implies y_2 = \frac{-3y_1 - 2y_3}{3} \) From (2): Substitute \( y_2 \): \[ 5y_1 + 4y_2 + 3y_3 = 0 \\ 5y_1 + 4\left( \frac{-3y_1 - 2y_3}{3} \right) + 3y_3 = 0 \\ 5y_1 - \frac{12y_1}{3} - \frac{

Let's carefully revisit each part of the problem with detailed steps and correct the mistake in part (c). --- Given the system: \[ \begin{cases} 3x_1 + 5x_2 = 3 \\ 3x_1 + 4x_2 = 4 \\ 2x_1 + 3x_2 = 5 \end{cases} \] we define: \[ A = \begin{bmatrix} 3 & 5 \\ 3 & 4 \\ 2 & 3 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix} \] --- ### **Part (a): Projection of \(\mathbf{b}\) onto the column space of \(A\)** The projection is given by: \[ \boxed{ \mathbf{b}_p = A \hat{\mathbf{x}}, \quad \text{where} \quad \hat{\mathbf{x}} = (A^T A)^{-1} A^T \mathbf{b} } \] **Step 1:** Compute \(A^T A\) \[ A^T = \begin{bmatrix} 3 & 3 & 2 \\ 5 & 4 & 3 \end{bmatrix} \] \[ A^T A = \begin{bmatrix} 3 & 3 & 2 \\ 5 & 4 & 3 \end{bmatrix} \begin{bmatrix} 3 & 5 \\ 3 & 4 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} (3)(3) + (3)(3) + (2)(2) & (3)(5) + (3)(4) + (2)(3) \\ (5)(3) + (4)(3) + (3)(2) & (5)(5) + (4)(4) + (3)(3) \end{bmatrix} \] Calculations: - Entry (1,1): \(9 + 9 + 4 = 22\) - Entry (1,2): \(15 + 12 + 6 = 33\) - Entry (2,1): same as (1,2): 33 - Entry (2,2): \(25 + 16 + 9 = 50\) \[ A^T A = \begin{bmatrix} 22 & 33 \\ 33 & 50 \end{bmatrix} \] **Step 2:** Compute \(A^T \mathbf{b}\) \[ A^T \mathbf{b} = \begin{bmatrix} 3 & 3 & 2 \\ 5 & 4 & 3 \end{bmatrix} \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix} \] Calculate: - First component: \(3*3 + 3*4 + 2*5 = 9 + 12 + 10 = 31\) - Second component: \(5*3 + 4*4 + 3*5 = 15 + 16 + 15 = 46\) \[ A^T \mathbf{b} = \begin{bmatrix} 31 \\ 46 \end{bmatrix} \] **Step 3:** Solve for \(\hat{\mathbf{x}}\) \[ \hat{\mathbf{x}} = (A^T A)^{-1} A^T \mathbf{b} \] Calculate the inverse: \[ \det(A^T A) = 22 \times 50 - 33 \times 33 = 1100 - 1089 = 11 \] \[ (A^T A)^{-1} = \frac{1}{11} \begin{bmatrix} 50 & -33 \\ -33 & 22 \end{bmatrix} \] Multiply: \[ \hat{\mathbf{x}} = \frac{1}{11} \begin{bmatrix} 50 & -33 \\ -33 & 22 \end{bmatrix} \begin{bmatrix} 31 \\ 46 \end{bmatrix} \] Compute numerator: - First component: \(50 \times 31 - 33 \times 46 = 1550 - 1518 = 32\) - Second component: \(-33 \times 31 + 22 \times 46 = -1023 + 1012 = -11\) Therefore: \[ \hat{\mathbf{x}} = \frac{1}{11} \begin{bmatrix

Let's carefully determine the projection of \(\mathbf{b}\) onto the left null-space of \(A\). --- ### Part (c): Projection of \(\mathbf{b}\) onto the **left null-space** of \(A\) **Recall:** - The **left null-space** of \(A\) is the null-space of \(A^T\), i.e., all vectors \(\mathbf{y}\) such that: \[ A^T \mathbf{y} = \mathbf{0} \] - The **projection** of \(\mathbf{b}\) onto this subspace is the component of \(\mathbf{b}\) lying in the null-space of \(A^T\). --- ### **Step 1: Find a basis for the null-space of \(A^T\)** Given: \[ A^T = \begin{bmatrix} 3 & 3 & 2 \\ 5 & 4 & 3 \end{bmatrix} \] We seek \(\mathbf{y} = (y_1, y_2, y_3)^T\) satisfying: \[ A^T \mathbf{y} = \mathbf{0} \implies \begin{cases} 3 y_1 + 3 y_2 + 2 y_3 = 0 \quad (1) \\ 5 y_1 + 4 y_2 + 3 y_3 = 0 \quad (2) \end{cases} \] --- ### **Step 2: Solve for \(\mathbf{y}\)** From equation (1): \[ 3 y_1 + 3 y_2 + 2 y_3 = 0 \] Express \( y_1 \): \[ 3 y_1 = -3 y_2 - 2 y_3 \Rightarrow y_1 = - y_2 - \frac{2}{3} y_3 \] Substitute into (2): \[ 5 y_1 + 4 y_2 + 3 y_3 = 0 \] \[ 5 \left( - y_2 - \frac{2}{3} y_3 \right) + 4 y_2 + 3 y_3 = 0 \] Simplify: \[ -5 y_2 - \frac{10}{3} y_3 + 4 y_2 + 3 y_3 = 0 \] Combine like terms: \[ (-5 y_2 + 4 y_2) + \left(- \frac{10}{3} y_3 + 3 y_3 \right) = 0 \] \[ - y_2 + \left( - \frac{10}{3} y_3 + \frac{9}{3} y_3 \right) = 0 \] \[ - y_2 - \frac{1}{3} y_3 = 0 \] Express \( y_2 \): \[ y_2 = - \frac{1}{3} y_3 \] Now, substitute \( y_2 \) back into the expression for \( y_1 \): \[ y_1 = - \left( - \frac{1}{3} y_3 \right) - \frac{2}{3} y_3 = \frac{1}{3} y_3 - \frac{2}{3} y_3 = - \frac{1}{3} y_3 \] Let \( y_3 = t \), a free parameter. Thus, the general solution: \[ \boxed{ \mathbf{y} = t \begin{bmatrix} - \frac{1}{3} \\ - \frac{1}{3} \\ 1 \end{bmatrix} } \] The **basis vector** for the null-space: \[ \mathbf{y}_0 = \begin{bmatrix} - \frac{1}{3} \\ - \frac{1}{3} \\ 1 \end{bmatrix} \] --- ### **Step 3: Normalize the basis vector** For simplicity, multiply by 3: \[ \mathbf{y}_0' = \begin{bmatrix} -1 \\ -1 \\ 3 \end{bmatrix} \] *Note:* The projection of \(\mathbf{b}\) onto the null space is in the direction of \(\mathbf{y}_0'\). --- ### **Step 4: Compute the projection of \(\mathbf{b}\) onto the subspace** The projection of a vector \(\mathbf{b}\) onto the subspace spanned