Let $A = \begin{bmatrix} -1 & 2 \\ 2 & 1 \\ 1 & 2 \end{bmatrix}$ and $b = \begin{bmatrix} 5 \\ -4 \\ 4 \end{bmatrix}$.

The orthogonal projection of $b$ onto $\text{Col}(A)$ and the least-squares solution of $A\mathbf{x} = \mathbf{b}$ are:

Neetesh Kumar | October 21, 2024

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This is the solution to Math2B Course: Linear Algebra
Final Exam Question Number 26
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Step-by-step solution:

Step 1: Compute the orthogonal projection of $\mathbf{b}$ onto $\text{Col}(A)$

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To compute the orthogonal projection of $\mathbf{b}$ onto the column space of $A$, we use the formula:

$\text{Proj}_{\text{Col}(A)}(\mathbf{b}) = A(A^T A)^{-1} A^T \mathbf{b}$

First, compute $A^T A$:

$A^T A = \begin{bmatrix} -1 & 2 & 1 \\ 2 & 1 & 2 \end{bmatrix} \begin{bmatrix} -1 & 2 \\ 2 & 1 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 2 & 9 \end{bmatrix}$

Next, compute $A^T \mathbf{b}$:

$A^T \mathbf{b} = \begin{bmatrix} -1 & 2 & 1 \\ 2 & 1 & 2 \end{bmatrix} \begin{bmatrix} 5 \\ -4 \\ 4 \end{bmatrix} = \begin{bmatrix} -5 + (-8) + 4 \\ 10 + (-4) + 8 \end{bmatrix} = \begin{bmatrix} -9 \\ 14 \end{bmatrix}$

Now, compute $(A^T A)^{-1}$:

$(A^T A)^{-1} = \begin{bmatrix} 6 & 2 \\ 2 & 9 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{9}{50} & \frac{-1}{25} \\ \frac{-1}{25} & \frac{3}{25} \end{bmatrix}$

Now, compute the projection:

$\text{Proj}_{\text{Col}(A)}(\mathbf{b}) = A \begin{bmatrix} \frac{9}{50} & \frac{-1}{25} \\ \frac{-1}{25} & \frac{3}{25} \end{bmatrix} \begin{bmatrix} -9 \\ 14 \end{bmatrix} = A \begin{bmatrix} -\frac{109}{50} \\ \frac{51}{25} \end{bmatrix} = A \begin{bmatrix} -2.18 \\ 2.04 \end{bmatrix}$

Multiplying $A$ with this result:

$\text{Proj}_{\text{Col}(A)}(\mathbf{b}) = \begin{bmatrix} -1 & 2 \\ 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} -\frac{109}{50} \\ \frac{51}{25} \end{bmatrix} = \text{Proj}_{\text{Col}(A)}(\mathbf{b}) = \begin{bmatrix} \frac{313}{50} \\ -\frac{58}{25} \\ \frac{19}{10} \end{bmatrix}$

Step 2: Solve the least-squares solution $A\mathbf{x} = \mathbf{b}$

The least-squares solution can be found by solving:

$\mathbf{x} = (A^T A)^{-1} A^T \mathbf{b}$

We already computed $A^T A = \begin{bmatrix} 6 & 2 \\ 2 & 9 \end{bmatrix}$, $(A^T A)^{-1} = \begin{bmatrix} \frac{9}{50} & \frac{-1}{25} \\ \frac{-1}{25} & \frac{3}{25} \end{bmatrix}$, and $A^T \mathbf{b} = \begin{bmatrix} -9 \\ 14 \end{bmatrix}$.

Now compute:

$\mathbf{x} = \begin{bmatrix} \frac{9}{50} & \frac{-1}{25} \\ \frac{-1}{25} & \frac{3}{25} \end{bmatrix} \begin{bmatrix} -9 \\ 14 \end{bmatrix} = \begin{bmatrix} \frac{-109}{50} \\ \frac{51}{25} \end{bmatrix} = \begin{bmatrix} -2.18 \\ 2.04 \end{bmatrix}$

Final Answer:

The correct answer is Option - B

• The orthogonal projection of $\mathbf{b}$ onto $\text{Col}(A)$ is $\begin{bmatrix} \frac{313}{50} \\ -\frac{58}{25} \\ \frac{19}{10} \end{bmatrix}$
• The least-squares solution is $\mathbf{x} = \begin{bmatrix} \frac{-109}{50} \\ \frac{51}{25} \end{bmatrix}$

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