Questions: Let A=[[2, 1], [1, 1], [2, 1]] and b=[[-3], [-3], [3]]. The QR factorization of the matrix A is given by: [[2, 1], [1, 1], [2, 1]]=[[2/3, -1/6 sqrt(2)], [1/3, 2/3 sqrt(2)], [2/3, -1/6 sqrt(2)]]*[[3, 5/3], [0, sqrt(2)/3]] (a) Applying the QR factorization to solving the least squares problem Ax=b gives the system: [[3, 5/3], [0, sqrt(2)/3]] x=[[square], [square]] (b) Use backsubstitution to solve the system in part (a) and find the least squares solution. x^=[[square], [square]]

Solution Steps

To solve the least squares problem \( A \mathbf{x} = \mathbf{b} \) using the given QR factorization, we can follow these steps:

1. Compute \( Q^T \mathbf{b} \) to transform the system into an upper triangular form.
2. Solve the resulting upper triangular system using back substitution.

Given: \[ Q = \begin{bmatrix} \frac{2}{3} & -\frac{1}{6} \sqrt{2} \\ \frac{1}{3} & \frac{2}{3} \sqrt{2} \\ \frac{2}{3} & -\frac{1}{6} \sqrt{2} \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} -3 \\ -3 \\ 3 \end{bmatrix} \]

We compute: \[ Q^T \mathbf{b} = \begin{bmatrix} \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ -\frac{1}{6} \sqrt{2} & \frac{2}{3} \sqrt{2} & -\frac{1}{6} \sqrt{2} \end{bmatrix} \begin{bmatrix} -3 \\ -3 \\ 3 \end{bmatrix} = \begin{bmatrix} -1 \\ -2.8284 \end{bmatrix} \]

Given: \[ R = \begin{bmatrix} 3 & \frac{5}{3} \\ 0 & \frac{\sqrt{2}}{3} \end{bmatrix}, \quad Q^T \mathbf{b} = \begin{bmatrix} -1 \\ -2.8284 \end{bmatrix} \]

We solve the system \( R \mathbf{x} = Q^T \mathbf{b} \):

\[ \begin{bmatrix} 3 & \frac{5}{3} \\ 0 & \frac{\sqrt{2}}{3} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} -1 \\ -2.8284 \end{bmatrix} \]

Using back substitution:

1. Solve for \( x_2 \): \[ \frac{\sqrt{2}}{3} x_2 = -2.8284 \implies x_2 = \frac{-2.8284 \cdot 3}{\sqrt{2}} = -6 \]

2. Solve for \( x_1 \): \[ 3 x_1 + \frac{5}{3} x_2 = -1 \implies 3 x_1 + \frac{5}{3} (-6) = -1 \implies 3 x_1 - 10 = -1 \implies 3 x_1 = 9 \implies x_1 = 3 \]

Final Answer