91Ó°ÊÓ

We need to calculate \(A^TA\) and \(A^T\mathbf{b}\) for the given matrix \(A=\left(\begin{array}{rr}1 & 2 \\\ 2 & 4 \\\ -1 & -2\end{array}\right)\) and vector \(\mathbf{b}=\left(\begin{array}{l}3 \\\ 2 \\\ 1\end{array}\right)\). First, let's find \(A^T\) by transposing the given matrix \(A\): \(A^T = \left(\begin{array}{rrr}1 & 2 & -1 \\\ 2 & 4 & -2\end{array}\right)\) Now, let's compute \(A^TA\) and \(A^T\mathbf{b}\): \(A^TA = \left(\begin{array}{rr}6 & 12 \\\ 12 & 24\end{array}\right)\) \(A^T\mathbf{b} = \left(\begin{array}{r}0 \\\ 0\end{array}\right)\)

The normal equation is \(A^TA\mathbf{x}=A^T\mathbf{b}\), with \(A^TA = \left(\begin{array}{rr}6 & 12 \\\ 12 & 24\end{array}\right)\) and \(A^T\mathbf{b} = \left(\begin{array}{r}0 \\ 0\end{array}\right)\). This results in the following system of equations: \(\begin{cases} 6x_1 + 12x_2 = 0 \\ 12x_1 + 24x_2 = 0 \end{cases}\) We can notice that the second equation in the system is just a multiple of the first one, resulting in an infinite number of solutions. Therefore, the least squares solution for part (a) is \(x_1 = -2x_2\), where \(x_2\) can be any real number.

We need to calculate \(A^TA\) and \(A^T\mathbf{b}\) for the given matrix \(A=\left(\begin{array}{rrr}1 & 1 & 3 \\\ -1 & 3 & 1 \\\ 1 & 2 & 4\end{array}\right)\) and vector \(\mathbf{b}=\left(\begin{array}{r}-2 \\\ 0 \\\ 8\end{array}\right)\). First, let's find \(A^T\) by transposing the given matrix \(A\): \(A^T = \left(\begin{array}{rrr}1 & -1 & 1 \\\ 1 & 3 & 2 \\\ 3 & 1 & 4\end{array}\right)\) Now, let's compute \(A^TA\) and \(A^T\mathbf{b}\): \(A^TA = \left(\begin{array}{rrr}3 & 0 & 0 \\\ 0 & 14 & 10 \\\ 0 & 10 & 26\end{array}\right)\) \(A^T\mathbf{b} = \left(\begin{array}{r}-6 \\\ 8 \\\ 16\end{array}\right)\)

The normal equation is \(A^TA\mathbf{x}=A^T\mathbf{b}\), with \(A^TA = \left(\begin{array}{rrr}3 & 0 & 0 \\\ 0 & 14 & 10 \\\ 0 & 10 & 26\end{array}\right)\) and \(A^T\mathbf{b} = \left(\begin{array}{r}-6 \\\ 8 \\\ 16\end{array}\right)\). This results in the following system of equations: \(\begin{cases} 3x_1 = -6 \\ 14x_2 + 10x_3 = 8 \\ 10x_2 + 26x_3 = 16 \end{cases}\) We can solve this system of linear equations to find the unique solution for \(\mathbf{x}\): 1. Divide the first equation by 3: \(x_1 = -2\) 2. Multiply the second equation by 5 and subtract the third equation: \(10x_2 + 26x_3 = 40\), then \(-16x_3 = 24\), so \(x_3 = -\frac{3}{2}\) 3. Substitute \(x_3 = -\frac{3}{2}\) in the second equation and solve for \(x_2\): \(x_2 = \frac{1}{2}\) Thus, the least squares solution for part (b) is: \(\mathbf{x}=\left(\begin{array}{r}-2 \\\ \frac{1}{2} \\\ -\frac{3}{2}\end{array}\right)\).