1. **Problem:** Show that the reflection of vector $v = (1,1)$ about the y-axis produces the same result as a rotation of $v$ through 90° counterclockwise about the origin. 2. **Reflection about the y-axis:** The reflection matrix about the y-axis is $$R_y = \begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}$$ Applying this to $v$: $$R_y v = \begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix} \begin{pmatrix}1 \\ 1\end{pmatrix} = \begin{pmatrix}-1 \\ 1\end{pmatrix}$$ 3. **Rotation by 90° counterclockwise:** The rotation matrix for 90° CCW is $$R_{90} = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$$ Applying this to $v$: $$R_{90} v = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} \begin{pmatrix}1 \\ 1\end{pmatrix} = \begin{pmatrix}-1 \\ 1\end{pmatrix}$$ 4. **Conclusion:** Both transformations produce the vector $(-1,1)$, so reflection about the y-axis equals a 90° CCW rotation of $v$. 2. **Problem:** Compute the inverse of matrix $$A = \begin{pmatrix}2 & 0 & -2 \\ 0 & 3 & 1 \\ 4 & 8 & 3\end{pmatrix}$$ using the augmented matrix method. 3. **Step 1:** Form augmented matrix $[A|I]$: $$\left(\begin{array}{ccc|ccc}2 & 0 & -2 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 & 1 & 0 \\ 4 & 8 & 3 & 0 & 0 & 1\end{array}\right)$$ 4. **Step 2:** Make leading 1 in row 1: Divide row 1 by 2: $$R1 \to \frac{1}{2} R1 = (1, 0, -1, \frac{1}{2}, 0, 0)$$ 5. **Step 3:** Eliminate below pivot in column 1: $$R3 \to R3 - 4 \times R1 = (0, 8, 7, -2, 0, 1)$$ 6. **Step 4:** Make leading 1 in row 2: Divide row 2 by 3: $$R2 \to \frac{1}{3} R2 = (0, 1, \frac{1}{3}, 0, \frac{1}{3}, 0)$$ 7. **Step 5:** Eliminate above and below pivot in column 2: $$R3 \to R3 - 8 \times R2 = (0, 0, \frac{13}{3}, -2, -\frac{8}{3}, 1)$$ $$R1 \to R1 - 0 \times R2 = R1$$ (no change) 8. **Step 6:** Make leading 1 in row 3: Multiply row 3 by $\frac{3}{13}$: $$R3 \to \frac{3}{13} R3 = (0, 0, 1, -\frac{6}{13}, -\frac{8}{13}, \frac{3}{13})$$ 9. **Step 7:** Eliminate above pivot in column 3: $$R1 \to R1 + R3 = (1, 0, 0, \frac{1}{2} - \frac{6}{13}, 0 - \frac{8}{13}, 0 + \frac{3}{13}) = (1, 0, 0, \frac{13}{26} - \frac{12}{26}, -\frac{8}{13}, \frac{3}{13}) = (1, 0, 0, \frac{1}{26}, -\frac{8}{13}, \frac{3}{13})$$ $$R2 \to R2 - \frac{1}{3} R3 = (0, 1, 0, 0 + \frac{2}{13}, \frac{1}{3} + \frac{8}{39}, 0 - \frac{1}{13}) = (0, 1, 0, \frac{2}{13}, \frac{13}{39} + \frac{8}{39}, -\frac{1}{13}) = (0, 1, 0, \frac{2}{13}, \frac{21}{39}, -\frac{1}{13})$$ 10. **Final inverse matrix:** $$A^{-1} = \begin{pmatrix} \frac{1}{26} & -\frac{8}{13} & \frac{3}{13} \\ \frac{2}{13} & \frac{21}{39} & -\frac{1}{13} \\ -\frac{6}{13} & -\frac{8}{13} & \frac{3}{13} \end{pmatrix}$$ 3. **Problem:** Solve the system $$\begin{cases} 2x + y - z = 8 \\ -3x + 4y + 2z = -2 \\ x + 5y + 3z = 12 \end{cases}$$ 4. **Step 1:** Use elimination or substitution. 5. Multiply equation 3 by 2 and add to equation 2 to eliminate $z$: $$2(x + 5y + 3z) = 24 \Rightarrow 2x + 10y + 6z = 24$$ Add to equation 2: $$(-3x + 4y + 2z) + (2x + 10y + 6z) = -2 + 24$$ $$-x + 14y + 8z = 22$$ 6. From equation 1: $$z = 2x + y - 8$$ 7. Substitute $z$ into the new equation: $$-x + 14y + 8(2x + y - 8) = 22$$ $$-x + 14y + 16x + 8y - 64 = 22$$ $$15x + 22y = 86$$ 8. Substitute $z$ into equation 3: $$x + 5y + 3(2x + y - 8) = 12$$ $$x + 5y + 6x + 3y - 24 = 12$$ $$7x + 8y = 36$$ 9. Solve system: $$\begin{cases} 15x + 22y = 86 \\ 7x + 8y = 36 \end{cases}$$ Multiply second by 2: $$14x + 16y = 72$$ Subtract from first: $$(15x - 14x) + (22y - 16y) = 86 - 72$$ $$x + 6y = 14$$ 10. From $x + 6y = 14$, express $x = 14 - 6y$. Substitute into $7x + 8y = 36$: $$7(14 - 6y) + 8y = 36$$ $$98 - 42y + 8y = 36$$ $$-34y = -62$$ $$y = \frac{62}{34} = \frac{31}{17}$$ 11. Find $x$: $$x = 14 - 6 \times \frac{31}{17} = 14 - \frac{186}{17} = \frac{238}{17} - \frac{186}{17} = \frac{52}{17}$$ 12. Find $z$: $$z = 2x + y - 8 = 2 \times \frac{52}{17} + \frac{31}{17} - 8 = \frac{104}{17} + \frac{31}{17} - \frac{136}{17} = \frac{-1}{17}$$ 13. **Solution:** $$x = \frac{52}{17}, \quad y = \frac{31}{17}, \quad z = -\frac{1}{17}$$ 4. **Problem:** Perform LU decomposition of $$A = \begin{pmatrix}2 & 3 & 1 \\ 4 & 7 & 2 \\ 6 & 18 & -1\end{pmatrix}$$ 5. **Step 1:** Initialize $$L = \begin{pmatrix}1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1\end{pmatrix}, \quad U = \begin{pmatrix}u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33}\end{pmatrix}$$ 6. **Step 2:** Set first row of $U$ as first row of $A$: $$u_{11} = 2, u_{12} = 3, u_{13} = 1$$ 7. **Step 3:** Compute $l_{21} = \frac{a_{21}}{u_{11}} = \frac{4}{2} = 2$, $l_{31} = \frac{a_{31}}{u_{11}} = \frac{6}{2} = 3$ 8. **Step 4:** Compute second row of $U$: $$u_{22} = a_{22} - l_{21} u_{12} = 7 - 2 \times 3 = 1$$ $$u_{23} = a_{23} - l_{21} u_{13} = 2 - 2 \times 1 = 0$$ 9. **Step 5:** Compute $l_{32}$: $$l_{32} = \frac{a_{32} - l_{31} u_{12}}{u_{22}} = \frac{18 - 3 \times 3}{1} = 9$$ 10. **Step 6:** Compute $u_{33}$: $$u_{33} = a_{33} - l_{31} u_{13} - l_{32} u_{23} = -1 - 3 \times 1 - 9 \times 0 = -4$$ 11. **Final LU decomposition:** $$L = \begin{pmatrix}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 9 & 1\end{pmatrix}, \quad U = \begin{pmatrix}2 & 3 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & -4\end{pmatrix}$$