1. **Matrix Addition and Scalar Multiplication** Given matrices: $$A=\begin{pmatrix}2 & -3 & 5 \\ 1 & 0 & 4 \\ 7 & 3 & 6\end{pmatrix}, \quad B=\begin{pmatrix}8 & 13 & 5 \\ 9 & 10 & 6 \\ 3 & 7 & 4\end{pmatrix}$$ (a) To find $A + B$, add corresponding elements: $$A+B=\begin{pmatrix}2+8 & -3+13 & 5+5 \\ 1+9 & 0+10 & 4+6 \\ 7+3 & 3+7 & 6+4\end{pmatrix}=\begin{pmatrix}10 & 10 & 10 \\ 10 & 10 & 10 \\ 10 & 10 & 10\end{pmatrix}$$ (b) To find $4A - 2B$, multiply each element of $A$ by 4 and each element of $B$ by 2, then subtract: $$4A=\begin{pmatrix}8 & -12 & 20 \\ 4 & 0 & 16 \\ 28 & 12 & 24\end{pmatrix}, \quad 2B=\begin{pmatrix}16 & 26 & 10 \\ 18 & 20 & 12 \\ 6 & 14 & 8\end{pmatrix}$$ $$4A - 2B=\begin{pmatrix}8-16 & -12-26 & 20-10 \\ 4-18 & 0-20 & 16-12 \\ 28-6 & 12-14 & 24-8\end{pmatrix}=\begin{pmatrix}-8 & -38 & 10 \\ -14 & -20 & 4 \\ 22 & -2 & 16\end{pmatrix}$$ 2. **Inverse of Matrix $A$** The inverse $A^{-1}$ exists if $\det(A) \neq 0$. Calculate determinant: $$\det(A) = 2(0\cdot6 - 4\cdot3) - (-3)(1\cdot6 - 4\cdot7) + 5(1\cdot3 - 0\cdot7)$$ $$= 2(0 - 12) + 3(6 - 28) + 5(3 - 0) = 2(-12) + 3(-22) + 5(3) = -24 - 66 + 15 = -75$$ Since $\det(A) = -75 \neq 0$, $A$ is invertible. Use the formula: $$A^{-1} = \frac{1}{\det(A)} \mathrm{adj}(A)$$ Calculate cofactors and adjugate matrix: $$\mathrm{adj}(A) = \begin{pmatrix}0\cdot6 - 4\cdot3 & -(1\cdot6 - 4\cdot7) & 1\cdot3 - 0\cdot7 \\ -( -3\cdot6 - 5\cdot3) & 2\cdot6 - 5\cdot7 & -(2\cdot3 - (-3)\cdot7) \\ -3\cdot4 - 0\cdot5 & -(2\cdot4 - 5\cdot1) & 2\cdot0 - (-3)\cdot1 \end{pmatrix}^T$$ Calculate each element: $$= \begin{pmatrix}-12 & 22 & 3 \\ 33 & -23 & -12 \\ -12 & -3 & 3 \end{pmatrix}^T = \begin{pmatrix}-12 & 33 & -12 \\ 22 & -23 & -3 \\ 3 & -12 & 3 \end{pmatrix}$$ Therefore: $$A^{-1} = \frac{1}{-75} \begin{pmatrix}-12 & 33 & -12 \\ 22 & -23 & -3 \\ 3 & -12 & 3 \end{pmatrix} = \begin{pmatrix}\frac{12}{75} & -\frac{33}{75} & \frac{12}{75} \\ -\frac{22}{75} & \frac{23}{75} & \frac{3}{75} \\ -\frac{3}{75} & \frac{12}{75} & -\frac{3}{75} \end{pmatrix}$$ 3. **Solve system using Cramer's Rule:** Equations: $$3x + 5y = 8$$ $$-2x + 5y = -3$$ Coefficient matrix: $$D = \begin{vmatrix}3 & 5 \\ -2 & 5\end{vmatrix} = 3\times5 - (-2)\times5 = 15 + 10 = 25$$ For $x$: $$D_x = \begin{vmatrix}8 & 5 \\ -3 & 5\end{vmatrix} = 8\times5 - (-3)\times5 = 40 + 15 = 55$$ For $y$: $$D_y = \begin{vmatrix}3 & 8 \\ -2 & -3\end{vmatrix} = 3\times(-3) - (-2)\times8 = -9 + 16 = 7$$ Solutions: $$x = \frac{D_x}{D} = \frac{55}{25} = \frac{11}{5} = 2.2$$ $$y = \frac{D_y}{D} = \frac{7}{25} = 0.28$$ 4. **Vector operations** Given: $$\mathbf{a} = 4\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}, \quad \mathbf{b} = 2\mathbf{i} + 5\mathbf{j} - 4\mathbf{k}$$ (a) Cross product $\mathbf{a} \times \mathbf{b}$: $$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 3 & 2 \\ 2 & 5 & -4 \end{vmatrix} = \mathbf{i}(3\times(-4) - 2\times5) - \mathbf{j}(4\times(-4) - 2\times2) + \mathbf{k}(4\times5 - 3\times2)$$ $$= \mathbf{i}(-12 - 10) - \mathbf{j}(-16 - 4) + \mathbf{k}(20 - 6) = -22\mathbf{i} + 20\mathbf{j} + 14\mathbf{k}$$ (b) Dot product $\mathbf{a} \cdot \mathbf{b}$: $$4\times2 + 3\times5 + 2\times(-4) = 8 + 15 - 8 = 15$$ (c) Angle $\theta$ between $\mathbf{a}$ and $\mathbf{b}$: Use formula: $$\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}$$ Calculate magnitudes: $$|\mathbf{a}| = \sqrt{4^2 + 3^2 + 2^2} = \sqrt{16 + 9 + 4} = \sqrt{29}$$ $$|\mathbf{b}| = \sqrt{2^2 + 5^2 + (-4)^2} = \sqrt{4 + 25 + 16} = \sqrt{45}$$ Therefore: $$\cos \theta = \frac{15}{\sqrt{29} \times \sqrt{45}} = \frac{15}{\sqrt{1305}}$$ $$\theta = \cos^{-1}\left(\frac{15}{\sqrt{1305}}\right) \approx 57.69^\circ$$