1. Solve the system of equations using Gauss Elimination method: Given equations: $$x - 3y - 2z = 6$$ $$2x - 4y - 3z = 8$$ $$-3x + 6y + 8 = -5$$ Rearranged third equation: $$-3x + 6y = -13$$ Augmented matrix: $$\left[\begin{array}{ccc|c}1 & -3 & -2 & 6 \\ 2 & -4 & -3 & 8 \\ -3 & 6 & 0 & -13\end{array}\right]$$ Perform row operations to get upper triangular matrix: R2 = R2 - 2*R1: $$\left[\begin{array}{ccc|c}1 & -3 & -2 & 6 \\ 0 & 2 & 1 & -4 \\ -3 & 6 & 0 & -13\end{array}\right]$$ R3 = R3 + 3*R1: $$\left[\begin{array}{ccc|c}1 & -3 & -2 & 6 \\ 0 & 2 & 1 & -4 \\ 0 & -3 & -6 & 5\end{array}\right]$$ R3 = R3 + (3/2)*R2: $$\left[\begin{array}{ccc|c}1 & -3 & -2 & 6 \\ 0 & 2 & 1 & -4 \\ 0 & 0 & -\frac{9}{2} & -1\end{array}\right]$$ Back substitution: From last row: $$-\frac{9}{2}z = -1 \Rightarrow z = \frac{2}{9}$$ From second row: $$2y + z = -4 \Rightarrow 2y + \frac{2}{9} = -4 \Rightarrow y = -\frac{19}{9}$$ From first row: $$x - 3y - 2z = 6 \Rightarrow x -3(-\frac{19}{9}) - 2(\frac{2}{9}) = 6 \Rightarrow x + \frac{57}{9} - \frac{4}{9} = 6 \Rightarrow x = 6 - \frac{53}{9} = \frac{1}{9}$$ Answer for (1): $$(x,y,z) = \left(\frac{1}{9}, -\frac{19}{9}, \frac{2}{9}\right)$$ 2. Express $(4,14,-9)$ as a linear combination of $(1,1,1),(1,-3,2),(5,-1,-4)$: Let $$a,b,c$$ be scalars: $$a(1,1,1) + b(1,-3,2) + c(5,-1,-4) = (4,14,-9)$$ Gives system: $$a + b + 5c = 4$$ $$a - 3b - c = 14$$ $$a + 2b - 4c = -9$$ Subtract first from third: $$(a+2b-4c) - (a+b+5c) = -9 -4 \Rightarrow b - 9c = -13$$ From second minus first: $$(a - 3b - c) - (a + b +5c) = 14 -4 \Rightarrow -4b -6c = 10$$ First equation: $$a = 4 - b - 5c$$ From $b - 9c = -13$: $$b = -13 + 9c$$ Substitute into $$-4b -6c = 10$$: $$-4(-13 + 9c) - 6c = 10 \Rightarrow 52 -36c -6c = 10 \Rightarrow -42c= -42 \Rightarrow c=1$$ Then $$b = -13 + 9 = -4$$ $$a = 4 - (-4) - 5(1) = 4 +4 -5 = 3$$ Answer: $$(a,b,c) = (3,-4,1)$$ 3. Reduce matrix and find rank: $$\begin{bmatrix}1 & 1 & 2 \\ 2 & 4 & 9 \\ 1 & 5 & 12\end{bmatrix}$$ R2 = R2 - 2*R1: $$\begin{bmatrix}1 & 1 & 2 \\ 0 & 2 & 5 \\ 1 & 5 & 12\end{bmatrix}$$ R3 = R3 - R1: $$\begin{bmatrix}1 & 1 & 2 \\ 0 & 2 & 5 \\ 0 & 4 & 10\end{bmatrix}$$ R3 = R3 - 2*R2: $$\begin{bmatrix}1 & 1 & 2 \\ 0 & 2 & 5 \\ 0 & 0 & 0\end{bmatrix}$$ Number of nonzero rows = 2, so rank = 2 4. Determine linear independence of vectors: $$(1,2,5), (2,5,1), (1,5,2)$$ Form matrix: $$\begin{bmatrix}1 & 2 & 1 \\ 2 & 5 & 5 \\ 5 & 1 & 2\end{bmatrix}$$ Calculate determinant: $$\det = 1(5*2 - 5*1) - 2(2*2 - 5*5) + 1(2*1 - 5*5) = 1(10-5) - 2(4 - 25) + 1(2 - 25) = 5 + 42 - 23 = 24 \neq 0$$ Vectors are linearly independent. 5. Determine if vectors form basis of $\mathbb{R}^3$: $$(1,1,1), (1,2,3), (2,-1,1)$$ Matrix: $$\begin{bmatrix}1 & 1 & 2 \\ 1 & 2 & -1 \\ 1 & 3 & 1\end{bmatrix}$$ Calculate determinant: $$1(2*1 - (-1)*3) - 1(1*1 - (-1)*1) + 2(1*3 - 2*1) = 1(2+3) - 1(1+1) + 2(3-2) = 5 - 2 + 2 = 5 \neq 0$$ Hence vectors form a basis of $\mathbb{R}^3$. 6. Show linearity of $f: \mathbb{R}^3 \to \mathbb{R}^2$ given by $f(x,y,z) = (x+y+z, 2x - 3y +4z)$ Check $f(u+v) = f(u) + f(v)$ and $f(cu) = cf(u)$ for arbitrary vectors and scalar. It follows from linearity of addition and scalar multiplication on each coordinate function. Thus $f$ is linear. 7. Find basis and dimension of image of $f: \mathbb{R}^4 \to \mathbb{R}^3$ with $$f(x,y,z,t) = (x - y + z + t, x + 2z - t, x + y + 3z - 3t)$$ Image is column space of transformation matrix: $$\begin{bmatrix}1 & -1 & 1 & 1 \\ 1 & 0 & 2 & -1 \\ 1 & 1 & 3 & -3\end{bmatrix}$$ Reduce matrix: Subtract first row from second and third: $$\begin{bmatrix}1 & -1 & 1 & 1 \\ 0 & 1 & 1 & -2 \\ 0 & 2 & 2 & -4\end{bmatrix}$$ R3 = R3 - 2*R2: $$\begin{bmatrix}1 & -1 & 1 & 1 \\ 0 & 1 & 1 & -2 \\ 0 & 0 & 0 & 0\end{bmatrix}$$ Pivot columns are 1 and 2, so basis of image consists of first two columns: $$(1,1,1), (-1,0,1)$$ Dimension = 2. 8. Show $S = \{(1,1,0,-1), (1,2,1,3), (1,1,-9,2), (16,-13,1,3)\}$ are orthogonal: Compute dot products for all pairs: All pairs have dot product 0, so vectors are orthogonal. 9. Area of triangle with vertices $A(1,2), B(3,-1), C(4,5)$ using determinant: $$\text{Area} = \frac{1}{2} |\det \begin{bmatrix} x_B - x_A & y_B - y_A \\ x_C - x_A & y_C - y_A \end{bmatrix}|$$ $$= \frac{1}{2} |\det \begin{bmatrix} 3-1 & -1-2 \\ 4-1 & 5-2 \end{bmatrix}| = \frac{1}{2} |(2)(3) - ( -3)(3)| = \frac{1}{2} |6 + 9| = \frac{15}{2} = 7.5$$ 10. Find eigenvalues and eigenvectors of matrix: $$\begin{bmatrix}2 & 2 \\ 1 & 3\end{bmatrix}$$ Characteristic polynomial: $$\det(\begin{bmatrix}2-\lambda & 2 \\ 1 & 3-\lambda\end{bmatrix}) = (2-\lambda)(3-\lambda) - 2 = \lambda^2 -5\lambda + 4=0$$ Solve: $$\lambda = \frac{5 \pm \sqrt{25 -16}}{2} = \frac{5 \pm 3}{2}$$ Eigenvalues: $$\lambda_1 = 4, \quad \lambda_2 = 1$$ For $\lambda_1=4$: $$\begin{bmatrix}-2 & 2 \\ 1 & -1\end{bmatrix} x = 0$$ Eigenvector proportional to $\begin{bmatrix}1 \\ 1\end{bmatrix}$ For $\lambda_2=1$: $$\begin{bmatrix}1 & 2 \\ 1 & 2\end{bmatrix} x = 0$$ Eigenvector proportional to $\begin{bmatrix}2 \\ -1\end{bmatrix}$ 11. Solve system using Gauss Elimination: $$x + 2y + 3z = 1$$ $$x + 3y +6z = 3$$ $$2x + 6y + 13z = 1$$ Augmented matrix: $$\begin{bmatrix}1 & 2 & 3 & 1 \\ 1 & 3 & 6 & 3 \\ 2 & 6 & 13 & 1 \end{bmatrix}$$ R2 = R2 - R1: $$\begin{bmatrix}1 & 2 & 3 & 1 \\ 0 & 1 & 3 & 2 \\ 2 & 6 & 13 & 1\end{bmatrix}$$ R3 = R3 - 2 * R1: $$\begin{bmatrix}1 & 2 & 3 & 1 \\ 0 & 1 & 3 & 2 \\ 0 & 2 & 7 & -1\end{bmatrix}$$ R3 = R3 - 2*R2: $$\begin{bmatrix}1 & 2 & 3 & 1 \\ 0 & 1 & 3 & 2 \\ 0 & 0 & 1 & -5\end{bmatrix}$$ Back substitution: $$z = -5$$ $$y + 3z = 2 \Rightarrow y - 15 = 2 \Rightarrow y = 17$$ $$x + 2y + 3z =1 \Rightarrow x + 34 - 15 = 1 \Rightarrow x = -18$$ Answer: $$(x,y,z) = (-18,17,-5)$$ 12. Express $(3,10,7)$ as linear combination of $(1,3,-2), (1,4,2), (2,8,1)$: Let $$a,b,c$$ be scalars $$a(1,3,-2) + b(1,4,2) + c(2,8,1) = (3,10,7)$$ Gives system: $$a + b + 2c = 3$$ $$3a + 4b + 8c = 10$$ $$-2a + 2b + c =7$$ Solve by substitution: From first: $$a = 3 - b - 2c$$ Plug into second: $$3(3-b-2c) + 4b + 8c = 10 \Rightarrow 9 -3b -6c +4b + 8c = 10 \Rightarrow b + 2c = 1$$ Plug into third: $$-2(3 - b - 2c) + 2b + c = 7 \Rightarrow -6 + 2b + 4c + 2b + c = 7 \Rightarrow 4b + 5c = 13$$ From $$b + 2c = 1$$, $$b = 1 - 2c$$ Substitute into $$4b + 5c = 13$$: $$4(1-2c) + 5c = 13 \Rightarrow 4 -8c +5c =13 \Rightarrow -3c = 9 \Rightarrow c = -3$$ Then $$b = 1 - 2(-3) = 1 +6 =7$$ $$a = 3 - 7 - 2(-3) = 3 -7 +6 = 2$$ Answer: $$(a,b,c) = (2,7,-3)$$ 13. Reduce matrix and find rank: $$\begin{bmatrix}1 & 2 & 1 \\ 2 & 4 & 3 \\ 3 & 6 & 4\end{bmatrix}$$ R2 = R2 - 2*R1: $$\begin{bmatrix}1 & 2 & 1 \\ 0 & 0 & 1 \\ 3 & 6 & 4\end{bmatrix}$$ R3 = R3 - 3*R1: $$\begin{bmatrix}1 & 2 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 1\end{bmatrix}$$ R3 = R3 - R2: $$\begin{bmatrix}1 & 2 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}$$ Rank = 2 14. Check linear independence of vectors: $$(1,2,3), (0,0,0), (1,5,6)$$ Since second vector is zero vector, vectors are linearly dependent.