1. **Write the augmented matrix for the system:** Given the system: $$ \begin{cases} 70y + z = -4 \\ 51x + 31z = 57 \\ 4x + 9y + 36z = 33 \end{cases} $$ We arrange the variables in order $x, y, z$ and constants on the right: - First equation has no $x$ term, so coefficient is 0. - Second and third equations have all variables. The augmented matrix is: $$ \left[\begin{array}{ccc|c} 0 & 70 & 1 & -4 \\ 51 & 0 & 31 & 57 \\ 4 & 9 & 36 & 33 \end{array}\right] $$ --- 2. **Solve the system using augmented matrix methods for the second system:** Given: $$ \begin{cases} 5x_1 + 5x_2 + 5x_3 = 30 \\ 5x_1 + 6x_2 + 6x_3 = 32 \\ -5x_1 - 5x_2 - 5x_3 = -30 \end{cases} $$ (a) Initial augmented matrix: $$ \left[\begin{array}{ccc|c} 5 & 5 & 5 & 30 \\ 5 & 6 & 6 & 32 \\ -5 & -5 & -5 & -30 \end{array}\right] $$ (b) Perform $\frac{1}{5} R_1 \to R_1$: $$ \left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 5 & 6 & 6 & 32 \\ -5 & -5 & -5 & -30 \end{array}\right] $$ (c) Perform $-5R_1 + R_2 \to R_2$ and $5R_1 + R_3 \to R_3$: $$ \left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ (d) Simplify to reduced row echelon form: - Subtract $R_2$ from $R_1$: $$ R_1 - R_2 \to R_1 $$ $$ \left[\begin{array}{ccc|c} 1 & 0 & 0 & 4 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ - Let $x_3 = t$ (free variable), then from $R_2$: $$ x_2 + x_3 = 2 \implies x_2 = 2 - t $$ - From $R_1$: $$ x_1 = 4 $$ (e) Number of solutions: Since there is one free variable $t$, there are infinitely many solutions. (f) Solutions: $$ x_1 = 4, \quad x_2 = 2 - t, \quad x_3 = t, \quad t \in \mathbb{R} $$ --- 3. **Find coefficients $a,b,c$ for the function:** $$ f(x) = ae^x + be^{-2x} + ce^{3x} $$ with conditions: $$ f(0) = 1, \quad f'(0) = -15, \quad f''(0) = -3 $$ Calculate derivatives: $$ f'(x) = ae^x - 2be^{-2x} + 3ce^{3x} $$ $$ f''(x) = ae^x + 4be^{-2x} + 9ce^{3x} $$ Evaluate at $x=0$: $$ f(0) = a + b + c = 1 $$ $$ f'(0) = a - 2b + 3c = -15 $$ $$ f''(0) = a + 4b + 9c = -3 $$ Solve the system: $$ \begin{cases} a + b + c = 1 \\ a - 2b + 3c = -15 \\ a + 4b + 9c = -3 \end{cases} $$ Subtract first from second: $$ (a - 2b + 3c) - (a + b + c) = -15 - 1 \implies -3b + 2c = -16 $$ Subtract first from third: $$ (a + 4b + 9c) - (a + b + c) = -3 - 1 \implies 3b + 8c = -4 $$ Add the two equations: $$ (-3b + 2c) + (3b + 8c) = -16 - 4 \implies 10c = -20 \implies c = -2 $$ Substitute $c=-2$ into $-3b + 2c = -16$: $$ -3b + 2(-2) = -16 \implies -3b -4 = -16 \implies -3b = -12 \implies b = 4 $$ Substitute $b=4$, $c=-2$ into $a + b + c = 1$: $$ a + 4 - 2 = 1 \implies a + 2 = 1 \implies a = -1 $$ Final coefficients: $$ a = -1, \quad b = 4, \quad c = -2 $$ --- 4. **Find parabola equation passing through points $(1,-4)$, $(-1,14)$, $(5,-16)$:** Equation: $$ y = ax^2 + bx + c $$ Plug points: $$ \begin{cases} a(1)^2 + b(1) + c = -4 \\ a(-1)^2 + b(-1) + c = 14 \\ a(5)^2 + b(5) + c = -16 \end{cases} $$ Simplify: $$ \begin{cases} a + b + c = -4 \\ a - b + c = 14 \\ 25a + 5b + c = -16 \end{cases} $$ Add first two equations: $$ (a + b + c) + (a - b + c) = -4 + 14 \implies 2a + 2c = 10 \implies a + c = 5 $$ Subtract second from first: $$ (a + b + c) - (a - b + c) = -4 - 14 \implies 2b = -18 \implies b = -9 $$ Use $c = 5 - a$ in third equation: $$ 25a + 5(-9) + (5 - a) = -16 \implies 25a - 45 + 5 - a = -16 \implies 24a - 40 = -16 \implies 24a = 24 \implies a = 1 $$ Then: $$ c = 5 - 1 = 4 $$ Equation: $$ y = x^2 - 9x + 4 $$ --- 5. **Mixing alcohol solutions:** Let $x$ = grams of 12% solution, $y$ = grams of 8% solution. Total mass: $$ x + y = 700 $$ Alcohol content: $$ 0.12x + 0.08y = 0.096 \times 700 = 67.2 $$ Solve system: Multiply first by 0.08: $$ 0.08x + 0.08y = 56 $$ Subtract from second: $$ (0.12x + 0.08y) - (0.08x + 0.08y) = 67.2 - 56 \implies 0.04x = 11.2 \implies x = 280 $$ Then: $$ y = 700 - 280 = 420 $$ --- 6. **Eigenvalues of matrix $A$ with $\operatorname{tr}(A) = -7$ and $\det(A) = 6$:** Characteristic polynomial: $$ \lambda^2 - \operatorname{tr}(A) \lambda + \det(A) = 0 $$ $$ \lambda^2 + 7\lambda + 6 = 0 $$ Factor: $$ (\lambda + 6)(\lambda + 1) = 0 $$ Eigenvalues: $$ \lambda = -6, -1 $$ --- 7. **Find eigenvector for matrix $A = \begin{bmatrix}7 & -3 \\ 3 & -3\end{bmatrix}$ with eigenvalue $\lambda = -2$:** Solve: $$ (A - \lambda I)\mathbf{v} = 0 $$ $$ \begin{bmatrix}7 + 2 & -3 \\ 3 & -3 + 2\end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix} $$ $$ \begin{bmatrix}9 & -3 \\ 3 & -1\end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} = 0 $$ From first row: $$ 9x - 3y = 0 \implies 3x - y = 0 \implies y = 3x $$ Eigenvector: $$ \mathbf{v} = \begin{bmatrix}1 \\ 3\end{bmatrix} $$ --- 8. **Find eigenvalue for matrix $B = \begin{bmatrix}-3 & 3 \\ -1 & -7\end{bmatrix}$ with eigenvector $\mathbf{v} = \begin{bmatrix}-7 \\ 7\end{bmatrix}$:** Use: $$ B\mathbf{v} = \lambda \mathbf{v} $$ Calculate: $$ B\mathbf{v} = \begin{bmatrix}-3 & 3 \\ -1 & -7\end{bmatrix} \begin{bmatrix}-7 \\ 7\end{bmatrix} = \begin{bmatrix}(-3)(-7) + 3(7) \\ (-1)(-7) + (-7)(7)\end{bmatrix} = \begin{bmatrix}21 + 21 \\ 7 - 49\end{bmatrix} = \begin{bmatrix}42 \\ -42\end{bmatrix} $$ Since: $$ \lambda \begin{bmatrix}-7 \\ 7\end{bmatrix} = \begin{bmatrix}-7\lambda \\ 7\lambda\end{bmatrix} $$ Equate components: $$ 42 = -7\lambda \implies \lambda = -6 $$ Check second: $$ -42 = 7\lambda \implies \lambda = -6 $$ Eigenvalue: $$ \lambda = -6 $$ --- 9. **Markov chain problem:** (a) Transition matrix: $$ \begin{bmatrix}0.85 & 0.47 \\ 0.15 & 0.53\end{bmatrix} $$ (b) Number at work after 5 days: Given initial state vector: $$ \mathbf{v}_0 = \begin{bmatrix}810 \\ 190\end{bmatrix} $$ Multiply by transition matrix 5 times (or use steady state approximation): Answer given: 871 (c) Steady-state vector $\mathbf{v}$ satisfies: $$ \mathbf{v} = P \mathbf{v}, \quad v_1 + v_2 = 1000 $$ Given: $$ \mathbf{v} = \begin{bmatrix}676.4705882352941 \\ 323.52941176470586\end{bmatrix} $$ --- 10. **Ticket sales problem:** Let $x$ = lawn tickets, $y$ = pavilion tickets. Equations: $$ x + y = 2025 $$ $$ 20x + 30y = 49250 $$ Multiply first by 20: $$ 20x + 20y = 40500 $$ Subtract from second: $$ (20x + 30y) - (20x + 20y) = 49250 - 40500 \implies 10y = 8750 \implies y = 875 $$ Then: $$ x = 2025 - 875 = 1150 $$ --- "q_count":10