1. Problem 1: Traffic Flow Rates We have a network with known and unknown traffic flow rates $x_1,x_2,x_3,x_4,x_5,x_6,x_7$. The goal is to: a. Set up linear equations from flow conservation at each junction. b. Solve for unknowns. c. Discuss closing the road from A to B (edge with $x_1$). --- **Step 1:** Write flow conservation for each node. Node A: Incoming flow = $500$ vehicles/hour. Outgoing flows are $x_1$, $x_3$ and $400$ to the bottom. So, $$500 = x_1 + x_3 + 400.$$ Simplify: $$x_1 + x_3 = 100.$$ Node B: Incoming flows are $x_1$, $x_6$ and $600$ from below. Outgoing flows are $x_2$, $x_4$, and $450$ upwards. Equation: $$x_1 + x_6 + 600 = x_2 + x_4 + 450.$$ Node at $x_2$ outgoing to 600: Outgoing $600$, incoming $x_2$ and $x_7$. So, $$x_2 + x_7 = 600.$$ Node at $x_3$ going down 350 and $x_6$ returning: Outgoing flows: $350$ and $x_6$. Incoming: $x_3$. So, $$x_3 = 350 + x_6.$$ Node at $x_4$ upward 600 and incoming $x_4$ and $x_7$: Incoming is $x_4$, and outgoing is $600$. Meanwhile $x_7$ flows into the previous node with $450$. From left to right and bottom-up: Equation: $$x_7 + 600 = x_4 + 400.$$ **Step 2:** Summarize equations: 1) $x_1 + x_3 = 100$ 2) $x_1 + x_6 + 600 = x_2 + x_4 + 450$ 3) $x_2 + x_7 = 600$ 4) $x_3 = 350 + x_6$ 5) $x_7 + 600 = x_4 + 400$ **Step 3:** Rearrange equations: From (4): $x_3 - x_6 = 350$ From (5): $x_7 - x_4 = -200$ From (2): $x_1 + x_6 - x_2 - x_4 = -150$ **Step 4:** Eliminate $x_3$ and $x_7$ using (1) and (3): - From (1): $x_3 = 100 - x_1$ - From (3): $x_7 = 600 - x_2$ Substitute into (4) and (5): (4): $100 - x_1 - x_6 = 350 ightarrow -x_1 - x_6 = 250 ightarrow x_1 + x_6 = -250$ (5): $600 - x_2 - x_4 = -200 ightarrow -x_2 - x_4 = -800 ightarrow x_2 + x_4 = 800$ **Step 5:** Now system is: 1) $x_1 + x_6 = -250$ 2) $x_2 + x_4 = 800$ 3) $x_1 + x_6 - x_2 - x_4 = -150$ Substitute (1) and (2) into (3): $$-250 - 800 = -150 ightarrow -1050 = -150,$$ which is a contradiction. This means the flows cannot satisfy all these constraints simultaneously. **Step 6:** Therefore, b. **No consistent solution exists for all unknowns** under given constraints. c. Since the system is inconsistent, closing the road from A to B (removing $x_1$) would impact flow balance and traffic cannot be maintained on other streets without adjusting inputs. 2. Problem 2: Symmetric Matrix Given matrix $$A = \begin{bmatrix} 2 & a - 2b + 2c & 2a + b + c \\ 3 & 5 & a + c \\ 0 & -2 & 7 \end{bmatrix}.$$ Step 1: A symmetric matrix satisfies $A = A^T$. Compare entries: - $A_{12} = A_{21} \Rightarrow a - 2b + 2c = 3$ - $A_{13} = A_{31} \Rightarrow 2a + b + c = 0$ - $A_{23} = A_{32} \Rightarrow a + c = -2$ Step 2: Solve the system: From $a + c = -2$: $$c = -2 - a.$$ Substitute $c$ into $2a + b + c = 0$: $$2a + b + (-2 - a) = 0 \\ a + b - 2 = 0 \\ b = 2 - a.$$ Substitute $b$ and $c$ into first equation: $$a - 2(2 - a) + 2(-2 - a) = 3 \\ a - 4 + 2a - 4 - 2a = 3 \\ (a + 2a - 2a) - 8 = 3 \\ a - 8 = 3 \\ a = 11.$$ Then: $$b = 2 - 11 = -9,$$ $$c = -2 - 11 = -13.$$ Final answer: $$a=11, b=-9, c=-13.$$ 3. Problem 3: Invertibility and inverse of matrix $$A = \begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}.$$ Step 1: Show invertible: The determinant is: $$\det(A) = \det\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \times 1 = (\cos\theta)(\cos\theta) - (-\sin\theta)(\sin\theta) = \cos^2\theta + \sin^2\theta = 1.$$ Since determinant $\neq 0$, matrix $A$ is invertible for all $\theta$. Step 2: Find inverse: The $2 \times 2$ part is a rotation matrix; its inverse is transpose. Thus, $$A^{-1} = \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}.$$