Happy Monday! Here's your Problem of the Week. Given the matrices: A = [2, 1 / 3, 4] and B = [1, -1 / 2, 0] (A is a 2×2 matrix with row 1: 2, 1 and row 2: 3, 4) (B is a 2×2 matrix with row 1: 1, -1 and row 2: 2, 0) Find: a) AB b) det(A) and det(B) c) A inverse d) Use A inverse to solve the system: 2x + y = 5 and 3x + 4y = 10 Rules: ✅ Show your full working ✅ Post your answer in the comments ✅ React to someone else's attempt Full worked solution posted Friday. Good luck! 💪

What a week! Here's everything we covered: ✅ Problem of the Week #7 — differential equations ✅ How to study smarter in the final weeks before the EA ✅ The binomial theorem — 3 question types mastered ✅ Mathematical induction — the 4 step structure ✅ Worked solution with common mistakes breakdown If you missed any of these — scroll back through the feed and catch up. Everything is there for you. Now I want to say something important. The EA is close. For most QLD students sitting the EA this year — you have weeks, not months. That means every study session from here counts more than it did in Term 1. But here's what I want you to remember: The content is not going to change between now and exam day. The syllabus is fixed. The topics are the same ones you have been studying all year. What CAN change between now and exam day is your confidence, your speed, and your exam technique. That's what past papers give you. That's what this community gives you. That's what showing up every day gives you. So here's your task for this weekend: Do one past paper. Time yourself. Mark it honestly. Then come back here and tell me your score and which topics you dropped marks on. I'll tell you exactly what to focus on this week. You've got this. 💪 Drop your score below when you're done 👇

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Here's the full worked solution to this week's problem! Part a) Solve dy/dx = 2xy, given y = 3 when x = 0 Separate variables: 1/y dy = 2x dx Integrate both sides: ln|y| = x squared + c y = Ae to the power of (x squared) where A = e to the power of c Apply y = 3 when x = 0: 3 = Ae to the power of 0 = A Therefore y = 3e to the power of (x squared) ✅ Part b) Solve dy/dx + 2y = 6, given y = 1 when x = 0 This is a first order linear equation. Use the integrating factor method. P(x) = 2, so integrating factor = e to the power of (2x) Multiply both sides by e to the power of (2x): d/dx [y × e to the power of (2x)] = 6e to the power of (2x) Integrate both sides: y × e to the power of (2x) = 3e to the power of (2x) + c y = 3 + ce to the power of (-2x) Apply y = 1 when x = 0: 1 = 3 + c → c = -2 Therefore y = 3 - 2e to the power of (-2x) ✅ Part c) General solution of d²y/dx² - 5(dy/dx) + 6y = 0 Write the characteristic equation: m squared - 5m + 6 = 0 (m - 2)(m - 3) = 0 m = 2 or m = 3 Two distinct real roots so the general solution is: y = Ae to the power of (2x) + Be to the power of (3x) ✅ where A and B are arbitrary constants. Common mistakes to avoid: Mistake 1 — Part a: after integrating, don't forget that e to the power of (ln|y|) = y, not ln|y|. The exponential and log cancel each other. Mistake 2 — Part b: the integrating factor method requires the equation to be in the form dy/dx + P(x)y = Q(x) first. Always check the coefficient of dy/dx is 1 before identifying P(x). Mistake 3 — Part c: two distinct real roots give y = Ae to the power of (m₁x) + Be to the power of (m₂x). A repeated root gives y = (A + Bx)e to the power of (mx). Make sure you use the right form. New Problem of the Week drops Monday! 💪

Mathematical induction is one of those topics that looks intimidating but is actually one of the most structured proofs in Specialist Maths. Once you learn the 4 step structure — every induction proof follows the same pattern. Here it is: Step 1 — Base case Show the statement is true for n = 1 (or whatever the starting value is). Substitute n = 1 into both sides and show they are equal. Always end this step with: "Therefore the statement is true for n = 1." ✅ Step 2 — Inductive hypothesis Assume the statement is true for n = k. Write out exactly what that assumption says. Always start this step with: "Assume the statement is true for n = k, that is..." Step 3 — Inductive step This is the heart of the proof. Using the assumption from Step 2, prove the statement is true for n = k + 1. The trick: start with the left hand side for n = k + 1, then use your Step 2 assumption somewhere in the working to reach the right hand side. Always end this step with: "Therefore the statement is true for n = k + 1." ✅ Step 4 — Conclusion Write a formal conclusion statement. Always write something like: "Since the statement is true for n = 1, and whenever it is true for n = k it is also true for n = k + 1, by the principle of mathematical induction the statement is true for all integers n ≥ 1." Then write the square symbol □ to signal the proof is complete. The most common mistake: Students skip the conclusion or write a vague one like "therefore it is true." The conclusion must reference ALL THREE things: ✅ The base case ✅ The inductive step ✅ The principle of mathematical induction Missing any one of these costs marks. Try this one and post your working below 👇 Prove by mathematical induction that 1 + 2 + 3 + ... + n = n(n+1)/2 for all integers n ≥ 1.