Algebra | HSC - Wyatt's Notes

HSC mathematics study notes - Algebra

flowchart TD
    A[Algebra] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

The general form is $ax^2 + bx + c = 0$ where $a \neq 0$.

Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Discriminant $\Delta = b^2 - 4ac$:

• $\Delta > 0$: two distinct real roots
• $\Delta = 0$: one repeated real root
• $\Delta < 0$: no real roots

Sum and product of roots: $x_1 + x_2 = -\frac{b}{a}, \quad x_1 \cdot x_2 = \frac{c}{a}$

Arithmetic: $T_n = a + (n-1)d$, $S_n = \frac{n}{2}(2a + (n-1)d)$

Geometric: $T_n = ar^{n-1}$, $S_n = \frac{a(r^n - 1)}{r - 1}$ ($r \neq 1$)

Infinite geometric series: $S_\infty = \frac{a}{1 - r}$ when $|r| < 1$

Properties: $\log_a(MN) = \log_a M + \log_a N$ $\log_a \frac{M}{N} = \log_a M - \log_a N$ $\log_a M^n = n \log_a M$

Change of base: $\log_a b = \frac{\log_c b}{\log_c a}$

Addition: $(A + B)_{ij} = A_{ij} + B_{ij}$

Multiplication: $(AB)_{ij} = \sum_k A_{ik} B_{kj}$

Determinant (2x2): $\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc$

Inverse (2x2): $\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$

Problem: Solve $2x^2 - 7x + 3 = 0$.

Solution:

Step 1: Identify $a = 2$, $b = -7$, $c = 3$

Step 2: Calculate the discriminant: $\Delta = (-7)^2 - 4(2)(3) = 49 - 24 = 25$

Step 3: Apply the quadratic formula: $x = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4}$

Step 4: $x = 3$ or $x = \frac{1}{2}$

Answer: $x = 3$ or $x = \frac{1}{2}$

Problem: Find the sum of the first 8 terms of the geometric series $3 + 6 + 12 + \ldots$

Solution:

Step 1: Identify $a = 3$, $r = \frac{6}{3} = 2$, $n = 8$

Step 2: Apply the sum formula: $S_8 = \frac{3(2^8 - 1)}{2 - 1} = \frac{3(256 - 1)}{1} = 3 \times 255 = 765$

Answer: $S_8 = 765$

Problem: Find the inverse of $A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}$.

Solution:

Step 1: Calculate the determinant: $\det(A) = (3)(2) - (1)(5) = 6 - 5 = 1$

Step 2: Apply the inverse formula: $A^{-1} = \frac{1}{1} \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}$

Step 3: Verify: $AA^{-1} = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

Answer: $A^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}$

1. Always check the discriminant before solving quadratics
2. For geometric series, verify $|r| < 1$ before using the infinite sum formula
3. Matrix multiplication is not commutative: $AB \neq BA$ as a rule
4. When finding matrix inverses, always verify your answer

1. Solve $x^2 - 5x + 6 = 0$ using the quadratic formula
2. Find the sum of the first 10 terms of $5 + 15 + 45 + \ldots$
3. Find the inverse of $\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$

Problem: Solve $\log_2(x + 3) + \log_2(x - 1) = 5$.

Solution:

Step 1: Determine the domain. Both arguments must be positive: $x + 3 > 0 \implies x > -3$ $x - 1 > 0 \implies x > 1$

Step 2: Combine logarithms: $\log_2[(x+3)(x-1)] = 5$

Step 3: Convert to exponential form: $(x+3)(x-1) = 2^5 = 32$

Step 4: Expand and solve: $x^2 + 2x - 3 = 32$ $x^2 + 2x - 35 = 0$ $(x+7)(x-5) = 0$ $x = -7 \quad \text{or} \quad x = 5$

Step 5: Check domain: $x = -7$ fails $x > 1$, so reject. $x = 5$ satisfies $x > 1$.

Answer: $x = 5$

Common mistake: Forgetting to check the domain. Always verify that the solution satisfies the original logarithmic constraints.

Problem: The first term of an arithmetic sequence is 5 and the common difference is 3. Find the sum of the first 20 terms.

Solution:

Step 1: Use the arithmetic series formula: $S_n = \frac{n}{2}[2a + (n-1)d]$

Step 2: Substitute $n = 20$, $a = 5$, $d = 3$: $S_{20} = \frac{20}{2}[2(5) + 19(3)] = 10[10 + 57] = 10 \times 67 = 670$

Answer: $S_{20} = 670$

Common mistake: Using the wrong formula. For arithmetic series, $S_n = \frac{n}{2}(2a + (n-1)d)$ or equivalently $S_n = \frac{n}{2}(a + l)$ where $l$ is the last term.

Problem: If $A$ is a $3 \times 3$ matrix with $\det(A) = 5$, find $\det(2A)$ and $\det(A^{-1})$.

Solution:

Step 1: For a scalar multiple of an $n \times n$ matrix: $\det(kA) = k^n \det(A)$

Step 2: For $n = 3$ and $k = 2$: $\det(2A) = 2^3 \times 5 = 8 \times 5 = 40$

Step 3: For the inverse: $\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{5} = 0.2$

Answer: $\det(2A) = 40$, $\det(A^{-1}) = 0.2$

Common mistake: Writing $\det(2A) = 2\det(A)$. For an $n \times n$ matrix, $\det(kA) = k^n \det(A)$ because each of the $n$ rows is multiplied by $k$.

Problem: Solve $x^2 - 5x + 6 < 0$.

Solution:

Step 1: Factor the quadratic: $x^2 - 5x + 6 = (x - 2)(x - 3)$

Step 2: Find the roots: $x = 2$ and $x = 3$

Step 3: Since the coefficient of $x^2$ is positive, the parabola opens upward. The quadratic is negative between the roots.

Step 4: Solution: $2 < x < 3$

Answer: $x \in (2, 3)$

Common mistake: Forgetting that the inequality sign determines whether the solution is between or outside the roots. For $ax^2 + bx + c < 0$ with $a > 0$, the solution is between the roots.

Problem: Find the sum of the infinite series $1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots$

Solution:

Step 1: Identify $a = 1$, $r = \frac{1}{3}$

Step 2: Check $|r| < 1$: $\left|\frac{1}{3}\right| = \frac{1}{3} < 1$ (converges)

Step 3: Apply the infinite sum formula: $S_\infty = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}$

Answer: $S_\infty = \frac{3}{2}$

Common mistake: Using the formula when $|r| \geq 1$. The infinite geometric series only converges when $|r| < 1$.

Problem: Solve for $X$ if $\begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix} X = \begin{pmatrix} 7 & 4 \\ 17 & 10 \end{pmatrix}$.

Solution:

Step 1: Let $A = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}$. We need $X = A^{-1} B$.

Step 2: Find $A^{-1}$: $\det(A) = 2 \times 3 - 1 \times 5 = 1$ $A^{-1} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$

Step 3: Multiply: $X = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} \begin{pmatrix} 7 & 4 \\ 17 & 10 \end{pmatrix} = \begin{pmatrix} 21 - 17 & 12 - 10 \\ -35 + 34 & -20 + 20 \end{pmatrix} = \begin{pmatrix} 4 & 2 \\ -1 & 0 \end{pmatrix}$

Step 4: Verify: $AX = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix} \begin{pmatrix} 4 & 2 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 7 & 4 \\ 17 & 10 \end{pmatrix}$ (correct)

Answer: $X = \begin{pmatrix} 4 & 2 \\ -1 & 0 \end{pmatrix}$

Common mistake: For matrix equations $AX = B$, multiply both sides by $A^{-1}$ on the left: $X = A^{-1}B$, not $X = BA^{-1}$. Matrix multiplication is not commutative.

• Calculus — Differentiation and integration build on the algebraic foundations of functions and equations covered here.
• Waves — Wave equations use logarithmic and trigonometric functions that extend the algebraic techniques in this topic.
• Mechanics — Kinematic equations and force calculations apply algebraic manipulation to physical systems.
• Inorganic — Electrochemistry and thermochemistry calculations use the logarithmic and matrix skills developed in this topic.

Problem: Solve the inequality $2x^2 - 5x - 3 < 0$.

Solution:

Step 1: Factor the quadratic: $2x^2 - 5x - 3 = (2x + 1)(x - 3)$

Step 2: Find the roots: $x = -\frac{1}{2}$ and $x = 3$

Step 3: Since the coefficient of $x^2$ is positive ($2 > 0$), the parabola opens upward. The quadratic is negative between the roots.

Step 4: Solution: $-\frac{1}{2} < x < 3$

Answer: $x \in \left(-\frac{1}{2}, 3\right)$

Common mistake: Forgetting that the inequality sign determines whether the solution is between or outside the roots. For $ax^2 + bx + c < 0$ with $a > 0$, the solution is between the roots. For $ax^2 + bx + c > 0$ with $a > 0$, the solution is outside the roots.

Problem: Solve $9^x - 4 \cdot 3^x + 3 = 0$.

Solution:

Step 1: Let $u = 3^x$ (note $u > 0$). Then $9^x = (3^2)^x = (3^x)^2 = u^2$.

Step 2: Substitute: $u^2 - 4u + 3 = 0$

Step 3: Factor: $(u - 1)(u - 3) = 0$

Step 4: $u = 1$ or $u = 3$

Step 5: Back-substitute:

• $3^x = 1 \implies x = 0$
• $3^x = 3 \implies x = 1$

Step 6: Both solutions are valid (both satisfy $u > 0$).

Answer: $x = 0$ or $x = 1$

Common mistake: Forgetting to check that the solutions satisfy the original domain constraints. Always substitute back to verify.

Problem: Find the eigenvalues of $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$.

Solution:

Step 1: Set up the characteristic equation $\det(A - \lambda I) = 0$: $\det \begin{pmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{pmatrix} = 0$

Step 2: Compute the determinant: $(4 - \lambda)(3 - \lambda) - (1)(2) = 0$ $12 - 7\lambda + \lambda^2 - 2 = 0$ $\lambda^2 - 7\lambda + 10 = 0$

Step 3: Factor: $(\lambda - 5)(\lambda - 2) = 0$

Step 4: Eigenvalues: $\lambda_1 = 5$ and $\lambda_2 = 2$

Answer: The eigenvalues are $\lambda = 5$ and $\lambda = 2$

Common mistake: The characteristic equation is $\det(A - \lambda I) = 0$, not $\det(A) - \lambda = 0$. The eigenvalues are the roots of the characteristic polynomial.

1. For logarithmic equations, always check the domain after solving
2. When solving matrix equations, always verify your answer by multiplying back
3. The sum of an arithmetic series can also be written as $S_n = \frac{n}{2}(a + l)$ where $l$ is the last term
4. For geometric series with $|r| > 1$, the series diverges (no finite sum)

Quadratic equations describe parabolas, and the discriminant tells you how many times the curve crosses the x-axis — the roots are where the curve meets the ground. Sequences and series are patterns of growth: arithmetic sequences add a constant like footsteps, while geometric sequences multiply like compound interest. The sum of an infinite convergent series is a finite number approached but never quite reached, like Zeno’s paradox resolved by mathematics. Matrices are machines that transform space, and their inverses reverse that transformation.

Forgetting to check the discriminant before solving quadratics. Always compute Delta = b^2 - 4ac first. If Delta < 0, there are no real roots. Students sometimes proceed with the quadratic formula and obtain complex numbers when the problem expects real solutions or no solution.

Confusing the geometric series sum formula for finite and infinite series. The finite sum Sn = a(1 - r^n)/(1 - r) applies for any r != 1. The infinite sum S_infinity = a/(1 - r) only converges when |r| < 1. Using the infinite formula when |r| >= 1 gives incorrect results because the series diverges.

Making sign errors in matrix inverses. For a 2x2 matrix, the inverse formula swaps the diagonal elements and negates the off-diagonal elements, all divided by the determinant. Students often forget to negate the off-diagonal elements or swap the diagonal elements, giving an incorrect inverse that does not satisfy AA^-1 = I.