Engineering Mathematics · Lesson 1 of 4
Algebra, Complex Numbers, and Matrices: Quadratics, Progressions, Phasors, and Determinants
Master the algebra toolkit the REE math group leans on hardest, quadratics, progressions, logarithms, complex-number phasor arithmetic with j, and matrix and determinant methods for solving simultaneous equations.
15 min read · Super EaFree sample lesson
Mathematics carries 25% of your REE grade, and a large share of that 25% is algebra, complex numbers, and matrices, three topics that show up again and again once you reach the circuits and machines subjects too. Impedance is a complex number, mesh and nodal analysis are systems of equations you solve with determinants, and progressions and logarithms quietly resurface in engineering economy and in decibel and gain calculations. Master this lesson and you bank points here, and you also make later lessons faster.
Quadratic equations and the algebra core
Every quadratic ax^2 + bx + c = 0 solves with the quadratic formula: x = [-b +/- sqrt(b^2 - 4ac)] / (2a). Two shortcuts save time on exam day: the sum of roots is -b/a and the product of roots is c/a, so you can check or even solve some items without fully expanding the formula.
Worked example: Solve 2x^2 - 7x + 3 = 0. Here a = 2, b = -7, c = 3, so the discriminant is (-7)^2 - 4(2)(3) = 49 - 24 = 25. Then x = [7 +/- sqrt(25)] / 4 = [7 +/- 5] / 4, giving x = 3 or x = 0.5. As a check, the sum of roots -b/a = 7/2 = 3.5, and indeed 3 + 0.5 = 3.5.
Progressions and logarithms
An arithmetic progression (AP) has a constant common difference d; its nth term is a_n = a1 + (n-1)d, and the sum of the first n terms is S_n = (n/2)[2a1 + (n-1)d]. A geometric progression (GP) has a constant common ratio r; its nth term is a_n = a1 * r^(n-1), and when |r| < 1 the sum to infinity is S = a1 / (1 - r).
| Progression | nth term | Sum of n terms |
|---|---|---|
| Arithmetic | a_n = a1 + (n-1)d | S_n = (n/2)[2a1 + (n-1)d] |
| Geometric | a_n = a1 r^(n-1) | S_n = a1(1 - r^n)/(1 - r), r != 1 |
Worked example (AP): First term 7, common difference 4. The 15th term is a_15 = 7 + (14)(4) = 63. Worked example (GP): First term 4, common ratio 3. The 6th term is a_6 = 4(3^5) = 4(243) = 972.
Logarithms undo exponents: log_b(x) = y means b^y = x. Solve log and exponential equations by rewriting one side to match the other's base. Worked example: Solve log_3(x - 2) = 3. This means x - 2 = 3^3 = 27, so x = 29.
Complex numbers and phasors
Electrical engineering uses j (not i) for the imaginary unit, since i is reserved for current. By definition j = sqrt(-1), so j^2 = -1, j^3 = -j, and j^4 = 1, then the cycle repeats. A complex number written Z = R + jX (rectangular form) has magnitude |Z| = sqrt(R^2 + X^2) and angle theta = arctan(X/R) (polar form, Z = |Z| at theta).
| Operation | Easiest form |
|---|---|
| Add or subtract | Rectangular: add/subtract real and imaginary parts separately |
| Multiply or divide | Polar: multiply/divide magnitudes, add/subtract angles |
Worked example (rectangular to polar): Z = 6 + j8 ohm has |Z| = sqrt(36 + 64) = sqrt(100) = 10 ohm, and theta = arctan(8/6) = 53.13 degrees, so Z = 10 ohm at 53.13 degrees. Worked example (series impedance, rectangular): Z1 = 3 + j4 ohm and Z2 = 2 - j1 ohm in series give Z = (3+2) + j(4-1) = 5 + j3 ohm. Worked example (multiplying phasors, polar): (5 at 30 degrees)(4 at 60 degrees) = (5)(4) at (30+60) = 20 at 90 degrees.
Matrices and determinants
A determinant of a 2x2 matrix [[a, b], [c, d]] is ad - bc. For a 3x3 matrix, expand along any row or column using cofactors (each cofactor is a signed 2x2 determinant). Matrices let you solve simultaneous equations with Cramer's rule: for a1x + b1y = c1 and a2x + b2y = c2, x = Dx/D and y = Dy/D, where D is the determinant of the coefficient matrix, and Dx (or Dy) replaces that column with the constants column.
Worked example (2x2 determinant): For [[6, 2], [5, 3]], det = (6)(3) - (2)(5) = 18 - 10 = 8. Worked example (Cramer's rule): Solve 2x + 3y = 13 and 4x - y = 5. D = (2)(-1) - (3)(4) = -14. Dx = (13)(-1) - (3)(5) = -28, so x = -28/-14 = 2. Dy = (2)(5) - (13)(4) = -42, so y = -42/-14 = 3. The solution is (2, 3), and you can confirm it by substituting back into either original equation.
Exam-day strategy
- Recompute the discriminant carefully; a single sign slip in b^2 - 4ac changes both roots.
- Keep j powers straight by remembering the four-step cycle j, -1, -j, 1, then repeat; reduce any exponent mod 4.
- Add and subtract complex numbers in rectangular form, but multiply and divide in polar form; converting first saves a line of algebra.
- For Cramer's rule, compute D first. If D = 0, the system has no unique solution, a useful sanity check before you trust Dx and Dy.
- On determinant items, write out the ad - bc pattern explicitly before multiplying; rushing the cross-multiplication is the single most common source of sign errors.
- Double-check a matrix inverse by multiplying it back against the original matrix; the result should be the identity matrix.
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Lesson quiz
Check you actually have it
20 items on this lesson alone, randomized each try, with the reasoning on every answer.
Algebra, complex numbers, and matrices: quick check
Item 01 / 20 · Score 0
What is the inverse of the matrix [[4, 3], [2, 2]]?
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