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MAT423 · chapter 3 · Question 3

Determinants

One number that tells you if a matrix can be inverted. 6 recipes for Q3 (≈ 17 marks).

RECIPE2×2 and 3×3 formulas

a b;c d = ad − bc
  1. 2×2: main diagonal product minus the other diagonal product. 3 2;1 -1 = 3(−1) − 1(2) = −5.
  2. 3×3 (Sarrus / diagonal method): copy columns 1 and 2 to the right. Add the 3 down-right diagonals, subtract the 3 up-right diagonals:
    |A| = aei + bfg + cdh − ceg − afh − bdi
  3. Singular ⇔ |A| = 0 ⇔ no inverse. Non-singular ⇔ |A| ≠ 0 ⇔ inverse exists.
Senang ceritaDeterminant = "area scale factor". Zero area = flat = can't undo = singular.

DRAG ITDeterminant = area (with a sign)

Try the three buttons. They are exactly the 3 row/column-operation rules of Recipe 4: swap → sign flips, × k → det × k, add a multiple → no change.

WATCH ITSarrus on a 3×3

CALCULATORdet button

ClassWizStore A in MatA, then OPTN ▼ Determinant → OPTN MatA ) =
ES PLUSSHIFT47 (det) SHIFT43 ) =
Great for checking. But "by cofactor expansion" questions need the expansion written out.

DO ITPractice

PROVE IT3 quick checks

RECIPEMinor → cofactor → |A|

  1. Minor Mij: cover row i and column j. The 2×2 determinant of what is left.
  2. Cofactor Cij = (−1)i+jMij: i + j even → keep the sign, i + j odd → change the sign. Or use the chessboard + − +;− + −;+ − +
  3. |A| by expansion along ANY row or column: |A| = ai1Ci1 + ai2Ci2 + ai3Ci3. Pick the line with the most zeros.
  4. Cofactor matrix C(A): put every Cij in its place. (Q3(c) gives C(A) with letters missing; find them with step 1–2.)
  5. Shortcut in Q3(c): once you have C(A), |A| = row 1 of A × row 1 of C(A). No new minors needed.
Senang cerita"Tutup baris, tutup lajur, kira yang tinggal, letak tanda papan catur."

TAP ITCofactor explorer (notes Example 1)

WATCH ITJul 2024 Q3(a): expand along row 1

DO ITPractice

TRAPThe middle term of row 1 has a MINUS: |A| = a(…) − b(…) + c(…). And with a negative b, −(−5) = +5.

PROVE IT3 quick checks

RECIPESpot it, save time

  1. Triangular (upper, lower, diagonal): |A| = product of the main diagonal. -5 2 3;0 4 0;0 0 3 = (−5)(4)(3) = −60.
  2. |A| = 0 immediately if A has a zero row/column, two equal rows/columns, or one row/column is a multiple of another.
  3. Elementary matrices: swap → |E| = −1 · scale by k → |E| = k · add → |E| = 1.
  4. "Find m so A is singular": write |A| in terms of m (expand along the row/column with m), set it = 0, solve.

WATCH ITJul 2025 Q3(a): find m

DO ITPractice

PROVE IT3 quick checks

RECIPEThree rules, track the effect

Swap
× (−1)
Ri ↔ Rj or Ci ↔ Cj
Factor out k
k outside
a row/column is k × something
Add k × another
no change
Ri → Ri + kRj
  1. Look at the messy determinant. Does a column contain "+ 2 × another column" (like i + 2h, f + 2e)? Subtract it away (no change).
  2. Does a whole row/column share a factor (−9, 2, −3)? Take it outside.
  3. Are rows/columns in a different order from the given one? Swap them back; each swap puts a (−1) outside.
  4. When you reach the given determinant, replace it with its value and multiply everything outside.
Senang cerita"Swap flip, Scale scale, Add nothing."

WATCH ITFeb 2026 Q3(a)

PLAY ITRow-op determinant tracker

Reduce A to triangular form with row operations. The number in front keeps |A| equal the whole time. That's how you compute |A| "by elementary row operations".

DO ITPractice (endless)

TRAPWhen you factor k out of a row, the determinant is k × (the smaller one). Don't divide by mistake.

PROVE IT3 quick checks

RECIPESplit and solve

|AB| = |A||B|  ·  |Aᵀ| = |A|  ·  |A⁻¹| = 1/|A|  ·  |Aᵏ| = |A|ᵏ  ·  |cA| = cⁿ|A|
  1. If there's a number in front OUTSIDE the bars (3|…|), divide it away first.
  2. A number INSIDE the bars comes out as cⁿ, where n = size of the matrices (2×2 → c², 3×3 → c³).
  3. Split the product into |X||Y||Z| and replace each with its value (|A⁻¹| = 1/|A|, |Bᵀ| = |B|).
  4. Solve the simple equation for |C|.
Senang ceritaThe number escapes once for every row: a 3×3 has 3 rows, so 2 comes out as 2 × 2 × 2 = 8.

PLAY ITProperty calculator

WATCH ITJul 2025 Q3(b)

DO ITPractice (endless)

TRAP #1 of the chapter|2A| for a 3×3 is 8|A|, NOT 2|A|.

PROVE IT3 quick checks

RECIPE 1Adjoint method (4 steps)

  1. All 9 cofactors → the cofactor matrix C(A). (In exams, most are given; you find the missing ones.)
  2. |A| = row 1 of A × row 1 of C(A).
  3. adj(A) = C(A)ᵀ: rows become columns.
  4. A⁻¹ = 1/|A| adj(A): divide every entry by |A|.

WATCH ITJul 2024 Q3(c)

RECIPE 2Cramer's rule

x = |A_x|/|A|,   y = |A_y|/|A|,   z = |A_z|/|A|
  1. A = coefficient matrix, b = right-hand column. Find |A| (must be ≠ 0).
  2. For the variable they ask (say y): copy A but replace the y-column (column 2) by b → Ay.
  3. y = |Ay| / |A|. Only find the one they ask for!
Senang ceritaCramer = "kick out the column of the one you want, put the answers in".

TAP ITCramer on Feb 2025 Q3(c)

DO ITPractice

TRAPadj(A) is the transpose of the cofactor matrix. Forgetting to transpose = wrong inverse.

PROVE IT3 quick checks

How to scoreQ3 ≈ 17 marks. The pattern never changes: (a) cofactor/singular OR letter determinant · (b) |…| properties · (c) cofactor matrix + adjoint inverse, or A⁻¹ by ERO + Cramer.