9.8Solving Systems with Determinants
Compute 2×2 and 3×3 determinants. Apply Cramer's Rule: xᵢ = det(Aᵢ)/det(A) where Aᵢ replaces column i with the constant vector. Identify when the system has no unique solution.
Cramer's Rule gives explicit formulas for each variable in a linear system using determinants. It is theoretically important in linear algebra and provides insight into when systems have unique solutions.
Essential Question
How does Cramer's Rule use determinants to express the solution to a linear system as a ratio of determinants, and when is this method most useful?
Lesson Overview
2×2 determinant: det([[a,b],[c,d]]) = ad − bc
3×3 determinant (cofactor expansion along row 1):
det([[a,b,c],[d,e,f],[g,h,i]]) = a(ei−fh) − b(di−fg) + c(dh−eg)
Cramer's Rule for 2×2 system ax+by=e, cx+dy=f:
- D = det([[a,b],[c,d]]) = ad−bc
- Dₓ = det([[e,b],[f,d]]) — replace column 1 with constants
- Dᵧ = det([[a,e],[c,f]]) — replace column 2 with constants
- x = Dₓ/D, y = Dᵧ/D (only when D ≠ 0)
Cramer's Rule for 3×3 system Ax = b:
- D = det(A)
- Dₓ: replace column 1 with b; Dᵧ: replace column 2 with b; D_z: replace column 3 with b
- x = Dₓ/D, y = Dᵧ/D, z = D_z/D
When D = 0: the system is either inconsistent or dependent — no unique solution exists.
Advantage: gives a formula for each variable independently — useful for theoretical work and when only one variable is needed.
Worked Examples
Solve using Cramer's Rule: 3x + y = 7, 5x + 2y = 12.
D = det([[3,1],[5,2]]) = 3·2 − 1·5 = 6 − 5 = 1
Dₓ = det([[7,1],[12,2]]) = 7·2 − 1·12 = 14 − 12 = 2
Dᵧ = det([[3,7],[5,12]]) = 3·12 − 7·5 = 36 − 35 = 1
x = Dₓ/D = 2/1 = 2
y = Dᵧ/D = 1/1 = 1
Solve using Cramer's Rule: 2x − 3y = 4, x + 5y = −1.
D = det([[2,−3],[1,5]]) = 2·5 − (−3)·1 = 10 + 3 = 13
Dₓ = det([[4,−3],[−1,5]]) = 4·5 − (−3)·(−1) = 20 − 3 = 17
Dᵧ = det([[2,4],[1,−1]]) = 2·(−1) − 4·1 = −2 − 4 = −6
x = 17/13
y = −6/13
Evaluate the 3×3 determinant: det([[1,2,3],[4,5,6],[7,8,9]]).
Expand along row 1:
= 1·det([[5,6],[8,9]]) − 2·det([[4,6],[7,9]]) + 3·det([[4,5],[7,8]])
= 1·(45−48) − 2·(36−42) + 3·(32−35)
= 1·(−3) − 2·(−6) + 3·(−3)
= −3 + 12 − 9 = 0
Solve using Cramer's Rule: x+y+z=6, 2x−y+z=3, x+2y−z=2.
D = det([[1,1,1],[2,−1,1],[1,2,−1]])
= 1·(1−2) − 1·(−2−1) + 1·(4+1) = −1 + 3 + 5 = 7
Dₓ = det([[6,1,1],[3,−1,1],[2,2,−1]])
= 6·(1−2) − 1·(−3−2) + 1·(6+2) = −6 + 5 + 8 = 7
x = Dₓ/D = 7/7 = 1
Dᵧ = det([[1,6,1],[2,3,1],[1,2,−1]])
= 1·(−3−2) − 6·(−2−1) + 1·(4−3) = −5 + 18 + 1 = 14
y = Dᵧ/D = 14/7 = 2
z = 6 − x − y = 6 − 1 − 2 = 3
Determine the solution type: 4x − 2y = 6, 2x − y = 3.
D = det([[4,−2],[2,−1]]) = 4·(−1) − (−2)·2 = −4 + 4 = 0
Dₓ = det([[6,−2],[3,−1]]) = 6·(−1) − (−2)·3 = −6 + 6 = 0
Dᵧ = det([[4,6],[2,3]]) = 4·3 − 6·2 = 12 − 12 = 0
D = 0 and Dₓ = Dᵧ = 0 → dependent system
Guided Practice
Solve using Cramer's Rule: 4x + 3y = 10, 2x − y = 4.
Hint: D = det([[4,3],[2,−1]]); Dₓ = det([[10,3],[4,−1]]); Dᵧ = det([[4,10],[2,4]]).
Evaluate: det([[3,−1],[2,4]]).
Hint: Use the formula ad − bc.
Evaluate the 3×3 determinant: det([[2,1,0],[3,−1,2],[1,4,−1]]).
Hint: Expand along row 1.
Solve using Cramer's Rule: 2x + y − z = 3, x − y + 2z = 1, 3x + 2y + z = 8.
Hint: Compute D, Dₓ, Dᵧ, D_z using cofactor expansion.
A system has D = 0 and D_x = 5. What can you conclude?
Hint: What does D = 0 mean for Cramer's Rule?
Key Vocabulary
Determinant
A scalar value computed from a square matrix; det(A) ≠ 0 if and only if A is invertible.
Example: det([[2,1],[3,4]]) = 8 − 3 = 5
2×2 Determinant
det([[a,b],[c,d]]) = ad − bc (main diagonal product minus anti-diagonal product).
Example: det([[3,2],[1,4]]) = 12 − 2 = 10
3×3 Determinant
Computed by cofactor expansion along any row or column using 2×2 minors.
Cramer's Rule
xᵢ = Dᵢ/D where D = det(A) and Dᵢ is det(A) with column i replaced by the constant vector b.
Cofactor Expansion
Expanding a determinant along a row or column using 2×2 minors with alternating signs (+−+).
Minor
The determinant of the submatrix obtained by deleting one row and one column from the original matrix.
Singular Matrix
A matrix with det(A) = 0. Cramer's Rule fails — the system has no unique solution.
Quick Check Quiz
Interactive Practice — 5 Questions
det([[3,2],[1,4]]) =
In Cramer's Rule, Dₓ is formed by:
If D = 0 and D_x ≠ 0, the system is:
Solve: 2x + y = 5, x − y = 1 using Cramer's Rule. x =
The 3×3 determinant is computed using:
Common Mistakes
Replacing the wrong column when forming Dₓ, Dᵧ, D_z.
For xᵢ, replace column i (the column of coefficients for variable xᵢ) with the constant vector b.
Forgetting the alternating signs in cofactor expansion: +, −, +.
The cofactor expansion has alternating signs: a₁₁M₁₁ − a₁₂M₁₂ + a₁₃M₁₃. The sign pattern is (+−+/−+−/+−+).
Using Cramer's Rule when D = 0.
Cramer's Rule only works when D ≠ 0. If D = 0, the system has no unique solution — use elimination to determine if it's inconsistent or dependent.
Computing det([[a,b],[c,d]]) as ab − cd.
The 2×2 determinant is ad − bc (main diagonal product minus anti-diagonal product), not ab − cd.
Math Tips
Cramer's Rule is elegant for 2×2 systems and for theoretical work. For 3×3 and larger, Gaussian elimination is usually faster computationally.
When expanding a 3×3 determinant, choose the row or column with the most zeros — it minimizes computation.
The sign pattern for cofactor expansion: top-left is +, then alternates. For a 3×3: [[+,−,+],[−,+,−],[+,−,+]].
Cramer's Rule is useful when you only need ONE variable — compute just that Dᵢ and D, not all three.
If D = 0 and all Dᵢ = 0, the system is dependent. If D = 0 and any Dᵢ ≠ 0, the system is inconsistent.