Chapter 4: Determinants
Key Formulas:
• |A| = a₁₁(a₂₂a₃₃ − a₂₃a₃₂) − a₁₂(a₂₁a₃₃ − a₂₃a₃₁) + a₁₃(a₂₁a₃₂ − a₂₂a₃₁)
• |AB| = |A| |B|
• A−1 = adj(A)/|A|, if |A| ≠ 0
• Cramer's Rule: x = D₁/D, y = D₂/D, z = D₃/D
Q1. Find the determinant of A = [2 3; 1 4].
Solution
|A| = 2(4) − 3(1) = 8 − 3 = 5
Q2. Find the determinant of: [1 2 3; 4 5 6; 7 8 9]
Solution
|A| = 1(45 − 48) − 2(36 − 42) + 3(32 − 35)
= 1(−3) − 2(−6) + 3(−3)
= −3 + 12 − 9 = 0
Since |A| = 0, the matrix is singular (non-invertible).
Q3. Find values of x for which: [x+1 4; 3 x−1] has determinant equal to 0.
Solution
|A| = (x+1)(x−1) − 4(3) = x² − 1 − 12 = x² − 13 = 0
x² = 13
x = ±√13
Q4. Find the area of the triangle with vertices A(1, 1), B(4, 2), C(3, 5).
Solution
Area = ½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|
= ½ |1(2−5) + 4(5−1) + 3(1−2)|
= ½ |−3 + 16 − 3|
= ½ |10| = 5 sq. units
Q5. Find the inverse of A = [1 2; 3 5].
Solution
|A| = 1(5) − 2(3) = 5 − 6 = −1
adj(A) = [5 −2]
[−3 1]
A−1 = (1/(−1)) [5 −2] = [−5 2]
[−3 1] [3 −1]
Q6. Solve using Cramer's rule:
x + y + z = 6
x − y + 2z = 5
2x + y − z = 1
Solution
D = |1 1 1; 1 −1 2; 2 1 −1|
= 1(1−2) − 1(−1−4) + 1(1+2)
= −1 + 5 + 3 = 7
D₁ = |6 1 1; 5 −1 2; 1 1 −1| = 6(1−2) − 1(−5−2) + 1(5+1) = −6+7+6 = 7
D₂ = |1 6 1; 1 5 2; 2 1 −1| = 1(−5−2) − 6(−1−4) + 1(1−10) = −7+30−9 = 14
D₃ = |1 1 6; 1 −1 5; 2 1 1| = 1(−1−5) − 1(1−10) + 6(1+2) = −6+9+18 = 21
x = 7/7 = 1, y = 14/7 = 2, z = 21/7 = 3