Where Determinants Are Used
Cramer's Rule
Solve a system of linear equations using ratios of determinants directly.
Area of Triangle
Find the area of a triangle given coordinates of its three vertices.
Collinearity of Points
Check if three points lie on the same line using the zero-area condition.
Consistency of Equations
Determine if a system has a unique solution, infinitely many, or none.
Cramer's Rule
For a system AX = B of n linear equations in n unknowns, if |A| β 0, then the system has a unique solution given by:
x = Dβ/D, y = Dβ/D, z = Dβ/D
where D = |A|, and Dβ, Dβ, Dβ are determinants obtained by replacing the 1st, 2nd, 3rd column respectively with the constants column B.
|aβ bβ cβ|
|aβ bβ cβ|
|dβ bβ cβ|
|dβ bβ cβ|
|aβ dβ cβ|
|aβ dβ cβ|
|aβ bβ dβ|
|aβ bβ dβ|
D = |2,1;1,β3| = (2)(β3) β (1)(1) = β6β1 = β7
Replace col 1 with [5, β4]: Dβ = |5,1;β4,β3| = (5)(β3)β(1)(β4) = β15+4 = β11
Replace col 2 with [5, β4]: Dβ = |2,5;1,β4| = (2)(β4)β(5)(1) = β8β5 = β13
x = Dβ/D = β11/β7 = 11/7
y = Dβ/D = β13/β7 = 13/7
Expand |1,1,1;2,β1,1;1,2,β1| = 1(1β2) β 1(β2β1) + 1(4+1) = β1+3+5 = 7
Replace col 1 with [6,3,2]: Dβ = 6(1β2) β 1(β3β2) + 1(6+2) = β6+5+8 = 7
Replace col 2 with [6,3,2]: Dβ = 1(β3β2) β 6(β2β1) + 1(4β3) = β5+18+1 = 14
Replace col 3 with [6,3,2]: Dβ = 1(β2β6) β 1(4β3) + 6(4+1) = β8β1+30 = 21
x = 7/7 = 1, y = 14/7 = 2, z = 21/7 = 3
Area of a Triangle
If a triangle has vertices at (xβ, yβ), (xβ, yβ), and (xβ, yβ), its area can be found using a determinant.
Area = (1/2) Β· |det([[xβ, yβ, 1],[xβ, yβ, 1],[xβ, yβ, 1]])|
The determinant can be negative depending on vertex ordering. We take the absolute value and then multiply by 1/2 to ensure area is positive.
Set up: Ξ = (1/2)|det([[1,0,1],[4,3,1],[2,5,1]])|
Expand along R1: 1Β·|3,1;5,1| β 0Β·|4,1;2,1| + 1Β·|4,3;2,5|
= 1Β·(3β5) β 0 + 1Β·(20β6) = β2 + 14 = 12
Area = (1/2)|12| = 6 sq. units
Collinearity of Three Points
Three points (xβ,yβ), (xβ,yβ), (xβ,yβ) are collinear (lie on the same line) if and only if:
det([[xβ,yβ,1],[xβ,yβ,1],[xβ,yβ,1]]) = 0
This is because if they're collinear, the triangle formed has zero area.
det = 1(3β5) β 1(3β5) + 1(15β15) = 1(β2) β 1(β2) + 0 = β2+2 = 0
Determinant = 0 β Points are collinear. (They all lie on y = x.)
Consistency of a System of Equations
For a system AX = B, the nature of solutions depends on the determinant of A:
|A| β 0 β Unique Solution
System is consistent with exactly one solution. Use Cramer's rule or find Aβ»ΒΉΒ·B.
|A| = 0, (adj A)Β·B = O β Infinitely Many Solutions
System is consistent but has infinite solutions (dependent equations).
|A| = 0, (adj A)Β·B β O β No Solution
System is inconsistent. The equations are contradictory with no solution.
| |A| | (adj A)Β·B | Type | Solutions |
|---|---|---|---|
| β 0 | any | Consistent | Unique β |
| = 0 | = O (zero matrix) | Consistent | Infinitely many |
| = 0 | β O | Inconsistent | No solution β |
A = [[1,1],[2,2]], |A| = 2β2 = 0
adj(A) = [[2,β1],[β2,1]]
(adj A)Β·B = [[2,β1],[β2,1]]Β·[3,7] = [6β7, β6+7] = [β1, 1] β O
Since |A| = 0 and (adj A)Β·B β O β System is inconsistent (no solution)
Chapter Summary β All Key Formulae
|3Γ3 matrix| = aββAββ + aββAββ + aββAββ (row expansion)
Cofactor Aα΅’β±Ό = (β1)^(i+j) Γ Mα΅’β±Ό
adj(A) = Transpose of cofactor matrix
Aβ»ΒΉ = (1/|A|) Γ adj(A) [when |A| β 0]
A Β· adj(A) = |A| Β· I
Area of β³ = (1/2)|det([[xβ,yβ,1],[xβ,yβ,1],[xβ,yβ,1]])|
Collinear βΊ above determinant = 0
Cramer's Rule: x = Dβ/D, y = Dβ/D, z = Dβ/D
|kA| = kβΏ|A| for nΓn matrix
|AB| = |A|Β·|B|
Exam Tips: Always (1) check if |A| β 0 before finding inverse or unique solution. (2) Use row operations to simplify determinants. (3) In Cramer's rule, replace the correct column with constant terms. (4) Take absolute value when computing area.