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Introduction Mastering determinants doesn’t just get you 4-6 marks in IIT JEE—it unlocks Cramer’s Rule for solving linear systems, area/volume calculations in coordinate geometry, and even differentiation tricks in calculus that save you 10+ minutes in the exam. One wrong sign, and your entire solution collapses. This guide ensures you never lose marks on determinants again.
(a + b)(c + d)
+
-
[[a, b], [c, d]]
ad - bc
a
d
(-1)^(i+j)
i
j
+d
(-1)^(1+1) = +1
[[1, 2], [2, 4]]
0
ax + by = e
cx + dy = f
x = D₁/D
y = D₂/D
u = f(x,y)
v = g(x,y)
|∂u/∂x ∂u/∂y|
Formula: For A = [[a, b], [c, d]], |A| = ad - bc.
A = [[a, b], [c, d]]
|A| = ad - bc
Variables: - a, b, c, d = elements of the matrix.
a, b, c, d
MEMORISE THIS (Not given in JEE sheet).
Formula: For A = [[a, b, c], [d, e, f], [g, h, i]], |A| = a(ei - fh) - b(di - fg) + c(dh - eg).
A = [[a, b, c], [d, e, f], [g, h, i]]
|A| = a(ei - fh) - b(di - fg) + c(dh - eg)
Variables: - a, b, c, ... = elements of the matrix.
a, b, c, ...
|[[1, 2], [3, 4]]| = -2
|[[3, 4], [1, 2]]| = +2
|[[1, 2], [1, 2]]| = 0
k
|[[2, 4], [1, 3]]| = 2
|[[4, 8], [1, 3]]| = 4
|[[1, 2], [4, 6]]| = -2
|[[1, 2, 3], [0, 4, 5], [0, 0, 6]]| = 1×4×6 = 24
|AB| = |A| × |B|
|A| = 2
|B| = 3
|AB| = 6
MEMORISE ALL PROPERTIES (Not given in JEE sheet).
Formula: For the system:
a₁x + b₁y + c₁z = d₁ a₂x + b₂y + c₂z = d₂ a₃x + b₃y + c₃z = d₃
Let D = |A| (determinant of coefficient matrix). Let D₁ = determinant of matrix where 1st column is replaced by [d₁, d₂, d₃]. Let D₂ = determinant of matrix where 2nd column is replaced by [d₁, d₂, d₃]. Let D₃ = determinant of matrix where 3rd column is replaced by [d₁, d₂, d₃].
D = |A|
D₁
[d₁, d₂, d₃]
D₂
D₃
Then:
x = D₁ / D, y = D₂ / D, z = D₃ / D
Condition: D ≠ 0 (otherwise, no unique solution).
D ≠ 0
Formula: If A(t) = [[a(t), b(t)], [c(t), d(t)]], then:
A(t) = [[a(t), b(t)], [c(t), d(t)]]
d/dt |A(t)| = |[a'(t), b'(t)], [c(t), d(t)]| + |[a(t), b(t)], [c'(t), d'(t)]|
For 3×3, differentiate one row at a time and sum the determinants.
Formula: If A(x) = [[a(x), b(x)], [c(x), d(x)]], then:
A(x) = [[a(x), b(x)], [c(x), d(x)]]
∫ |A(x)| dx = ∫ (a(x)d(x) - b(x)c(x)) dx
(Integrate term by term.)
(-1)^(i+j) × minor
AX = B
D
A
D = 0
D₁, D₂, D₃
B
x, y, z
z = D₃/D
t
Problem: Find |A| where A = [[1, 2, 3], [4, 5, 6], [7, 8, 9]].
|A|
A = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
Solution: 1. Expand by first row: |A| = 1 × |[5, 6], [8, 9]| - 2 × |[4, 6], [7, 9]| + 3 × |[4, 5], [7, 8]| 2. Compute 2×2 determinants: - |[5, 6], [8, 9]| = (5×9) - (6×8) = 45 - 48 = -3 - |[4, 6], [7, 9]| = (4×9) - (6×7) = 36 - 42 = -6 - |[4, 5], [7, 8]| = (4×8) - (5×7) = 32 - 35 = -3 3. Substitute back: |A| = 1(-3) - 2(-6) + 3(-3) = -3 + 12 - 9 = 0
|A| = 1 × |[5, 6], [8, 9]| - 2 × |[4, 6], [7, 9]| + 3 × |[4, 5], [7, 8]|
|[5, 6], [8, 9]| = (5×9) - (6×8) = 45 - 48 = -3
|[4, 6], [7, 9]| = (4×9) - (6×7) = 36 - 42 = -6
|[4, 5], [7, 8]| = (4×8) - (5×7) = 32 - 35 = -3
|A| = 1(-3) - 2(-6) + 3(-3) = -3 + 12 - 9 = 0
What we did and why: We expanded along the first row because it’s straightforward. The determinant is 0 because the rows are linearly dependent (R₃ = 2R₂ - R₁).
R₃ = 2R₂ - R₁
Problem: Solve using Cramer’s Rule:
2x + y = 5 x - 3y = -12
Solution: 1. Write coefficient matrix A and constant matrix B: A = [[2, 1], [1, -3]], B = [5, -12] 2. Compute D: |A| = (2 × -3) - (1 × 1) = -6 - 1 = -7 3. Compute D₁ (replace 1st column with B): D₁ = |[5, 1], [-12, -3]| = (5 × -3) - (1 × -12) = -15 + 12 = -3 4. Compute D₂ (replace 2nd column with B): D₂ = |[2, 5], [1, -12]| = (2 × -12) - (5 × 1) = -24 - 5 = -29 5. Solve for x and y: x = D₁/D = -3 / -7 = 3/7 y = D₂/D = -29 / -7 = 29/7
A = [[2, 1], [1, -3]]
B = [5, -12]
|A| = (2 × -3) - (1 × 1) = -6 - 1 = -7
D₁ = |[5, 1], [-12, -3]| = (5 × -3) - (1 × -12) = -15 + 12 = -3
D₂ = |[2, 5], [1, -12]| = (2 × -12) - (5 × 1) = -24 - 5 = -29
x
y
x = D₁/D = -3 / -7 = 3/7
y = D₂/D = -29 / -7 = 29/7
What we did and why: We used Cramer’s Rule because the system is 2×2 and D ≠ 0. Always check D first—if it’s zero, Cramer’s Rule fails.
Problem: If A(t) = [[t, t²], [1, t³]], find d/dt |A(t)|.
A(t) = [[t, t²], [1, t³]]
d/dt |A(t)|
Solution: 1. Compute |A(t)| first: |A(t)| = (t × t³) - (t² × 1) = t⁴ - t² 2. Differentiate directly (easier here): d/dt |A(t)| = 4t³ - 2t 3. Verify using row differentiation (for practice): - Differentiate 1st row: |[1, 2t], [1, t³]| = (1 × t³) - (2t × 1) = t³ - 2t - Differentiate 2nd row: |[t, t²], [0, 3t²]| = (t × 3t²) - (t² × 0) = 3t³ - Sum: (t³ - 2t) + 3t³ = 4t³ - 2t (matches direct differentiation).
|A(t)|
|A(t)| = (t × t³) - (t² × 1) = t⁴ - t²
d/dt |A(t)| = 4t³ - 2t
|[1, 2t], [1, t³]| = (1 × t³) - (2t × 1) = t³ - 2t
|[t, t²], [0, 3t²]| = (t × 3t²) - (t² × 0) = 3t³
(t³ - 2t) + 3t³ = 4t³ - 2t
What we did and why: We first expanded the determinant to simplify differentiation. The row-wise method is useful for larger matrices where expansion is tedious.
+ - +
|kA| = k|A|
|kA| = kⁿ|A|
n×n
k³|A|
R₃ = R₁ + R₂
"Listen up—this is your 60-second determinant survival guide for JEE.
You’ve got this. Go crush those determinants!
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