🏆 Daily Math Board Challenge (2026 Edition)
Check whether (x + 1)² = 2(x − 3) is a quadratic equation.
🎯 Chapter Readiness Score35%
⚡ Key Formulas at a Glance
Formula (Hover/Click)
Inverse Formula
A⁻¹ = (1 / |A|) × adj(A), provided |A| ≠ 0
Formula (Hover/Click)
Adjoint Properties
A × adj(A) = |A| I, |adj(A)| = |A|ⁿ⁻¹
Formula (Hover/Click)
Determinant Product
|AB| = |A| × |B|, |k A| = kⁿ |A| for n×n matrix
Formula (Hover/Click)
Area of Triangle
Area = ½ | x ¹(y ²−y ³) + x ²(y ³−y ¹) + x ³(y ¹−y ²) |
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📊 Chapter Weightage Analytics: High Yield Topic in Board Papers
Years Included: Year 2024, Year 2023
📅 Year 2024 Board Examination Paper
Q1NCERT Ex 4.1 Q2 • 1 Marks1 Marks
Evaluate the determinant |[x, x+1], [x-1, x]|.
▶ Show Detailed Solution
Step 1 (Expansion):
Det = (x)(x) - (x+1)(x-1) = x² - (x² - 1) = 1.
✔ Final Answer: 1
Q2NCERT Ex 4.2 Q11 • 4 Marks4 Marks
Using properties of determinants, prove that |[a-b-c, 2a, 2a], [2b, b-c-a, 2b], [2c, 2c, c-a-b]| = (a + b + c)³.
▶ Show Detailed Solution
Step 1 (R1 -> R1 + R2 + R3):
Row 1 becomes [a+b+c, a+b+c, a+b+c]. Factor out (a+b+c).
Step 2 (Columns C2 -> C2 - C1, C3 -> C3 - C1):
Expanding gives (a+b+c) · (a+b+c)² = (a+b+c)³ = RHS.
✔ Hence Proved.
Q3NCERT Ex 4.3 Q1 • 2 Marks2 Marks
Find the area of the triangle with vertices (2, 7), (1, 1), and (10, 8) using determinants.
▶ Show Detailed Solution
Step 1 (Determinant Formula):
Area = (1/2) | [ [2, 7, 1], [1, 1, 1], [10, 8, 1] ] |.
Step 2 (Calculate Value):
= (1/2) | 2(1-8) - 7(1-10) + 1(8-10) | = (1/2) | -14 + 63 - 2 | = (1/2) |47| = 47/2 sq units.
✔ Final Answer: 47/2 sq units
Q4CBSE Board 2024 • 5 Marks5 Marks
Find A⁻¹ for A = [[2, -3, 5], [3, 2, -4], [1, 1, -2]] and solve system: 2x - 3y + 5z = 11, 3x + 2y - 4z = -5, x + y - 2z = -3.
▶ Show Detailed Solution
Step 1 (Determinant):
|A| = 2(-4+4) + 3(-6+4) + 5(3-2) = -6 + 5 = -1 ≠ 0.
Step 2 (Solve X = A⁻¹B):
A⁻¹ = [[0, -1, 2], [2, -9, 23], [1, -5, 13]].
X = A⁻¹ [[11], [-5], [-3]] = [[1], [2], [3]].
✔ Final Answer: x = 1, y = 2, z = 3
Q5NCERT Ex 4.1 Q2 (Variant #5) • 1 Marks1 Marks
Evaluate the determinant |[x, x+1], [x-1, x]|.
▶ Show Detailed Solution
Step 1 (Expansion):
Det = (x)(x) - (x+1)(x-1) = x² - (x² - 1) = 1.
✔ Final Answer: 1
📅 Year 2023 Board Examination Paper
Q1NCERT Ex 4.2 Q11 (Variant #6) • 4 Marks4 Marks
Using properties of determinants, prove that |[a-b-c, 2a, 2a], [2b, b-c-a, 2b], [2c, 2c, c-a-b]| = (a + b + c)³.
▶ Show Detailed Solution
Step 1 (R1 -> R1 + R2 + R3):
Row 1 becomes [a+b+c, a+b+c, a+b+c]. Factor out (a+b+c).
Step 2 (Columns C2 -> C2 - C1, C3 -> C3 - C1):
Expanding gives (a+b+c) · (a+b+c)² = (a+b+c)³ = RHS.
✔ Hence Proved.
Q2NCERT Ex 4.3 Q1 (Variant #7) • 2 Marks2 Marks
Find the area of the triangle with vertices (2, 7), (1, 1), and (10, 8) using determinants.
▶ Show Detailed Solution
Step 1 (Determinant Formula):
Area = (1/2) | [ [2, 7, 1], [1, 1, 1], [10, 8, 1] ] |.
Step 2 (Calculate Value):
= (1/2) | 2(1-8) - 7(1-10) + 1(8-10) | = (1/2) | -14 + 63 - 2 | = (1/2) |47| = 47/2 sq units.
✔ Final Answer: 47/2 sq units
Q3CBSE Board 2024 (Variant #8) • 5 Marks5 Marks
Find A⁻¹ for A = [[2, -3, 5], [3, 2, -4], [1, 1, -2]] and solve system: 2x - 3y + 5z = 11, 3x + 2y - 4z = -5, x + y - 2z = -3.
▶ Show Detailed Solution
Step 1 (Determinant):
|A| = 2(-4+4) + 3(-6+4) + 5(3-2) = -6 + 5 = -1 ≠ 0.
Step 2 (Solve X = A⁻¹B):
A⁻¹ = [[0, -1, 2], [2, -9, 23], [1, -5, 13]].
X = A⁻¹ [[11], [-5], [-3]] = [[1], [2], [3]].
✔ Final Answer: x = 1, y = 2, z = 3
Q4NCERT Ex 4.1 Q2 (Variant #9) • 1 Marks1 Marks
Evaluate the determinant |[x, x+1], [x-1, x]|.
▶ Show Detailed Solution
Step 1 (Expansion):
Det = (x)(x) - (x+1)(x-1) = x² - (x² - 1) = 1.
✔ Final Answer: 1
Q5NCERT Ex 4.2 Q11 (Variant #10) • 4 Marks4 Marks
Using properties of determinants, prove that |[a-b-c, 2a, 2a], [2b, b-c-a, 2b], [2c, 2c, c-a-b]| = (a + b + c)³.
▶ Show Detailed Solution
Step 1 (R1 -> R1 + R2 + R3):
Row 1 becomes [a+b+c, a+b+c, a+b+c]. Factor out (a+b+c).
Step 2 (Columns C2 -> C2 - C1, C3 -> C3 - C1):
Expanding gives (a+b+c) · (a+b+c)² = (a+b+c)³ = RHS.
✔ Hence Proved.