🏆 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) |
🔍
1NCERT 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
2NCERT 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.
3NCERT 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
4CBSE 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
5NCERT 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
6NCERT 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.
7NCERT 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
8CBSE 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
9NCERT 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
10NCERT 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.
11NCERT Ex 4.3 Q1 (Variant #11) • 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
12CBSE Board 2024 (Variant #12) • 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
13NCERT Ex 4.1 Q2 (Variant #13) • 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
14NCERT Ex 4.2 Q11 (Variant #14) • 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.
15NCERT Ex 4.3 Q1 (Variant #15) • 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
16CBSE Board 2024 (Variant #16) • 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
17NCERT Ex 4.1 Q2 (Variant #17) • 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
18NCERT Ex 4.2 Q11 (Variant #18) • 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.
19NCERT Ex 4.3 Q1 (Variant #19) • 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
20CBSE Board 2024 (Variant #20) • 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
21NCERT Ex 4.1 Q2 (Variant #21) • 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
22NCERT Ex 4.2 Q11 (Variant #22) • 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.
23NCERT Ex 4.3 Q1 (Variant #23) • 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
24CBSE Board 2024 (Variant #24) • 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
25NCERT Ex 4.1 Q2 (Variant #25) • 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
26NCERT Ex 4.2 Q11 (Variant #26) • 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.
27NCERT Ex 4.3 Q1 (Variant #27) • 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
28CBSE Board 2024 (Variant #28) • 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
29NCERT Ex 4.1 Q2 (Variant #29) • 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
30NCERT Ex 4.2 Q11 (Variant #30) • 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.
31NCERT Ex 4.3 Q1 (Variant #31) • 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
32CBSE Board 2024 (Variant #32) • 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
33NCERT Ex 4.1 Q2 (Variant #33) • 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
34NCERT Ex 4.2 Q11 (Variant #34) • 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.
35NCERT Ex 4.3 Q1 (Variant #35) • 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
36CBSE Board 2024 (Variant #36) • 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
37NCERT Ex 4.1 Q2 (Variant #37) • 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
38NCERT Ex 4.2 Q11 (Variant #38) • 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.
39NCERT Ex 4.3 Q1 (Variant #39) • 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
40CBSE Board 2024 (Variant #40) • 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
41NCERT Ex 4.1 Q2 (Variant #41) • 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
42NCERT Ex 4.2 Q11 (Variant #42) • 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.
43NCERT Ex 4.3 Q1 (Variant #43) • 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
44CBSE Board 2024 (Variant #44) • 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
45NCERT Ex 4.1 Q2 (Variant #45) • 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
46NCERT Ex 4.2 Q11 (Variant #46) • 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.
47NCERT Ex 4.3 Q1 (Variant #47) • 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
48CBSE Board 2024 (Variant #48) • 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
49NCERT Ex 4.1 Q2 (Variant #49) • 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
50NCERT Ex 4.2 Q11 (Variant #50) • 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.