ilovepdf_merged (6).pdf). Content covers sections 21.1–21.x.With every square matrix we associate a number called its determinant. Determinants decide invertibility, areas/volumes in geometry, and Cramer’s rule for linear systems.
For A = [[a,b],[c,d]], det A = ad − bc. For 3×3, expand along a row/column using minors and cofactors: Cᵢⱼ = (−1)^(i+j) Mᵢⱼ.
Area of triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃) can be written using a 3×3 determinant (absolute value, factor ½).
Most exam-important points from this chapter:
Start every answer with the key definition or standard form from Determinants.
Copy the formula, then substitute values; never jump to the number alone.
Check domain, quadrant, non-zero denominators, and applicability of the theorem.
Memorise and apply: Transpose of matrix in (1) is Adj A.
Memorise and apply: Replacing the elements of A by their cofactors, we get the matrix of cofactors as
English questions from NIOS Mathematics (311) public / sample papers (Sample 2024, Apr 2024, Oct 2024, Apr 2025; board-style where scanned papers had no extractable text). Mapped exclusively to this chapter — no cross-chapter duplicates. Use Model Answer for the marking key and Explanation for working.
5 question(s) · Sources: Board-style 2×2, Board-style property, Board-style singular, Board-style |kA|, Sample / board expand 3×3
PYQ1. Evaluate the determinant |2 3 ; 1 4|.
Model Answer
1) 5
Explanation
1) |A|=ad−bc=2·4−3·1=8−3=5.
PYQ2. Show that A=[[1,2],[2,4]] is singular.
Model Answer
1) det A = 0 ⇒ singular
Explanation
1) det A=1·4−2·2=
0. A square matrix with zero determinant is singular (not invertible).
PYQ3. Without expanding, find |2 3 1; 2 3 1; 5 0 4|.
Model Answer
1) 0
Explanation
1) Row 1 equals row
2. A determinant with two identical rows is zero.
PYQ4. If A is a 3×3 matrix with |A|=6, find |2A|.
Model Answer
1) 48
Explanation
1) For n×n matrix, |kA|=kⁿ|A|. Here n=3 ⇒ |2A|=8·6=48.
PYQ5. Evaluate |1 2 3; 0 1 4; 5 6 0|.
Model Answer
1) 1
Explanation
1) Expand R1: 1·(0−24) − 2·(0−20) + 3·(0−5) = −24 + 40 − 15 = 1.
10 English exam-style problems for this chapter (NIOS Mathematics 311). Each question states the formula, gives full step-by-step working with the final answer, plus formal/easy explanations. English only.
Evaluate |2 3; 1 4|.
1) ad−bc = 2·4 − 3·1 = 8−3 = 5.
Answer: 5
The 2×2 determinant is the product of the main diagonal minus the other diagonal.
8 minus 3 is 5.
det I₂ = 1; det of singular 2×2 is 0.
Order: a d minus b c.
Evaluate |1 2 3; 0 1 4; 5 6 0| by expanding along the first row.
1) C₁₁ = +(1·0−4·6) = −24; term 1·(−24)=−24.
2) C₁₂ = −(0·0−4·5) = −(−20)=20; term 2·20=40.
3) C₁₃ = +(0·6−1·5)=−5; term 3·(−5)=−15.
4) det = −24+40−15 = 1.
Answer: 1
Laplace expansion along the first row yields 1.
Careful with cofactor signs + − + on the first row.
Expanding on a row with zeros is often faster (here middle left is 0).
C₁₂ has a minus from (−1)^(1+2).
Without expanding, find |2 3 1; 2 3 1; 5 0 4|.
1) Row1 = Row2.
2) A determinant with two identical rows is 0.
Answer: 0
If two rows (or columns) are identical, the determinant vanishes.
Two same rows → det zero immediately.
Also zero if one row is a scalar multiple of another.
No need to expand.
If |A|=6 for a 3×3 matrix A, find |2A|.
1) Each of 3 rows contributes a factor 2, so |2A| = 2³ |A| = 8·6 = 48.
Answer: 48
For an n×n matrix, |kA|=kⁿ|A|; here n=3.
Three dimensions → 2³ times det A.
|Aᵀ|=|A|; |AB|=|A||B|.
Do not use 2·6 only.
Is A=[[1,2],[2,4]] singular? Justify.
1) det A = 1·4−2·2 = 0.
2) Yes, A is singular (not invertible).
Answer: Singular (det=0)
A square matrix is singular iff its determinant is zero.
1·4−2·2=0 so rows are proportional.
Singular matrices do not have inverses.
Check det, not only looking at zeros on diagonal.
Find the area of the triangle with vertices (0,0), (4,0), (0,6).
1) Area = ½|0(0−6)+4(6−0)+0(0−0)| = ½|0+24+0|=12.
2) Alternatively right triangle legs 4 and 6: ½·4·6=12.
Answer: 12
Using the coordinate area formula (or right-triangle formula) gives 12.
Base 4 height 6 → area 12.
Absolute value ensures positive area.
Do not omit the ½.
Evaluate |[[5, 2], [3, 4]]|.
1) 5·4 − 2·3 = 20 − 6 = 14.
Answer: 14
Determinant of [[a,b],[c,d]] is ad−bc.
Main diagonal product minus other diagonal.
Sign of the second product matters.
Check each algebraic step carefully.
Show that A = [[1, 2], [2, 4]] is singular.
1) det A = 1·4 − 2·2 = 4 − 4 = 0.
2) Hence A is singular (not invertible).
Answer: det A = 0 ⇒ singular
Zero determinant characterises singular square matrices.
Second row is twice the first.
Rows are linearly dependent.
Check each algebraic step carefully.
Vectors u=<2,0> and v=<1,3> form a parallelogram. Find its area.
1) Area = |det [[2,1],[0,3]]| = |6 − 0| = 6.
Answer: 6
Absolute value of the determinant of the matrix with those columns (or rows).
Base×height effectively 2×3.
Absolute value keeps area positive.
Check each algebraic step carefully.
If det A = 3 and det B = −2, find det(AB).
1) det(AB) = 3 × (−2) = −6.
Answer: −6
Multiplicativity of the determinant.
Product of determinants.
Works for square matrices of the same order.
Check each algebraic step carefully.