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CBSE · Class 12 · Mathematics

Determinants

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
The value of −6177 is:
  1. 49
  2. −35
  3. −55
  4. −49
2.
The value of 54353−1−45−3 is:
  1. 172
  2. 164
  3. −167
  4. 167
3.
The area of the triangle with vertices (1, −6), (4, 1), (4, −1) is:
  1. 3 square units
  2. 4 square units
  3. 3/2 square units
  4. 6 square units
4.
If x−16x = 45213, then x is:
  1. ±7
  2. 6 only
  3. ±36
  4. ±6
5.
A determinant is defined only for:
  1. diagonal matrices only
  2. row matrices
  3. any matrix
  4. square matrices
6.
A is a square matrix of order 3 with |A| = −5. Then |A−1| equals:
  1. 25
  2. −5
  3. 1/5
  4. −1/5
7.
The cofactor of a23 in 021−41432−4 is:
  1. 6
  2. −5
  3. −6
  4. 24
8.
Assertion (A): The points (1, 4), (3, 5) and (−5, 1) are collinear.
Reason (R): Three points are collinear when the triangle they form has zero area.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Find adj A and A−1 for A = 5301.
10.
Find the inverse of A = 23111−100−2.
11.
Solve by the matrix method: −5x − 3y = −23, 5x + 4y = 29.
12.
Find k if the area of the triangle with vertices (k, 0), (2, 0) and (0, −5) is 5 square units.
13.
For A = −162232135, find the minors and cofactors of the elements of the first row, and hence |A|.
14.
For A = −67−2−1, verify that A (adj A) = |A| I.
15.
Use determinants to find the equation of the line through (1, 6) and (−5, 1).
16.
Find x if the matrix −186x3 is singular.
Section C — Word problems4 × 4 = 16 marks
17.
Solve by the matrix method: −2x + y + 4z = −14, 2x − 2y = −2, 3x − y + 4z = −31.
18.
Arjun buys 5 kg sugar, 2 kg potatoes, 2 kg onions for Rs 190; Priya buys 1 kg sugar, 5 kg potatoes, 1 kg onions for Rs 105; Kabir buys 5 kg sugar, 3 kg potatoes, 3 kg onions for Rs 235. Use matrices to find the price of 1 kg of each.
19.
Find A−1 for A = 33−12−112−12, and verify that A A−1 = I.
20.
For A = 40−32, show that A2 − 6A + 8I = O, and use it to find A−1.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 12 · Mathematics

Answer key

Determinants · AW-12M-DET-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) −49(−6) × 7 − 1 × 7
2(d) 167expand along the first row: 5 × (−4) + 4 × (19) + 3 × (37)
3(a) 3 square units½ |x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)| = ½ × 6
4(d) ±6x2 − (−6) = 42, so x2 = 36
5(d) square matrices—
6(d) −1/5|A−1| = 1/|A|
7(a) 6(−1)5 × 0232 = (−1)5 × (−6)
8(a) Both A and R are true, and R is the correct explanation of A.A is true: the area determinant is 0. R is true and explains A.
9adj A = 1−305; A−1 = 1/5−3/501|A| = 5; A−1 = (1/|A|) adj A
10A−1 = −13−21−23/200−1/2|A| = 2; adj A = −26−42−4300−1
11x = 1, y = 6|A| = −5; X = A−1B
12k = 0 or k = 4½ |−5(2 − k)| = 5 ⇒ |2 − k| = 2
13M11 = 9, M12 = 8, M13 = 3; A11 = 9, A12 = −8, A13 = 3; |A| = −51|A| = (−1)(9) + 6(−8) + 2(3)
14adj A = −1−72−6; A (adj A) = 200020 = 20I|A| = 20
155x − 6y = −31xy1161−511 = 0
16x = −9−54 − 6x = 0
17x = −5, y = −4, z = −5|A| = 24; adj A = −8−88−8−208412; X = (1/|A|)(adj A)B
18sugar: Rs 20 per kg; potatoes: Rs 10 per kg; onions: Rs 35 per kg|A| = 20; X = A−1B
19|A| = −9; A−1 = 1/95/9−2/92/9−8/95/90−11adj A = −1−52−28−509−9
20A2 = 160−184; A−1 = 1/403/81/2multiply A2 − 6A + 8I = O by A−1: A−1 = (1/8)(6I − A)