Determinants — Class 12 MCQs with Answers

Determinants — important questions & MCQs with answers (Class 12 Mathematics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   det of [[1,2],[3,4]]:

   A10B2C−2D0
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   Answer: −2

   Explanation: det = ad − bc = 1×4 − 2×3 = 4 − 6 = −2. Mixing up the diagonal order (bc − ad) gives +2, the tempting wrong-sign distractor.

2. Q2 of 6Easy

   Matrix is singular if det =

   A0B1C−1Dany
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   Answer: 0

   Explanation: A matrix is singular exactly when det(A) = 0, since then adj(A)/det(A) is undefined and no inverse exists — not det = 1 or any other value.

3. Q3 of 6Medium

   If the matrix [[x, 2], [3, 4]] is singular, then x is:

   A3/2B6C2/3D4
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   Answer: 3/2

   Explanation: Singularity requires 4x − 6 = 0, so x = 3/2.

4. Q4 of 6Medium

   Assertion (A): If A is a square matrix of order 3 with |A| = 4, then |adj A| = 16. Reason (R): If A is a square matrix of order n, then |adj A| = |A|ⁿ.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: A is true but R is false

   Explanation: A is true: |adj A| = |A|ⁿ⁻¹ = 4² = 16 for n = 3. R is false: the power is n − 1, not n.

5. Q5 of 6Hard

   If A is 3×3 with det(A) = 8, det(adj(A)) =

   A8B64C16D1
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   Answer: 64

   Explanation: For an n×n matrix, det(adj A) = (det A)ⁿ⁻¹; here n = 3, so det(adj A) = 8² = 64, not just det(A) = 8 itself.

6. Q6 of 6Hard

   Assertion (A): The matrix [[2, λ, −3], [0, 2, 5], [1, 1, 3]] is invertible for every real value of λ. Reason (R): A square matrix is invertible if and only if its determinant is non-zero.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: A is false but R is true

   Explanation: Expanding along R1: 2(6 − 5) − λ(0 − 5) − 3(0 − 2) = 8 + 5λ, which is zero when λ = −8/5. For that λ the matrix is singular, so A is false. R is true (Theorem 4).

Determinants notes: key concepts (Class 12 Mathematics)

The determinant assigns a single number to a square matrix that decides invertibility. This chapter covers 2×2 and 3×3 determinants, minors and cofactors, the properties that simplify them, the adjoint and inverse, and three applications: area of a triangle, testing collinearity, and Cramer's rule.

Determinant

A scalar attached to a square matrix; for 2×2 it is ad − bc, and a 3×3 is expanded along any row or column using cofactors.

Minor

Mᵢⱼ is the determinant of the submatrix left after deleting row i and column j of A.

Cofactor

Aᵢⱼ = (−1)^(i+j) Mᵢⱼ; the sign follows a checkerboard pattern starting with + at the top-left.

Cofactor expansion

|A| = Σ aᵢⱼ Aᵢⱼ along one row or column; the sum along one row with cofactors of another row is zero.

Properties of determinants

Swapping two rows changes the sign; identical rows give 0; multiplying a row by k multiplies the determinant by k.

Singular vs non-singular

A is singular if |A| = 0 (no inverse) and non-singular if |A| ≠ 0 (invertible).

Adjoint (adjugate)

adj A is the transpose of the cofactor matrix and satisfies A·(adj A) = (adj A)·A = |A| I.

Inverse via adjoint

A⁻¹ = (1/|A|) adj A, defined only when |A| ≠ 0; this is how a 3×3 inverse is computed.

Determinant of a product

|AB| = |A||B|, |Aᵀ| = |A|, |A⁻¹| = 1/|A|, and |kA| = kⁿ|A| for an n×n matrix.

Area of a triangle

Area = ½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|, written as a determinant of the three vertices.

Collinearity test

Three points are collinear exactly when that triangle-area determinant is zero.

Cramer's rule

For a linear system with coefficient determinant D ≠ 0, x = Dₓ/D, y = D_y/D, replacing a column by the constants.

Homogeneous systems

AX = 0 has a non-trivial solution iff |A| = 0; otherwise only the trivial X = 0 works.