Previous year questions: Determinants

Q1 (CBSE Board 2023, 1 mark)

A square matrix A has order 3 and its determinant is 6. Work out the determinant of the adjoint of A. Answer: For order 3, the determinant of the adjoint equals the square of the determinant of A. That is 6 squared, which is 36. The answer is 36. The rule used is that the adjoint determinant is a power of the original determinant, and the power depends on the order. Writing the order and the power in one line is the whole method, and the square follows from the rule for order three. Order and power are the only two inputs needed, so the answer is one line of reasoning. The adjoint power follows directly from the order of the matrix.

Q2 (CBSE Board 2023, 1 mark)

For a 2 by 2 matrix A the determinant is 2. Evaluate the determinant of 4 times the inverse of A. Answer: For a 2 by 2 matrix, multiplying by 4 multiplies the determinant by 4 squared, which is 16. The determinant of A inverse is the reciprocal of the determinant of A, so it is 1/2. Their product is 16 times 1/2, which is 8. The answer is 8. Scaling a matrix scales its determinant by a power of the scalar, and inverting a matrix inverts its determinant. The two separate rules, one for scaling and one for inversion, multiply to give the answer. The square of 4 is 16 and the reciprocal of 2 is one half, and the two factors multiply to 8. The scaling rule and the inverse rule combine into one multiplication.

Q3 (CBSE Board 2024, 1 mark)

Compute the determinant whose rows are [3, 4, 16], [5, 3, 12] and [7, 2, 8]. Answer: Expand along the first row: 3(24 minus 24) minus 4(40 minus 84) plus 16(10 minus 21). The first term is 0, the second is minus 4 times minus 44, which is 176, and the third is 16 times minus 11, which is minus 176. The sum is 0, so the answer is 0. The second and third terms are large but opposite in sign, which is what makes the total vanish. The paired terms cancel, which is why the arithmetic can be done mentally after the first two products. Pairing the positive and negative products makes the cancellation visible without a calculator. Writing the three products in a row makes the cancellation obvious. The two large products cancel, which leaves zero.

Q4 (CBSE Board 2024, 1 mark)

Three points (2, 3), (0, 4) and (p, 2) lie on one straight line. Find p. Answer: Collinearity makes the area determinant zero. Expanding the determinant with rows [2, 3, 1], [0, 4, 1] and [p, 2, 1] along the second row gives 0 minus 4(2 minus p) plus 1(4 minus 3p), which simplifies to minus 8 plus 4p plus 4 minus 3p, that is p minus 4. Setting it to zero gives p = 4. The determinant condition for three collinear points is the single idea behind this answer. Only three entries of the determinant are non-zero, so the expansion is shorter than it looks. Substituting p equal to 4 in the determinant gives zero, which confirms collinearity. A single unknown remains after the determinant is expanded.

Q5 (CBSE Board 2025, 1 mark)

A triangle with corners (3, 0), (0, k) and (minus 3, 0) covers 9 square units, and k is positive. Find k.

Answer: Place the three points in a determinant with a third column of ones: rows [3, 0, 1], [0, k, 1] and [minus 3, 0, 1]. Expanding along the first row gives 3(k minus 0) minus 0(0 minus minus 3) plus 1(0 minus minus 3k), which is 3k plus 0 plus 3k, equal to 6k. The area is half the absolute value, so the area is 3 times the absolute value of k. With k positive and area 9, 3k equals 9 and k equals 3.

Q6 (CBSE Board 2026, 1 mark)

For an order 3 matrix A the product A times the adjoint of A equals 2026 times the identity. Find the determinant of the adjoint of A. Answer: The product of a matrix with its adjoint equals the determinant of the matrix times the identity. Comparing with 2026 times the identity gives the determinant of A equal to 2026. For order 3, the determinant of the adjoint is the square of the determinant, so it is 2026 squared. Reading the scalar from the identity product is the fastest route and needs no cofactor work. The scalar on the identity is the determinant itself, and the order fixes the power for the adjoint. The scalar in front of the identity is 2026, so the adjoint determinant is that number squared. The scalar on the identity is the determinant, and the power for the adjoint is fixed by the order.

Q7 (CBSE Board 2026, 1 mark)

A matrix A is non-singular. Pick the statement that does not hold: the adjoint is singular, the inverse of the adjoint equals the adjoint of the inverse, the determinant is non-zero, or the inverse exists. Answer: For a non-singular matrix the determinant is non-zero, so the inverse exists. The adjoint is also non-singular because its determinant is a power of the determinant of A, and the inverse of the adjoint equals the adjoint of the inverse. The first statement says the adjoint is singular, which is the false one. The adjoint of an invertible matrix stays invertible because its determinant is a non-zero power. The false statement is the only one that tries to distribute an inverse over a sum. The statement that fails is the only one that uses addition inside the inverse. The false statement is the one that spreads an inverse over a sum.

Q8 (CBSE SQP 2021-22, 1 mark)

The determinant of an order 3 square matrix is minus 4. Compute the determinant of its adjoint. Answer: For order 3 the determinant of the adjoint is the determinant of A raised to the power 2. Here that is minus 4 squared, which is 16. The answer is 16. Squaring a negative determinant makes the sign disappear, which is why the answer is positive. The square of a negative number is positive, so the sign disappears at the end. The fourth power is not involved because the rule is |A| to the power n minus 1. Squaring removes the minus sign, so the answer is positive.

Q9 (CBSE SQP 2024-25, 1 mark)

An order 3 square matrix A satisfies A times adjoint of A equal to 2025 times the identity. Find the determinant of A. Answer: The identity A times adjoint of A equals determinant times identity holds for every square matrix. Comparing the scalar factor gives the determinant of A equal to 2025. The order confirms the same value, and no square or square root is involved here. No square or square root appears here because the identity is linear in the determinant. No exponent appears because the identity is linear in the determinant. The scalar in front of the identity is exactly the determinant of the matrix. The identity is linear in the determinant, so no exponent appears.

Q10 (CBSE SQP 2023-24, 1 mark)

For invertible matrices A and B of the same order, spot the wrong statement: the inverse of AB is the inverse of B times the inverse of A, the determinant of AB is the product of the determinants, the inverse of the transpose is the transpose of the inverse, or the inverse of A plus B is the inverse of A plus the inverse of B. Answer: The first three are standard identities for invertible matrices. The fourth is false because the inverse of a sum is not the sum of the inverses, which can be checked with A and B both equal to the identity matrix. The incorrect statement is the fourth. A single pair of identity matrices is enough to disprove the false statement. A single counterexample with identity matrices is enough, and it is worth writing both sides explicitly. The right side of the false statement is finite and the left side is not, which settles it in one counterexample.