Matrices and Determinants: JEE Advanced PYQ, 2007–2026

Answer: (C)

When Δ = 0, Cramer's rule x = Δx/Δ cannot be used. If the system had any solution (x, y, z), then Δx = xΔ = 0, and similarly Δy = Δz = 0. Here Δx = 5 ≠ 0, so no solution can exist and the system is inconsistent.

Answer: (B)

Δ = (k − 1)²(k + 2). For k ≠ 1 and k ≠ −2, Δ ≠ 0 and the solution is unique. For k = 1, all three equations reduce to x + y + z = 1, so there are infinitely many solutions. For k = −2, adding the equations gives (k + 2)(x + y + z) = 3, i.e. 0 = 3, so the system is inconsistent. The answer is k = −2.

Answer: (A)

A⁻¹ = adj A / det A, and the (i,j) entry of adj A is the cofactor C_ji. Here det A = 1(1·1 − 1·0) − 2(0·1 − 1·1) + x(0·0 − 1·1) = 3 − x. The (1,3) entry of A⁻¹ needs C₃₁ = (+1)·(2·1 − x·1) = 2 − x. So (2 − x)/(3 − x) = 2, which gives 2 − x = 6 − 2x and x = 4. Check: det A = −1 and C₃₁ = −2, so the entry is 2.

Answer: (D)

Every column adds up to 1 + 2 − 3 = 0, so adding the second and third rows to the first makes a row of zeros: the determinant is 0. (In general it equals −(a³ + b³ + c³ − 3abc), which contains the factor a + b + c.)

Answer: (A)

Since I and A commute, (I + A)⁵ = Σ C(5,r)Aʳ. For r = 0 the term is I. For every r ≥ 1, A² = A gives Aʳ = A. So (I + A)⁵ = I + [C(5,1) + C(5,2) + C(5,3) + C(5,4) + C(5,5)]A = I + (32 − 1)A = I + 31A. Hence k = 31.

Answer: (A)

Take determinants on both sides of AB = −BA. Use det(XY) = det X · det Y and det(kM) = kⁿ det M for an n × n matrix M. Here n = 3, so det(−BA) = (−1)³ det(BA) = −det B · det A. Therefore det A · det B = −det A · det B, which gives det A · det B = 0. So at least one of A and B is singular. The example A = diag(1, −1, 1), B with 1's at (1,2) and (2,1) shows that exactly one may be singular and that neither need be zero, with AB ≠ O.

Answer: (A)

A − λI has rows (4 − λ, 1) and (2, 3 − λ). Its determinant is (4 − λ)(3 − λ) − 1 · 2 = λ² − 7λ + 10. Setting it to zero gives (λ − 2)(λ − 5) = 0, so the roots are 2 and 5 and λ₁² + λ₂² = 4 + 25 = 29. Check with identities: (sum)² − 2(product) = 49 − 20 = 29. The wrong answers come from using the sum, the product or the square of the sum directly.

Answer: (A)

A symmetric 3×3 matrix is fixed by its diagonal (a11, a22, a33) and by a12, a13, a23, because a_ji = a_ij. The trace condition a11 + a22 + a33 = 2 with entries in {0,1,2} allows only the patterns {2,0,0} and {1,1,0}, each with 3 arrangements, so 6 diagonals. The off-diagonal entries are unrestricted, giving 3^3 = 27 choices. The count is 6 × 27 = 162.

Answer: (D)

The entries of adj A are cofactors, which are 2 × 2 minors of A up to sign. adj A ≠ O means at least one 2 × 2 minor is non-zero, so rank A ≥ 2. Since det A = 0, rank A < 3, so rank A = 2. Three unknowns minus rank 2 leaves one free parameter, so the solutions of AX = O form a line through the origin.

Answer: (A)

First, area ABC = ½|1(5 − 11) + 7(11 − 2) + 3(2 − 5)| = ½|−6 + 63 − 9| = 24. By the section formula D = (3, 3), E = (17/3, 7), F = (7/3, 8). Then area DEF = ½|3(7 − 8) + (17/3)(8 − 3) + (7/3)(3 − 7)| = ½|−3 + 85/3 − 28/3| = ½|−3 + 19| = 8. In general, dividing each side in ratio t : (1 − t) in the same cyclic sense gives an area ratio of 1 − 3t + 3t², which is 1/3 for t = 1/3. Hence the area is 24/3 = 8. The inner triangle is not similar to ABC, so squaring the ratio is wrong.