Matrices and Determinants: JEE Main PYQ, 2020–2025

Answer: (A)

A product XY is defined only when the number of columns of X equals the number of rows of Y. For AB with A of order 3 × 4, B must have 4 rows. For BA, the columns of B must equal the rows of A, which is 3. So B has 4 rows and 3 columns, and its order is 4 × 3. Checking only one product gives 4 × 4 or 3 × 3. Assuming B has the same order as A gives 3 × 4.

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: (A)

The area is ½ times the absolute value of the determinant. The sign of the determinant only shows whether the vertices are listed clockwise or anticlockwise. So Area = ½|−12| = 6 square units. Check: the base from (1,1) to (5,1) is 4 and the height is 3, so ½·4·3 = 6.

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: (B)

A homogeneous system AX = O is always consistent because X = O is a solution. The coefficient matrix here is 2 × 3, so its rank is at most 2, less than the number of unknowns (3). Hence there are 3 − rank ≥ 1 free variables and infinitely many solutions. Indeed, subtracting gives 3x + 3y + 3z = 0, i.e. x + y + z = 0, and with the first equation y + 2z = 0. Taking z = t gives y = −2t and x = t, so (t, −2t, t) is a solution for every real t.

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)

adj(A) is the transpose of the cofactor matrix, so its (1,3) entry is the cofactor C₃₁ of A. Delete row 3 and column 1 to get the rows (2, 0) and (1, 3). Their determinant is 2·3 − 0·1 = 6. The sign is (−1)^(3+1) = +1, so the entry is 6.