JEE Advanced Previous Year Questions (2018 - 2023): Matrices and Determinants | Mathematics (Maths) for JEE Main & Advanced PDF Download

2023

Q1: Let .
Then the number of invertible matrices in R is :   [JEE Advanced 2023 Paper 2]
Ans: 3780

Q2: Let α,β and γ be real numbers. Consider the following system of linear equations
x + 2y + z = 7
x + αz = 11
2x − 3y + βz = γ
Match each entry in List-I to the correct entries in List-II.              [JEE Advanced 2023 Paper 1]

For unique solution Δ ≠ 0
For infinite solution
Δ = Δx = Δy = Δz = 0
For no solution Δ=0 and atleast one in Δx, Δy, Δz is non zero.

∴ Infinite solution

⇒ No solution.]

∴ x = 11, y = -2, z = 0 is the solution.

Q3: Let M = (a_ij),i,j ∈ {1, 2, 3}, be the 3 × 3 matrix such that a_ij = 1 if j + 1 is divisible by i, otherwise  a_ij = 0. Then which of the following statements is(are) true?                   [JEE Advanced 2023 Paper 2]
(a) M is invertible
(b) There exists a nonzero column matrix  such that
(c) The set {X∈ 𝕽³: MX = 0} ≠ {0}, where 0 =
(d) The matrix (M −2I) is invertible, where I is the 3 × 3 identity matrix
Ans: (b) & (c)

|M| = −1 + 1 = 0 ⇒M is singular so non-invertible ⇒ [A] is wrong. 

infinite solutions exists [B] is correct. 

Option (D) ;

 is wrong

Option (C);

∴ Infinite solution
Option (C) is correct

2022

Q1: Let β be a real number. Consider the matrix

If   is a singular matrix, then the value of 9β is _________.     [JEE Advanced 2022 Paper 2]
Ans: 3
 is a singular matrix. So determinant of this matrix equal to zero.

Now given,

∴ |A| = 2 - 3 = -1

= -4

We get |A| ≠ 0 and |A + I| ≠ 0

∴ |A|⁵|A − βI| |A + I| = 0 is possible only when |A − βI| = 0

= 2 - 3 - 3β
∴ 2 - 3 +3β
⇒ 3β = 1
⇒ 9β = 3

Q2: If , then which of the  following matrices is equal to M²⁰²²?    [JEE Advanced 2022 Paper 2]
(a)

(b)
(c)

(d)
Ans: (a)

and so on

Now,

=

∴ Option (A) is correct

Q3: Let p,q,r be nonzero real numbers that are, respectively, the 10th ,100th  and 1000th terms of a harmonic progression. Consider the system of linear equations
x + y + z = 1
10x + 100y + 1000z = 0
qrx + pry + pqz = 0

2021

Q1: For any 3 × 3 matrix M, let |M| denote the determinant of M. Let I be the 3 × 3 identity matrix. Let E and F be two 3 × 3 matrices such that (I − EF) is invertible. If G = (I − EF)⁻¹, then which of the following statements is (are) TRUE?           [JEE Advanced 2021 Paper 1]
(a) | FE | = | I − FE| | FGE |
(b) (I − FE)(I + FGE) = I
(c) EFG = GEF
(d) (I − FE)(I − FGE) = I
Ans: (a), (b) & (c)
∵ I − EF = G⁻¹ 
⇒ G − GEF = I ..... (i)
and G − EFG = I ..... (ii)
Clearly, GEF = EFG → option (c) is correct.
Also, (I − FE) (I + FGE)
= I − FE + FGE − FEFGE
= I − FE + FGE − F(G − I) E
= I − FE + FGE − FGE + FE
= I → option (b) is correct but option (d) is incorrect.
∵ (I − FE) (I − FGE) = I − FE − FGE + F(G − I) E
= I − 2FE
Now, (I − FE) (− FGE) = − FE
⇒ | I − FE | | FGE | = | FE |
→ option (a) is correct.

Q2: For any 3 × 3 matrix M, let | M | denote the determinant of M. Let

If Q is a nonsingular matrix of order 3 × 3, then which of the following statements is(are) TRUE?
(a) F = PEP and  
(b) | EQ + PFQ⁻¹ | = | EQ | + | PFQ⁻¹ |
(c) | (EF)³ | > | EF |²
(d) Sum of the diagonal entries of P⁻¹EP + F is equal to the sum of diagonal entries of E + P⁻¹FP   [JEE Advanced 2021 Paper 1 ]
Ans: (a), (b) & (d)
For Option (a):

= F
and

Hence, option (a) is correct.
For option (b)

∵ | E | = 0 and | F | = 0 and | Q | ≠ 0

Let,

From Eqs. (ii) and (iii), we get Eq. (i) is true.
Hence, option (b) is correct.
For option (c)

i.e. 0 > 0 which is false.
For option (d)

From Eqs. (iv) and (v) option (d) is also correct. 

2020

So,

=

∴

Q2: Let M be a 3 × 3 invertible matrix with real entries and let I denote the 3 × 3 identity matrix. If M⁻¹ = adj(adj M), then which of the following statements is/are ALWAYS TRUE?   [JEE Advanced 2020 Paper 1]
(a) M = I
(b) det M = 1
(c) M² = I
(d) (adj M)² = I
Ans: (b), (c) & (d)
It is given that matrix M be a 3 × 3 invertible matrix, such that
M⁻¹ = adj(adj M) ⇒ M⁻¹ = |M| M
(∵ for a matrix A of order 'n' adj(adjA) = |A|ⁿ⁻² A}
⇒ M⁻¹ M = |M|M² 
⇒ M²|M| = I .....(i)
∵ det(M² |M|) = det(I) = 1
⇒ |M|³|M|² = 1
⇒ |M| = 1 .....(ii)
from Eqs. (i) and (ii), we get
M² = I
As, adj M = |M|M⁻1 = M
⇒ (adj M)² = M² (adj M)² = I

2019

Q1: Suppose  holds for some positive integer n. Then  equals ____________.       [JEE Advanced 2019 Paper 2]
Ans: 6.20
It is given that,

⇒ n = 4

= 1/5 (32 - 1) = 31/5
= 6.20

Q2: Let  , where α = α(θ) and β = β(θ) are real numbers, and I is the 2 × 2 identity matrix. If α^* is the minimum of the set {α(θ) : θ ∈ [0, 2π)} and {β(θ) : β ∈ [0, 2π)}, then the value of α^* + β^* is 
(a) -17/16
(b) -31/16
(c) -37/16
(d) -29/16
Ans: (d)
It is given that matrix

 , where α = α(θ) and β = β(θ) are real numbers, and I is the 2 × 2 identity matrix.
Now,

and

Now,

and,
∵ α is minimum at sin²(2θ) = 1 and β is minimum at sin2(2θ) = 1
So,  

2018

Q1: Let P be a matrix of order 3 × 3 such that all the entries in P are from the set {−1, 0, 1}. Then, the maximum possible value of the determinant of P is ______ .       [JEE Advanced 2018 Paper 2 ]
Ans: 4
Let

=

Now, maximum value of Det (P) = 6
If  and
But it is not possible as

Similar contradiction occurs when

Now, for value to be 5 one of the terms must be zero but that will make 2 terms zero which means answer cannot be 5
Now,

Hence, maximum value is 4