JEE Advanced Previous Year Questions (2018 - 2023): Matrices and Determinants

## 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]

The correct option is:
(a) (P)→(3)  (Q)→(2)  (R)→(1)  (S)→(4)
(b) (P)→(3)  (Q)→(2)  (R)→(5)  (S)→(4)
(c) (P)→(2)  (Q)→(1)  (R)→(4)  (S)→(5)
(d) (P)→(2)  (Q)→(1)  (R)→(1)  (S)→(3)
Ans:
(a)
x + 2y + z = 7
x + αz = 11
2x − 3y + βz = γ
Using Cramer's rule

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 = (aij),i,j ∈ {1, 2, 3}, be the 3 × 3 matrix such that aij = 1 if j + 1 is divisible by i, otherwise  aij = 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  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

The correct option is:
(a) (I) → (T); (II) → (R); (III) → (S); (IV) → (T)
(b) (I) → (Q); (II) → (S); (III) → (S); (IV) → (R)
(c) (I) →(Q); (II) → (R); (III) →(P); (IV) → (R)
(d) (I) → (T); (II) → (S); (III) → (P); (IV) → (T)
Ans:
(b)
Given
x + y + z = 1  ----(1)
10x + 100y + 1000z = 0  ----(2)
qrx + pry + pqz = 0  ----(3)
Now equation (3) can be re-written as

Now given p,q and r are th th 10th ,100th  and th 1000th  term of an. H.P.,
So let

Now from equation (3)

Now from and (1), (2) and (3) we get

(I) If

And equation (1) and equation (2) represents non-parallel plane equation (2) and equation (3) represents same plane
⇒ Infinitely many solutions.
Now finding solution by taking z = λ then

So P is not valid for any value of  λ → Q

So no solution.
(II)  II → S
(IV) If
So infinitely many solutions.
IV → R

## 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

Q1: The trace of a square matrix is defined to be the sum of its diagonal entries. If A is a 2 × 2 matrix such that the trace of A is 3 and the trace of A3 is −18, then the value of the determinant of A is ______.    [JEE Advanced 2020 Paper 2]
Ans:
5
Let a square matrix 'A' of order 2 × 2, such that tr(A) = 3, is

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
Ans:
(b), (c) & (d)
It is given that matrix M be a 3 × 3 invertible matrix, such that
(∵ for a matrix A of order 'n' adj(adjA) = |A|n−2 A}
⇒ M−1 M = |M|M2
⇒ M2|M| = I .....(i)
∵ det(M|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|M1 = M

## 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

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

The document JEE Advanced Previous Year Questions (2018 - 2023): Matrices and Determinants | Mathematics (Maths) for JEE Main & Advanced is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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## Mathematics (Maths) for JEE Main & Advanced

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