Determinants
JEE Main 2023 (April) Chapter-wise Qs Bank
Questions with Solutions
1.
MathonGo
If the system of equations
x + y + az = b
2x + 5y + 2z = 6
x + 2y + 3z = 3
has infinitely many solutions, then 2a + 3b is equal to
[2023 (06 Apr Shift 1)]
2.
(1)
25
(2)
20
(3)
23
(4)
28
For the system of equations
x + y + z = 6
x + 2y + αz = 10
x + 3y + 5z = β
, which one of the following is NOT true?
[2023 (06 Apr Shift 2)]
(1) System has no solution for α = 3,
β = 24
(2) System has a unique solution for α = −3,
β = 14
(3) System has infinitely many solutions for α = 3,
(4) System has a unique solution for α = 3,
3.
β = 14
β ≠ 14
Let S be the set of all values of θ ∈ [−π, π] for which the system of linear equations
x + y + √3z = 0
−x + (tan θ)y + √7z = 0
x + y + (tan θ)z = 0
has non-trivial solution.Then
120
π
∑
0∈S
θ
is equal to
[2023 (08 Apr Shift 2)]
4.
(1)
20
(2)
40
(3)
30
(4)
10
For the system of linear equations
2x − y + 3z = 5
3x + 2y − z = 7
4x + 5y + αz = β
,
which of the following is NOT correct?
[2023 (10 Apr Shift 1)]
(1) The system has infinitely many solutions for α =– 5 and β = 9
(2) The system has infinitely many solutions for α = −6 and β = 9
(3) The system in inconsistent for α =– 5 and β = 8
(4) The system has a unique solution for α ≠– 5 and β = 8
5.
Let S be the set of values of λ, for which the system of equations
6λx − 3y + 3z = 4λ
2
, 2x + 6λy + 4z = 1 and 3x + 2y + 3λz = λ has no solution. Then,12 ∑
λ∈S
|λ|
is equal to _______.
[2023 (10 Apr Shift 2)]
6.
Let A be a 2 × 2 matrix with real entries such that A
′
= αA + 1,
where α ∈ R −{−1, 1}., If det (A
2
− A)= 4
, the sum of all possible values of α is equal to
[2023 (11 Apr Shift 1)]
(1)
(2)
(3)
(4)
0
3
2
2
5
2
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Determinants
JEE Main 2023 (April) Chapter-wise Qs Bank
Questions with Solutions
7.
MathonGo
If the system of linear equations
7x + 11y + αz = 13
5x + 4y + 7z = β
175x + 194y + 57z = 361
has infinitely many solutions, then α + β + 2 is equal to
[2023 (11 Apr Shift 2)]
(1)
4
(2)
3
(3)
5
(4)
6
8.
If
∣x + 1
∣
∣
x
∣
∣
x
x
x + λ
x
x
x
x + λ
∣
∣
∣=
∣
2
9
8
(103x + 81)
, then λ,
are the roots of the equation
λ
3
∣
[2023 (11 Apr Shift 2)]
(1)
4x
(2)
4x
(3)
4x
(4)
4x
2
2
2
2
9.
Let D
k
+ 24x − 27 = 0
− 24x − 27 = 0
+ 24x + 27 = 0
− 24x + 27 = 0
∣ 1
∣
=∣ n
∣
2k
n
∣ n
2
2k − 1
+ n + 2
n
2
+ n
n
n
2
2
∣
∣
∣
∣
. If ∑
n
k=1
Dk = 96
, then n is equal to _________.
+ n + 2 ∣
[2023 (12 Apr Shift 1)]
10. For the system of linear equations
2x + 4y + 2az = b
x + 2y + 3z = 4
2x + 5y + 2z = 8
which of the following is NOT correct?
[2023 (13 Apr Shift 1)]
(1) It has unique solution if a = b = 6
(2) It has infinitely many solutions if a = 3,
b = 6
(3) It has infinitely many solutions if a = 3,
b = 8
(4) It has unique solution if a = b = 8
11. If the system of equations
2x + y − z = 5
2x − 5y + λz = μ
x + 2y − 5z = 7
has infinitely many solutions, then(λ + μ)
2
2
+ (λ − μ)
is equal to
[2023 (13 Apr Shift 2)]
(1)
904
(2)
916
(3)
912
(4)
920
12. Let the system of linear equations
– x + 2y − 9z = 7
−x + 3y + 7z = 9
−2x + y + 5z = 8
−3x + y + 13z = λ
has a unique solution x = α,
y = β, z = γ
. Then the distance of the point (α,
β, γ)
from the plane 2x − 2y + z = λ is
[2023 (15 Apr Shift 1)]
(1)
11
(2)
7
(3)
9
(4)
13
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Determinants
JEE Main 2023 (April) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
ANSWER KEYS
1. (3)
2. (4)
3. (1)
4. (2)
9. (6)
10. (2)
11. (2)
12. (2)
1.
5. (24)
6. (4)
7. (1)
(3)
Given system of equations are
x + y + az = b. . . . .(i)
2x + 5y + 2z = 6. . . . .(ii)
x + 2y + 3z = 3. . . . .(iii)
It is given that the system of equations have infinite solutions.
⇒ Δ = 0
∣1
1
a∣
∣
⇒ 2
∣
5
∣
2 = 0
∣
∣1
2
3∣
⇒ 1(15 − 4)−1(6 − 2)+a(4 − 5)= 0
⇒ a = 7
Also, we have ⇒ Δ
∣ b
1
a∣
∣
⇒ 6
∣
5
∣
2 = 0
∣
∣3
2
3∣
1
= 0
⇒ b(15 − 4)−1(18 − 6)+7(12 − 15)= 0
⇒ b = 3
∴ a = 7, b = 3
∴ 2a + 3b= 14 + 9 = 23
Hence, this is the correct option.
2.
(4)
Given, the system of equations
x + y + z = 6
x + 2y + αz = 10
x + 3y + 5z = β
Now finding,
∣1
1
1 ∣
∣
△= 1
∣
2
α
∣1
3
5 ∣
∣
∣
⇒△= 1(10 − 3α)−(5 − α)+(3 − 2)
⇒△= 6 − 2α
Now for unique solution, △≠ 0 ⇒ α ≠ 3
So, for unique solution ⇒ α ≠ 3
But in option (D), system has unique solution for α = 3 which is completely wrong.
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8. (4)
Determinants
JEE Main 2023 (April) Chapter-wise Qs Bank
Questions with Solutions
3.
MathonGo
(1)
Given,
S
be the set of all values of θ ∈ [−π, π] for which the system of linear equations
x + y + √3z = 0
−x + (tan θ)y + √7z = 0
x + y + (tan θ)z = 0
has non-trivial solution,
So, by condition of non-trivial solution we get,
∣
1
∣
∣ −1
∣
∣
1
√3
tan θ
√7
1
⇒(tan
⇒ tan
2
∣
∣= 0
∣
tan θ ∣
1
2
∣
θ − √7)−(− tan θ − √7)+√3(−1 − tan θ)= 0
θ +(1 − √3)tan θ − √3 = 0
⇒(tan θ + 1)(tan θ − √3)= 0
⇒ tan θ = √3, −1
π
⇒ θ =
3
,
−2π
3
−π
,
3π
,
4
4
Hence,
120
π
4.
∑ θ = 120(
1
2
−
3
3
−
1
4
+
3
4
)= 20
(2)
Given,
For the system of linear equations
2x − y + 3z = 5
3x + 2y − z = 7
4x + 5y + αz = β
,
Now finding,
∣2
∣
−1
3
∣
D = 3
∣
2
−1
∣4
5
∣
∣
α ∣
⇒ D = 2(2α + 5)+(3α + 4)+3(7)
⇒ D = 7α + 35
Now we know that, for unique solutionD ≠ 0 ⇒ α ≠– 5
So, in option (B) α =– 6 corresponds to infinitely many solution, which is completely wrong hence, this is answer.
5.
(24)
Given,
System of equations,
6λx − 3y + 3z = 4λ
2
2x + 6λy + 4z = 1
3x + 2y + 3λz = λ
For no solution △= 0,
∣ 6λ
So, ∣
∣
2
6λ
∣ 3
⇒ 9λ
3
3 ∣
−3
2
4
∣
∣
= 0
3λ ∣
− 7λ − 2 = 0
⇒ (λ − 1)(3λ + 1)(3λ + 2) = 0
Hence, λ ∈{1,
−1
3
,
−2
3
}
Now putting the value of λ in 12 ∑
12 ∑
λ∈S
|λ|= 12 ×(1 +
1
3
+
2
3
λ∈S
|λ|
we get,
)= 24
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Determinants
JEE Main 2023 (April) Chapter-wise Qs Bank
Questions with Solutions
6.
MathonGo
(4)
Given,
be a 2 × 2 matrix with real entries such that A
A
Now let A =[
a
b
c
d
c
b
d
⇒[
= αA + 1,
]
Now putting the value in A
a
′
′
we get,
= αA + I
αa + 1
αb
αc
αd + 1
]=[
]
Now equating both side we get,
1
a = αa + 1 ⇒ a =
b = αc . . . . (ii)
. . . . . . . . .(i)
1−α
and c = αb . . . . . . . .(iii)
Now from equation (ii) & (iii) we get,
c = 0
or α = ±1 (not possible as mentioned in question)
∴ c = 0
⇒ c = 0, b = 0
Also d = αd + 1 ⇒ d =
Also given, ∣∣A
2
1
1−α
− A∣
∣= 4
⇒ |A||A − I | = 4
2
⇒ (
1
1−α
Now let
So, t
2
2
) (
1
− 1)
= 4
= t
1−α
(t − 1)
1
1−α
2
= 4
⇒ t(t − 1)= ±2
⇒ t = 2 or t = −1
1
⇒ α =
2
,2
So, sum of all possible value will be
7.
1
2
+ 2 =
5
2
(1)
Given,
System of linear equations,
7x + 11y + αz = 13
5x + 4y + 7z = β
175x + 194y + 57z = 361
Now we know that, for infinite solution D = D
1
= D2 = D3 = 0
Now solving D = 0 we get,
∣
∣
∣
7
11
5
4
∣ 175
α ∣
7
194
∣
∣
= 0
57 ∣
Now applying operation R
∣7
∣
∣
5
∣0
11
α
4
7
−81
3
→ R3 − 25R1
we get,
∣
∣
∣
= 0
57 − 25α ∣
⇒ 81(49 − 5α) + (57 − 25α)(−27) = 0
⇒ 270α = −2430 ⇒ α = −9
And now solving D
∣ 13
∣
∣ β
∣
11
∣ 361
194
4
1
= 0
we get,
−9 ∣
∣
7 ∣= 0
∣
57 ∣
⇒ 13(4 × 57 − 7 × 194)−β(11 × 57 + 9 × 194)+361(11 × 7 + 9 × 4)= 0
⇒ 13(−1130)−β(2373)+361(113)= 0
⇒ 13(−10)−β(21)+361 = 0
⇒ 21β = 231
⇒ β = 11
∴ α + β + 2 = 4
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Determinants
JEE Main 2023 (April) Chapter-wise Qs Bank
Questions with Solutions
8.
MathonGo
(4)
x
x
∣
Given that, ∣
∣x + 1
x
x + λ
x
∣
∣=
∣
∣
x
x
∣
∣
x + λ
2
9
8
(103x + 81)
∣
Put x = 0 as x ∈ R
∣1
0
0 ∣
∣
⇒∣ 0
∣
λ
∣
0 ∣=
∣
∣0
0
λ
2
9
8
(103(0)+81)
∣
On expanding the determinant we get,
⇒ λ
⇒ λ
3
9
=
3
=
2
⇒ λ =
× 81
8
9
3
3
and
9
2
λ
9
=
3
6
Now the required quadratic equation is x
⇒ x
2
27
−(6)x +
⇒ 4x
2
4
2
−(
2
+
9
6
)x +
9
2
×
9
6
= 0
= 0
− 24x + 27 = 0
Hence, the required quadratic equation is 4x
9.
9
2
− 24x + 27 = 0
(6)
∣ 1
Given that D
k
2k
∣
=∣ n
∣
n
2
∣ n
⇒ ∑
n
k=1
Dk
∣
∣
⇒ ∑
n
k=1
n
∣∑1
∣
=∣ n
∣
Dk =
n
k=1
Dk
n
k=1
n
2
n
2
n
∣n
∣
=∣ n
∣
2
+ n
n
∣
Dk = ∣ 0
∣
∣n
2(
2
2
+ n
n
2
2
2
n
n
2
2
+ n + 2
n
+ n
)−n
2
n
2
∣
∣
∣
∣
∣
∣
∣
∣
∣
∣
2
+ n + 2∣
→ R2 − R3
−2
0
∣
2
−n − 2
∣
∣
∣
2
∣
+ n + 2
n
+ n + 2
n
and R
2
n ( n+1 )
n(n + 1)
2
+ n + 2 ∣
+ n + 2
n
n
2
∣
∣
∣
2
2∑k − ∑1∣
∣
2
∣
n
∣
n ( n+1 )
∣ 0
n
+ n
+ n + 2
n
∣n
⇒ ∑
2
n
∣
∣
∣n
∣
R1 → R1 − R2
n
2∑k
∣n
⇒ ∑
2
∣
2k − 1
+ n + 2
+ n
n
2
+ n + 2∣
Expand the determinant along R .
1
⇒ ∑
⇒ n
2
n
k=1
Dk = 2((−n)(−n − 2))= 96
+ 2n = 48
⇒ n = 6,
− 8
So, n = 6
Hence this is the required answer.
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Determinants
JEE Main 2023 (April) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
10. (2)
Given,
System of linear equations,
2x + 4y + 2az = b
x + 2y + 3z = 4
x + 2y + 3z = 4
Now finding,
∣2
4
∣
△= 1
∣
2
∣2
2a ∣
3
∣
∣
2 ∣
−5
⇒△= −18(a − 3)
Hence, for a = 3,
△= 0
Now finding,
∣ b
2
4
∣
△3 = 4
∣
1
2
∣8
2
∣
∣
∣
−5 ∣
⇒△3 = 9(b − 8)
So, for b = 8,
△3 = 0
Now solving options we get,
If a = 3 and b = 8 we'll have infinite solution
For a = 8, b = 8,
For a = b = 6,
∴
△≠ 0,
△≠ 0,
△3 = 0 ⇒
△3 ≠ 0 ⇒
unique solution
unique solution
For a = 3, b = 6 we'll have no solution
11. (2)
Given system of equations.
2x + y − z = 5
2x − 5y + λz = μ
x + 2y − 5z = 7
For infinite solution Δ = 0 = Δ
∣2
∣
⇒ Δ = 2
∣
−5
∣1
1
= Δ2 = Δ3
−1 ∣
1
λ
∣
∣
= 0
−5 ∣
2
⇒ 51 − 3λ = 0
⇒ λ = 17
∣ 5
∣
⇒ Δ3 =∣ μ
∣
2
∣ 7
1
2
1 ∣
∣
−5 ∣= 0
∣
2
∣
⇒ −3(μ + 13) = 0
⇒ μ = −13
Now (λ + μ)
2
2
+ (λ − μ)
2
= (17 + 13)
2
+ (17 − 13)
= 900 + 16
= 916
Hence this is the correct option.
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Determinants
JEE Main 2023 (April) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
12. (2)
Given,
System of linear equation,
– x + 2y − 9z = 7 . . . . . . .(i)
– x + 3y + 7z = 9 . . . . . . .(ii)
– 2x + y + 5z = 8 . . . . . .(iii)
– 3x + y + 13z = λ . . . . . .(iv)
Now solving, equations (i),
(ii) & (iii)
we get,
x =– 3, y = 2, z = 0
Now substituting in equation (iv) we get,
3 × 3 + 2 = λ
⇒ λ = 11
Now finding the distance of Point (– 3,
∣
d =∣
∣
−6−4−11
√
2
2
2 +2 +1
∣
∣=
∣
21
3
2, 0)
from the plane 2x − 2y + z = 11 we get,
= 7
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