Determinants - 1 - Free MCQ Practice Test with solutions, JEE Maths

Apply C₂ → C₂ + C_3,

Apply , R₁ → R₁+R₂+R₃,

Apply , C₃→ C₃ - C₁, C₂^→C₂ - C₁,

=(a+b+c)³

The determinant of a matrix A and its transpose always same.

⇒ (1-x)(2-x)(3-x) = 0 ⇒x = 1,2,3

If A is a symmetric matrix then by definition AT=A
Option A is correct.

Apply , C1→C1 - C3, C₂→C₂-C₃

= 10 - 12 = -2

If A is anon singular matrix of order , then 

Step 1: Compute the determinant

det(A) =
= 2(2·3 − 5·1) − λ(0·3 − 5·1) + (−3)(0·1 − 2·1)

= 2(6 − 5) − λ(0 − 5) + (−3)(0 − 2)

= 2(1) − λ(−5) + (−3)(−2)

= 2 + 5λ + 6

= 5λ + 8

Step 2: Condition for invertibility
det(A) ≠ 0 ⇒ 5λ + 8 ≠ 0
⇒ λ ≠ −8/5.

The correct option is C.

Option B is correct.

Expanding the determinant along the third column gives the equation (α + 3/2)·(7α/3) - (α + 1/3)·(-7/2) = 0.

Multiplying both sides by 6 to clear denominators gives 14α² + 42α + 7 = 0.

Dividing by 7 yields 2α² + 6α + 1 = 0.

Solving the quadratic gives α = (-3 ± √7)/2.

Numerically, α ≈ -2.8229 and α ≈ -0.1771.

Both values lie in the interval (-3, 0); therefore the correct option is B.

∵ a, b, c and in A.P
⇒ 2b = a + c ⇒ a - 2b + c = 0
∴ ax + by + c passes through fixed point (1, -2)
∴ P = (1, -2)
For infinite solution,
D = D₁ = D₂ = D₃ = 0

⇒ α = 8

∴ Q = (8, 6)
∴ Q² = 113

Apply , C₁ → C₁ - C₂, C₂ → C₂ - C₃,

Because here row 1 and 2 are identical

R₂ → R₂ - R₁, R₃ → R₃ - R₁

On applying determinant formula, we get

2cos⁴x * (0-(-9)) - 2sin⁴x * (-9-0) + (3 + sin²2x) * (9-0)

2cos⁴x * 9 - 2sin⁴x * (-9) + (3 + sin²2x) * 9

18cos⁴x +18 sin⁴x + (3 + sin²2x) * 9

18cos⁴x + 18sin⁴x + 27 + 9sin²2x
 

We are to find:

(1/5)·f'(0)

We differentiate each term with respect to x:

Use the chain rule:
d/dx [cos⁴x] = 4cos³x * (−sinx) = −4cos³x sinx
So:
d/dx [18cos⁴x] = 18 × (−4cos³x sinx) = −72cos³x sinx

d/dx [sin⁴x] = 4sin³x * cosx
So:
d/dx [18sin⁴x] = 18 × 4sin³x cosx = 72sin³x cosx

Use chain rule:
d/dx [sin²(2x)] = 2sin(2x) * cos(2x) * 2 = 4sin(2x)cos(2x)
So:
d/dx [9sin²(2x)] = 9 × 4sin(2x)cos(2x) = 36sin(2x)cos(2x)

So,

f′(x) = −72cos³x sinx + 72sin³x cosx + 36sin(2x)cos(2x)

Plug in x = 0:

• sin(0) = 0

• cos(0) = 1

• sin(2×0) = 0

• cos(2×0) = 1

Now evaluate each term:

1. −72cos³(0) sin(0) = −72(1)(0) = 0

2. 72sin³(0) cos(0) = 72(0)(1) = 0

3. 36sin(0)cos(0) = 36(0)(1) = 0

So,

f′(0) = 0

(1/5) × f′(0) = (1/5) × 0 = 0

x + y + z = 4μ,
x + 2y + 2z = 10μ,
x + 3y + 4λ²z = μ² + 15
Δ =  = = (2λ - 1)²

Hence, the system has a unique solution for all real values of μ.
Therefore, the system is consistent.

Therefore, Option (d) will be an answer.

Because , the inverse of an identity matrix is an identity matrix.

 = 0
⇒ 2λ² - λ - 1 = 0
λ = 1, -1/2
 = 0 = μ = 13
Infinite solution λ = 1 & μ = 13
For unique solution λ ≠ 1
For no solution λ = 1 & μ ≠ 13
If λ ≠ 1 and μ ≠ 13
Considering the case when λ = -1/2 and μ ≠ 13, this will generate no solution case.

D = 
= 1(αβ + 3) + 2(2β - 9) + 1(-2 - 3α)
= αβ + 3 + 4β - 18 - 2 - 3α
For infinite solutions D = 0, D₁ = 0, D₂ = 0 and D₃ = 0
D = 0
αβ - 3α + 4β = 17 ...... (1)

⇒ 1(5β - 9) + 4(2β - 9) + 1(6 - 15) = 0
13β - 9 - 36 - 9 = 0
13β = 54, β = 54/13 put in (1)
(54/13)α - 3α + 4(54/13) = 17
54α - 39α + 216 = 221
15α = 5, α = 1/3
Now, 12α + 13β = 12 * (1/3) + 13 * (54/13)
= 4 + 54 = 58

For a square matrix A to be invertible, its determinant must satisfy a specific condition:

• Invertibility Condition: A matrix A is invertible if and only if the determinant of A, denoted as det(A), is non-zero.

f(0) =  = 0
f'(x) = 
∴  f'(0) = 
= 24 - 6 = 18
∴ 2f(0) + f'(0) = 42

Using family of planes
2x + 3y - z - 5 = k₁(x + αy + 3z + 4) + k₂(3x - y + βz - 7)
2 = k₁ + 3k₂, 3 = k₁α - k₂, -1 = 3k₁ + βk₂, -5 = 4k₁ - 7k₂
On solving we get
k₂ = 13/19, k₁ = -1/19, α = -70, β = -16/13
13αβ = 13(-70)(-16/13) = 1120

x + 2y + 3z = 3 .... (i)
4x + 3y - 4z = 4 .... (ii)
8x + 4y - λz = 9 + μ .... (iii)
(i) × 4 - (ii) → 5y + 16z = 8 .... (iv)
(ii) × 2 - (iii) → 2y + (λ - 8)z = -1 - μ .... (v)
(iv) × 2 - (iii) × 5 → (32 - 5(λ - 8))z = 16 - 5( - 1 - μ)
For infinite solutions → 72 - 5λ = 0 → λ = 72/5
21 + 5μ = 0 → μ = -21/5
⇒ (λ, μ) ≡ (72/5, -21/5)

Δ = ​​​​​​
= a(15a² + 31a + 36) = 0 ⇒ a = 0
Δ ≠ 0 for all a ∈ ℝ - {0}
Hence S₁ = ℝ - {0}, S₂ = ∅

To determine the number of decomposed determinants, we start by considering the linearity property of determinants over columns. Each column in the given third-order determinant is a sum of terms: the first column has 2 terms per element, the second column has 3 terms per element, and the third column has 4 terms per element.

1. First Column (2 terms per element): Each element in the first column can be split into two terms, leading to 2 determinants.

2. Second Column (3 terms per element): Each element in the second column can be split into three terms. For each of the 2 determinants from the first column, splitting the second column results in 2×3=6 determinants.

3. Third Column (4 terms per element): Each element in the third column can be split into four terms. For each of the 6 determinants from the previous step, splitting the third column results in 6×4=24 determinants.

Thus, the total number of decomposed determinants is 2×3×4=24 

The value of n is 24 

Eliminate variables from (1) and (2), we get

Substitute (A) in (1), we get

Substitute in eqn 3, we get

Option (a) is correct.
Option (b) is incorrect (since solution exists when β=2)

Option (c) is correct.

Unique solution exists for all β

Thus option (d) is wrong because it restricts unnecessarily to β=2.

So, option (b) is the correct answer.