Matrices and Determinants Test - 3

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  

x ₁= 1, y ₁= -1, z ₁= -1

x ₂= 2, y ₂= 1, z ₂= -2

x ₃= 2, y ₃= 0, z ₃= -1

α= |B| = 3
β= 1
α³+ β³= 27 + 1 = 28

Correct option is D.

 = 0
⇒(2 α+ 3) { 7 α/ 6 } - (3 α+ 1) { -7 / 6 } = 0
⇒(2 α+ 3) * (7 α/ 6) + (3 α+ 1) * (7 / 6) = 0
⇒2 α²+ 3 α+ 3 α+ 1 = 0
⇒2 α²+ 6 α+ 1 = 0
⇒α= (-3 + √7) / 2 , (-3 - √7) / 2
Hence option (2) is correct.

∵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 = D ₁= D ₂= D ₃= 0

⇒ α= 8

∴Q = (8, 6)
∴Q ²= 113

Apply , C₁  →C₁  - C₂ , C₂  →C₂  - C₃ ,

Because here row 1 and 2 are identical

R ₂→R ₂- R ₁, R ₃→R ₃- R ₁

f(x) = 45
f '(x) = 0

x + y + z = 4 μ,
x + 2y + 2z = 10 μ,
x + 3y + 4 λ²z = μ²+ 15
Δ=   = = (2 λ- 1)²
For unique solution Δ≠0, 2 λ- 1 ≠0, ( λ≠1/2 )
Let Δ= 0, λ= 1/2
Δy = 0, Δx = Δz = 
= (μ- 15)(μ- 1)
For infinite solution, λ= 1/2, μ= 1 or 15

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

 = 0
⇒2 λ²- 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(2 β- 9) + 1(-2 - 3 α)
= αβ+ 3 + 4 β- 18 - 2 - 3 α
For infinite solutions D = 0, D ₁= 0, D ₂= 0 and D ₃= 0
D = 0
αβ- 3 α+ 4 β= 17 ...... (1)

⇒1(5 β- 9) + 4(2 β- 9) + 1(6 - 15) = 0
13 β- 9 - 36 - 9 = 0
13 β= 54, β= 54/13 put in (1)
(54/13)α- 3 α+ 4(54/13) = 17
54 α- 39 α+ 216 = 221
15 α= 5, α= 1/3
Now, 12 α+ 13 β= 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 = k ₁(x + αy + 3z + 4) + k ₂(3x - y + βz - 7)
2 = k ₁+ 3k ₂, 3 = k ₁α- k ₂, -1 = 3k ₁+ βk ₂, -5 = 4k ₁- 7k ₂
On solving we get
k ₂= 13/19, k ₁= -1/19, α= -70, β= -16/13
13 αβ= 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 - 5 λ= 0 →λ= 72/5
21 + 5 μ= 0 →μ= -21/5
⇒(λ, μ) ≡(72/5, -21/5)

Δ=  ​​​​​​
= a(15a ²+ 31a + 36) = 0 ⇒a = 0
Δ≠0 for all a ∈ℝ- {0}
Hence S ₁= ℝ- {0}, S ₂= ∅

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  

The given system is represented in terms of determinants:
D = | α2 1 |
| 2 α3 1 |
| 3 α2 |
which gives D = 0 when α= -1 or α= 3.
Dx = | 2 1 1 |
| 3 1 1 |
| α2 β|
which gives
Dx = 0 when β= 2.|
Dy = | α1 1 || 2 α3 1 |
| 3 2 β|
which gives Dy = 0 (determinant is zero).
Dz = | α2 1 |
| 2 α3 1 |
| 3 αβ|
which gives Dz = 0.
The solution states that when β= 2 and α= -1, the system has an infinite solution.