Determinants

For a system of linear homogeneous equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero.
Δ = | 4 λ 2 |
| 2 -1 | = 0
| μ 2 3 |
To simplify, perform the row operation R? → R? - 2R? :
Δ = | 0 λ+2 0 |
| 2 -1 | = 0
| μ 2 3 |
Expand the determinant along the first row:
- (λ+2) * det (| 2 1 |, | μ 3 |) = 0.
- (λ+2) (2*3 - 1*μ) = 0.
(λ+2) (μ-6) = 0.
This implies that either λ+2 = 0 or μ-6 = 0.
So, the conditions are λ = -2 (for any μ) or μ = 6 (for any λ).

Given f(x) = e^x sin(x).
Let F(x) = ∫[0 to x] f(t) dt.
By the Fundamental Theorem of Calculus, F'(x) = f(x) = e^x sin(x).

The integral I = ∫[0 to 1] (F'(x) + f(x))e^x dx
= ∫[0 to 1] (e^x sin(x) + e^x sin(x))e^x dx = ∫[0 to 1] 2e^(2x) sin(x) dx.
The text computes I = ∫[0 to 1] 2 sin(x) dx = [-2cos(x)] from 0 to 1 = -2cos(1) - (-2cos(0)) = 2(1 - cos(1)). This assumes an error in the problem statement where the integral was (F'(x)+f(x))dx, not with an extra e^x term.
Using the series expansion for cos(1) = 1 - 1/2! + 1/4! - .
2(1 - cos(1)) = 2(1 - (1 - 1/2 + 1/24 - .)) = 1 - 1/12 + . ≈ 11/12 ≈ 0.916.
The inequality 330/360 < I < 331/360 (i.e., 0.9166 < I < 0.9194) is checked

We need to evaluate the sum:
Σ (100 - r) (100 + r) for r from 0 to 99.

Σ (100² - r²) for r from 0 to 99
= Σ 100² - Σ r² for r from 0 to 99
= 100 * 100² - [99 (99+1) (2*99+1)]/6
= 100³ - [99 * 100 * 199]/6
= 100³ - (1650 * 199)

Comparing this with (100)³ – 199β, we get:
β = 1650
If we consider a comparison form α = 3β, this part of the solution seems to have a typo, but based on the final calculation, Slope = α/β = 1650 / 3 = 550 seems to be intended.

Kindly consider the following figure

A = [0, sin α], [sin α, 0]
A² = A ⋅ A = [0, sin α], [sin α, 0] ⋅ [0, sin α], [sin α, 0]
= [00 + sinαsinα, 0sinα + sinα0], [sinα0 + 0sinα, sinαsinα + 00]
= [sin²α, 0], [0, sin²α] = (sin²α)I
A² - (1/2)I = [sin²α, 0], [0, sin²α] - [1/2, 0], [0, 1/2]
= [sin²α - 1/2, 0], [0, sin²α - 1/2]
det (A² - (1/2)I) = (sin²α - 1/2)² - 0 = 0
sin²α - 1/2 = 0
sin²α = 1/2
sin α = ±1/√2
A possible value for α is π/4.

Here D = | 2 -4 λ |
| 1 -6 1 | = (λ-3) (3λ+2)
| λ -10 4 |
D = 0 ⇒ λ = 3, -2/3

Let A =
| a? |
| b? |
| c? |

Ax? = B?
a? + a? + a? = 1
b? + b? + b? = 0
c? + c? + c? = 0
Similar 2a? + a? = 0 and a? = 0
2b? + b? = 2, b? = 0
2c? + c? = 0, c? = 2
∴ a? = 0, b? = 1, c? = -1,
a? = 1, b? = -1, c? = -1
A =
| 1 0 |
| -1 0 |
| -1 -1 2 |
∴ |A| = 2

Here,
| 1 |
| 2 4 -1 | = 0 ⇒ λ = 9/2
| 3 2 λ |

Also,
| 1 2 |
| 2 4 6 | = 0 ⇒ μ = 5
| 3 2 μ |

∴ Option B is correct.