Maths Determinants

$|\mathrm{A}\mathrm{d}\mathrm{j}?A|=64=|A{|}^{n-1}\Rightarrow |A|=\pm 8$ $\alpha =\pm 8$

$\beta =|\mathrm{a}\mathrm{d}\mathrm{j}?(\mathrm{a}\mathrm{d}\mathrm{j}?A\left)\right|=|A{|}^{(n-1{)}^2}=8^4\mathrm{}\text{Also}\left|\frac{\sqrt[]{\beta }}{\alpha }\right|=\frac{8^2}{8}=8$

Kindly consider the following figure

$\underset{x\to 0}{lim}\frac{{\displaystyle \underset{0}{\overset{x^2}{\int }}\left(sin\sqrt[]{t}\right)dt}}{x^3}$ $\left(\frac{0}{0}\right)$   aplying L' Hospital rule

$=\underset{x\to 0}{lim}\frac{sin\left(\sqrt[]{x^2}\right).2x}{3x^2}$

$=\frac{2}{3},forx>0$

| 1+x² |
| x 1+x | = − (x - α? ) (x − α? ) (x – α? ) (x – α? )
| x² x 1+x |

Put x = 0
| 1 0 |
| 0 1 0 | = - (-α? ) (-α? ) (-α? ) (-α? )
| 0 1 |
α ⇒? α? α? α? = −1

|adj (adj (A)| = |A|? ¹² = |A|?
|A| = | (14, 28, -14), (-14, 28), (25, -14, 14) |
= (14)² | (1, 2, -1), (-1, 2), (25/14, -1, 1) |
= (14)² (3–2 (–5)–1 (–1)
|A|? = (14)? => |A| = 14²

(P? ¹AP - I)²
= (P? ¹AP - I) (P? ¹AP - I)
= P? ¹A (PP? ¹)AP - P? ¹AP - P? ¹AP + I
= P? ¹A²P - 2P? ¹AP + I
= P? ¹ (A² - 2A + I)P = P? ¹ (A - I)²P
| (P? ¹AP - I)²| = |P? ¹ (A - I)²P| = |P? ¹| | (A - I)²| |P| = | (A - I)²| = |A - I|²
A - I = [1, 7, w²], [-1, w², 1], [0, -w, -w]
|A - I| = 1 (-w³ + w) - 7 (w) + w² (w) = -w³ + w - 7w + w³ = -6w.
|A - I|² = (-6w)² = 36w².

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).

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.