Determinants

x-2y+5z=0
-2x+4y+z=0
-7x+14y+9z=0
2x (i)+ (ii) => z=0
=> x=2y
=> 15 ≤ x²+y²+z² ≤ 150
=> 15 ≤ 4y²+y² ≤ 150
=> 3 ≤ y² ≤ 30
=> y = ±2, ±3, ±4, ±5
=> 8 solutions

First, express area as a function of t. Suppose there is a triangle whose vertices are A (0,0), B (t,0) and C (0, t). Here, we can use the determinant formula for the area of a triangle. $\text{Area}=\frac{1}{2}{x}_1(y_2$
$\left|{\mathrm{-\; y}}_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$ Let us substitute the coordinates in the above equation: $\begin{array}{l}\text{Area}=\frac{1}{2}\left|0(0-t)+t(t-0)+0(0-0)\right|\\ =\frac{1}{2}\left|t*t\right|\end{array}$

$=\frac{1}{2}{t}^2$

$\text{So, the area as a function of}t\text{is:}$

$\mathrm{Area}\left(t\right)=\frac{1}{2}{t}^2$ Now, let us calculate area when t = 4 and substitute t = 4 into function: $\mathrm{Area}\left(4\right)=\frac{1}{2}*4^2=\frac{1}{2}*16=8$

? 2 adj (3 A adj (2A)|
= 23.? 3 A adj (2A)|
|2
= 23 ⋅ (33)2 ⋅ | A|2 ⋅ |adj (2 A)|2
= 23 ⋅ 36 ⋅ | A|2 ⋅ (|2 A|2)2
= 23 ⋅ 36 ⋅ | A|2 [ (23)2 ⋅ | A|2]2
= 23. 36. |A|2. 212. |A|4
= 215. 36. |A|6
= 215 ⋅ 36 ⋅ 56 = 2? ⋅ 3? ⋅ 5?
? ? = 15? = 6? = 6
? +? +? = 27

Δ = | x-2 2x-3 3x-4 |
| 2x-3 3x-4 4x-5 |
| 3x-5 5x-8 10x-17|
= Ax³ + Bx² + Cx + D.
R? → R? - R? , R? → R? - R?
Δ = | x-2 2x-3 3x-4 |
| x-1 |
| x-2 (x-2) 6 (x-2) |
= (x-1) (x-2) | 1 2x-3 3x-4 |
| 1 |
| 1 2 6 |
= -3 (x - 1)² (x - 2) = -3x³ + 12x² - 15x + 6
∴ B + C = 12 - 15 = -3

Find the area of a triangle with vertices A(?2,?3), B(4,0), and C(1,5).

x+z=h. x=80cos30, y=80sin30. tan75= (h-y)/z. h=80.

 Δ = |1 1 1| = 0
|1 2 3|
|3 2 λ|
⇒ 1(2λ - 6) - 1(λ - 9) + 1(-4) = 0
⇒ λ = 1
Δx = |6 1 1|
|10 2 3| = 0
|μ 2 λ|
⇒ 2λ + μ = 16
⇒ μ = 14
μ - λ² = 14 - 1 = 13

So D = 0 → |, [1,3, k²], | = 0 ⇒ k² = 9
x + y + 3z = 0
x + 3y + 9z = 0
3x + y + 3z = 0
(1)- (3)
x = 0 ⇒ y + 3z = 0
y/z = -3
So x + (y/z) = -3