Maths Determinants

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.

|λ-1 3λ+1 2λ|
|λ-1 4λ-2 λ+3| = 0
|2 3λ+1 3 (λ-1)|
R? → R? - R? and R? → R? - R? (from a similar matrix setup, applying operations to simplify)
The provided solution uses a slightly different matrix but let's follow the subsequent steps.
A different matrix from the image is used in the calculation:
|λ-1 3λ+1 2λ|
|0 λ-3 -λ+3|
|3-λ 0 λ-3 |
C? → C? + C?
|3λ-1 3λ+1 2λ |
|3-λ λ-3-λ | = 0
|0 λ-3 |
⇒ (λ-3) [ (3λ-1) (λ-3) - (3λ+1) (3-λ)] = 0
⇒ (λ-3) [ (λ-3) (3λ-1) + (λ-3) (3λ+1)] = 0
⇒ (λ-3)² [3λ-1 + 3λ+1] = 0
⇒ (λ-3)² [6λ] = 0 ⇒ λ = 0, 3
Sum of values of λ = 3

A = [cosθ, sinθ], [-sinθ, cosθ]
A² = [cos2θ, sin2θ], [-sin2θ, cos2θ]
⇒ A? = [cos4θ, sin4θ], [-sin4θ, cos4θ]
B = [cos4θ, sin4θ], [-sin4θ, cos4θ] + [cosθ, sinθ], [-sinθ, cosθ]
= [cos4θ + cosθ, sin4θ + sinθ], [- (sin4θ + sinθ), cos4θ + cosθ]
det (B) = (cos4θ + cosθ)² + (sin4θ + sinθ)²
= (cos²4θ + sin²4θ) + (cos²θ + sin²θ) + 2 (cos4θcosθ + sin4θsinθ)
= 1 + 1 + 2cos (4θ - θ)
= 2 + 2cos3θ
Given 3θ = 3π/5
|B| = 2 + 2cos (3π/5)
= 2 + 2 (- (√5-1)/4) = 2 - (√5-1)/2 = (4-√5+1)/2 = (5-√5)/2 ∈ (1,2)

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.

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