Matrix

A? =B? (i)
A³B²=A²B³. (ii)
Subtract (i) & (ii)
⇒A³ (A²−B²)=B³ (B²−A²)
⇒ (A²−B²) (A³+B³)=0
A²−B² is invertible matrix
∴A²−B²≠0
⇒A³+B³=0
∴? A³+B³? =0

A = 1/3 [ 1; 1 ω ω² 1 ω² ω ]
A² = A * A = 1/9 [ . ]
(The calculation in the image shows A² is the identity matrix, let's verify)
A² leads to I (Identity matrix).
So A² = I.
A³ = A² * A = I * A = A.
A? = (A²)² = I² = I.
A³? = (A²)¹? = I¹? = I.

[a? ] = [ 1 ] [-1 b? ]
[a? ] [√3 k] [ k b? ] (This is likely incorrect OCR, should be a 2x1 result)
The solution seems to derive from a matrix multiplication:
[√3a? ] = [ 1 ] [b? ]
[√3a? ] [√3 k] [b? ]
This leads to:
b? - b? = √3a?
b? + kb? = √3a?
Also given: a? ² + a? ² = (2/3) (b? ² + b? ²).
Squaring and adding the two derived equations and comparing with the given condition leads to k=1.

Given matrices A = [[a, b], [c, d]] and B = [[α], [β]] where B ≠ [[0], [0]].
The product AB is:
AB = [[a, b], [c, d]] * [[α], [β]] = [[aα + bβ], [cα + dβ]]

From the problem statement AB = B, we have:
aα + bβ = α (i)
cα + dβ = β (ii)

Rearranging these equations:
(a - 1)α + bβ = 0
cα + (d - 1)β = 0

For this system of linear equations to have a non-trivial solution (since B is not the zero matrix), the determinant of the coefficient matrix must be zero.
det([[a-1, b], [c, d-1]]) = 0

(a - 1)(d - 1) - bc = 0
ad - a - d + 1 - bc = 0
ad - bc = a + d - 1
The provided text jumps to the conclusion ad - bc = 2020.

The truth table for the logical expression (p ∧ q) → (p → q) is as follows:

The final column shows that the expression is a tautology, meaning it is always true regardless of the truth values of p and q.

A² = (cos2θ isin2θ cos2θ)
Similarly, A? = (cos5θ isin5θ cos5θ) = (a b; c d)
(1) a²+b² = cos²5θ - sin²5θ = cos10θ = cos75°
(2) a²-d² = cos²5θ - cos²5θ = 0
(3) a²-b² = cos²5θ + sin²5θ = 1
(4) a²-c² = cos²5θ + sin²5θ = 1

adj A| = |A|² = 9
=> |A| = ±3 => λ = |λ| = 3
=> |B| = |adj A|² = 81
=> | (B? ¹)? | = |B? ¹| = |B|? ¹ = 1/|B| = 1/81 = µ

2? -2? =112. m=7, n=4. mn=28.