Let A be a square matrix with real entries, and let λ be its complex eigenvalue. Suppose v = (v1, v2, . . . , vn) T is a corresponding eigenvector, Av = λv. Prove that the λ is an eigenvalue of A and Av = λv. Here v is the complex conjugate of the vector v, v := (v1, v2, . . . , vn) T .

72 is the answer to this problem

7.9m and 4.3°

step-by-step explanation:

let east be +x and north be +y, x = distance * cos θ, y = distance * sin θ  

for 3.5 south, y = -3.5  

for 8.2, x = 8.2 * cos 30 and y = 8.2 * sin 30  

for 15 west, x = -15  

total x = 8.2 * cos 30 – 15 ≈ -7.9  

total y = -3.5 + 8.2 * sin 30 = 0.6  

magnitude = [(8.2 * cos 30 – 15)^2 + 0.6^2)]^0.5  

the magnitude is approximately 7.9 m  

tan θ = y/x = 0.6/-7.9  

the angle is approximately 4.3˚ north of west

using pemdas, (order of operations) you'll see the answer is 24.