Linear Algebra: Permutation Matrices Let M be the matrix {{0,0,0,1,0}, {0,0,1,0,0}, {0,1,0,0,0}, {0,0,0,0,1}{1,0,0,0,0}}. What is the smallest positive integer n such that M^n is the identity matrix? a. 2. b. 3. c. 4.d. 6.

Multiplying M by any matrix A would return new matrix, B, in which

• the 1st row of B is equal to the 4th row of A,

• the 2nd row of B is equal to the 3rd row of A,

• the 3rd row of B is equal to the 2nd row of A,

• the 4th row of B is equal to the 5th row of A, and

• the 5th row of B is equal to the 1st row of A.

The pattern here is

1 => 4 => 5 => 1

2 => 3 => 2

Let {4, 3, 2, 5, 1} denote the matrix M, where each number refers to the row of the identity matrix, I.

Using this notation, the pattern above gives

M² = {5, 2, 3, 1, 4}

M³ = {1, 3, 2, 4, 5}

M⁴ = {4, 2, 3, 5, 1}

M⁵ = {5, 3, 2, 1, 4}

M⁶ = {1, 2, 3, 4, 5}

so that n = 6.

(Notice that the first cycle has length 3 and the second one has length 2; the minimum n needed here is then LCM(2, 3) = 6.)

first, you divide the triangle in two, tô form a retangle triangle. after, using pythagoream theorem:

18^2 = 14^2 + (x/2)^2

(x^2/4) + 196 = 324

x^2/4 = 324 - 196

x^2/4 = 128

x^2 = 512

x = √512

x = √2^9

x = 16√2 cm

i hope i you.