(22) let a be a 100 x 90 matrix, representing an image with 9000 pixels. how many numbers do we need to store for a truncated singular value decompo- sition (tsvd) a = ak with k = 10 ? (a) 1090 (b) 1900 (c) 1910 (d) 1990 (e) 2018

The correct option is c.

Step-by-step explanation:

Let A be a 100 x 90 matrix.

The order of a matrix is m x n, where, m is number of rows and n is number of columns.

It is given that for a truncated singular value decompo- sition (TSVD) with k=10.

The formula for the number of numbers do we need to store for a truncated singular value decompo- sition (TSVD) is

Substitute k=10, m=100 and n=90 in the above formula.

The numbers that are needed to store for a truncated singular value decompo- sition (TSVD) is 1910.

Therefore the correct option is c.

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!