Question

The matrix A=[-17-51-85-21 is row equivalent to R=「1 3 5 15 45 75 1 -4 -12 -20 0 1. a. Find a basis for the row space of A, row(A) b. Write the sum of the 1st and 3rd row of A as a linear combination of your basis for row(A). 2. a. Find a basis for the column space of A, col(A) b. Write the difference if the 2nd and 4th column of A as a linear combination of your basis for col(A) 3. Find a basis for the null space of A, null(A).

Answer 1

a = [-17 15 -4 -51 45 -12 -35 75 -20 -2 1 0] is row equivalent to r = [1 0 0 0 3 0 0 0 5 0 0 0 0 1 0 0] therefore parts basis for the row parts of a ,row a { (1,3,5,0),(0,0,01)} r_1 = (-17,-51,-35,-2) r_3 = (-4,-12,-20,0) r_1 = (-17,-51,-35,0)+(-2(0,0,0,1) r_1 = -1z(1,3,5,0) + -2(0,0,01) r_3 = (-4,12,-20,0) r_3 = -4(1,3,5,0) therefore r_1 and r_2 of a is linear combination f and y null spolls of a [1 0 0 3 0 0 5 0 0 0 1 9]_3into4 [x_1 x_2 x_3 x_4]_4into1 = [0 0 0] x_4 + 3x_2 + 5x_3 = 0 x_4 = 0 x_1 = -(3x_2 + 5x_3) therefore [x_1 x_2 x_3 x_4] = [-3x_2 -5x_3 x_2 x_3 0] = x_2 [-3 1 0 0] + x_3 [-5 0 1 0] therefore ker a = {x elementof r^4: ax = 0} therefore basis of kera = null a = {(-3 1 0 0) ,(-5 0 1 0)}