Question 11, 2.8.1
Part 1 of 2
A = [[5, 2, 0, 16], [-10, -2, 10, -36], [15, -2, -40, 64]]
First row of the reduced echelon form gives the equation 0 = 0. To find the row echelon form, the equation x1 - 2x3 + 4x4 = 0 is obtained. The second row of the reduced echelon form gives the first equation for x1.
The second equation for x2 is +5x3 - 2x4 = 0
x2 = 2x4 - 5x3
General solution is x1 = 2x3 - 4x4
Obtained values for each variable, such that Ax = 0. 2x3 - 4x4 problem, let x3 = 3 and x4 = 2. Solve for the value of x1.
Question 11, 2.8.1 Part 1 of 2
For the matrix A below, find a nonzero vector in Nul A and a nonzero vector in Col A 5 A 16 -10 -2 10 -36 15 -2 -40 64
Solve the equation Ax = 0. Augment matrix A with a column of zeros. Then row reduce the augmented matrix to find its re -10 -2 160 10 -360 10 15 -2 -40 40 640 00 5 0 -20 00 the first equation for x first row of the reduced echelon form gives the equation -2xg + 4x4 = 0. The second row of the reduced echelon form gives 1 - 2x + 44 = 0 as the equation 0 = 0. To find the general solution, solve the first equation for x1 and the second equation for x2 in terms of th x1 = 2 - 44. The second equation for x2 + 5x - 2x = 0 x = 2x - 5x obtained values for each variable, such that Ax = 0 problem. Let x = 3 and x = 2. Solve for the value of x.