What is the size of matrix C(= AB), found by multiplying a non-zero column matri

Question

What is the size of matrix C(= AB), found by multiplying a non-zero column matrix A of size 3 times 1 and a nonzero row m matrix B of size 1 times 3, is 3 times 3 1 times 3 1 times 1 3 times 1 What is the answer to the below system of equations: L_1: x - 3y = 4 L_2:-2x + 6y = -8 u=(10, 2) u=(4, 0) u=(7, 1) All of the above What arc the pivot of the below echelon matrix? [0 2 3 4 5 9 0 7 0 0 0 3 4 1 2 5 0 0 0 0 0 5 7 2 0 0 0 0 0 0 8 6 0 0 0 0 0 0 0 0] [2 4 5 9 0 7 0 0 0 3 4 1 2 5 0 0 0 0 0 5 7 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0] [0 3 4 5 9 0 7 0 0 0 4 1 2 5 0 0 0 0 0 7 2 0 0 0 0 0 0 6 0 0 0 0 0 0 0 0] [0 4 5 9 0 7 0 0 0 1 2 5 0 0 0 0 0 2 0 0 0 0 0 0 8 0 0 0 0 0 0 0 0] [0 2 4 5 9 0 7 0 0 0 3 1 2 5 0 0 0 0 0 5 7 2 0 0 0 0 0 0 8 0 0 0 0 0 0 0 0] What is the value of the following matrix products: E_1 times E_2; E_2 times E_3? E_1 = [1 0 0 0 0 1 0 1 0] E_2 = [1 0 0 0 -6 0 0 0 1], E_3 = [1 0 0 0 1 0 -4 0 1] [1 0 0 0 0 1 0 -6 0][1 0 0 0 -6 0 0 1 1] [1 0 0 0 -6 1 1 0 0][1 0 0 0 1 0 -4 0 1] [1 0 0 0 0 1 0 -6 0][1 0 0 0 -6 0 0 0 1] [1 0 0 0 1 0 0 0 1], [-1 0 0 0 -1 0 0 0 1]

Explanation / Answer

8) The answer is d (all of the above)

L1: x-3y=4

L2: -2x+6y=-8

case 1: for (10,2)

L1: 10-3*2= 4

L2: -2*10+6*2 = -8

Therefore LHS = RHS

case 2: for (4,0)

L1: 4-3*0= 4

L2: -2*4+6*0 = -8

Therefore LHS = RHS

case 3: for (7,1)

L1: 7-3*1= 4

L2: -2*7+6*1 = -8

Therefore LHS = RHS

Therefore above all condition satisfies.

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