Question

Find the condition on a, b and c so that the following system of linear
equations has one parameter family of functions : x + y + z = a,
x + 2y + 3z = b, 3x + 5y + 7z = C​

Answer 1

Answer:

Given as x + y + z = 6 x + 2y + 3z = 14 x + 4y + 7z = 30 The equation can be written as |A| = 1(2) – 1(4) + 1(2) = 2 – 4 + 2 |A| = 0 Therefore, A is singular. So, the given system is either inconsistent or it is consistent with infinitely many solution according to as: (Adj A) x B ≠ 0 or (Adj A) x B = 0 Co-factors of A are C11 = (– 1)1 + 1 14 – 12 = 2 C21 = (– 1)2 + 1 7 – 4 = – 3 C31 = (– 1)3 + 1 3 – 2 = 1 C12 = (– 1)1 + 2 7 – 3 = – 4 C22 = (– 1)2 + 1 7 – 1 = 6 C32 = (– 1)3 + 1 3 – 1 = 2 C13 = (– 1)1 + 2 4 – 2 = 2 C23 = (– 1)2 + 1 4 – 1 = – 3 C33 = (– 1)3 + 1 2 – 1 = 1 AX = B has infinite many solution Suppose z = k x + y = 6 – k x + 2y = 14 – 3k The equation can be written as Here value of x, y and z satisfy the third equation. So, x = k - 2, y = 8 - 2k, z = k.