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in Matrices & determinants by (54.8k points)

If the system of linear equations x + ay + z = 3; x + 2y + 2z = 6; x + 5y + 3z = b has no solution, then

(A) a = −1, b = 9 

(B) a = −1, b ≠ 9 

(C) a ≠ −1, b = 9 

(D) a = 1, b ≠ 9

1 Answer

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Best answer

Answer is (B) a = −1, b ≠ 9

The given system of linear equations is

x + ay + z = 3

x + 2y + 2z = 6

x + 5y + 3z = b

Therefore, if the given system of equations has no solution, then a = −1 and b ≠ 9.

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