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03/02/2022 7:09 pm
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If the following system of linear equations
2x + y + z = 5
x – y + z = 3
x + y + az = b
has no solution, then:
(1) a = -\(\frac{1}{3}\), b ≠ \(\frac{7}{3}\)
(2) a ≠ \(\frac{1}{3}\), b = \(\frac{7}{3}\)
(3) a ≠ -\(\frac{1}{3}\), b = \(\frac{7}{3}\)
(4) a = \(\frac{1}{3}\), b ≠ \(\frac{7}{3}\)
1 Answer
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03/02/2022 7:15 pm
Correct answer: (4) a = \(\frac{1}{3}\), b ≠ \(\frac{7}{3}\)
Explanation:
Here D = \(\begin{vmatrix} 2 & 1 & 1 \\[0.3em] 1 & -1 & 1 \\[0.3em] 1 & 1 & a \end{vmatrix}\)
= 2(-a - 1) -1 (a - 1) + 1 + 1
= 1 - 3a
D3 = \(\begin{vmatrix} 2 & 1 & 5 \\[0.3em] 1 & -1 & 3 \\[0.3em] 1 & 1 & b \end{vmatrix}\)
= 2(-b - 3) - 1(b - 3) + 5(1 + 1)
= 7 - 3b
for a = \(\frac{1}{3}\), b ≠ \(\frac{7}{3}\), system has no solutions