Question:
If the system of linear equations :
\(x_1 + 2x_2 + 3x_3 = 6\)
\(x_1 + 3x_2 + 5x_3 = 9\)
\(2x_1 + 5x_2 + ax_3 = b\)
is consistent and has infinite number of solutions, then :
If the system of linear equations :
\(x_1 + 2x_2 + 3x_3 = 6\)
\(x_1 + 3x_2 + 5x_3 = 9\)
\(2x_1 + 5x_2 + ax_3 = b\)
is consistent and has infinite number of solutions, then :
\(x_1 + 2x_2 + 3x_3 = 6\)
\(x_1 + 3x_2 + 5x_3 = 9\)
\(2x_1 + 5x_2 + ax_3 = b\)
is consistent and has infinite number of solutions, then :
Show Hint
Row-reduce the augmented matrix; infinitely many solutions need a zero row in both the coefficient and the augmented part.
Updated On: Oct 31, 2026
- \(a = 8\), \(b\) can be any real number
- \(b = 15\), \(a\) can be any real number
- \(a \in R - \{8\}\) and \(b \in R - \{15\}\)
- \(a = 8, b = 15\)
Show Solution
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The Correct Option is D
Solution and Explanation
Augmented matrix \(\left[\begin{array}{ccc|c}1&2&3&6\\1&3&5&9\\2&5&a&b\end{array}\right]\).
\(R_2 \to R_2 - R_1\): \((0, 1, 2 \mid 3)\). \(R_3 \to R_3 - 2R_1\): \((0, 1, a-6 \mid b-12)\).
\(R_3 \to R_3 - R_2\): \((0, 0, a-8 \mid b-15)\).
The last row reads \((a-8)x_3 = b - 15\). Infinitely many solutions need \(a - 8 = 0\) and \(b - 15 = 0\).
So \(a = 8\) and \(b = 15\).
Answer: (4) \(a = 8,\ b = 15\)
\(R_2 \to R_2 - R_1\): \((0, 1, 2 \mid 3)\). \(R_3 \to R_3 - 2R_1\): \((0, 1, a-6 \mid b-12)\).
\(R_3 \to R_3 - R_2\): \((0, 0, a-8 \mid b-15)\).
The last row reads \((a-8)x_3 = b - 15\). Infinitely many solutions need \(a - 8 = 0\) and \(b - 15 = 0\).
So \(a = 8\) and \(b = 15\).
Answer: (4) \(a = 8,\ b = 15\)
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