The system \(x+y+z=1,\ x+2y+3z=2,\ x+3y+kz=3\) has infinitely many solutions when \(k=\)
A. 4
B. 5
C. 3
D. 6
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Best answer
Answer: B. 5
\(\Delta=k-5=0\Rightarrow k=5\). Then \(E_2-E_1\) and \(E_3-E_2\) both give \(y+2z=1\), so the system is consistent.
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· 23 Oct 2026
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