Discussion
The following system of equations
\(x + y + z = 1\)
\(2x + 3y - mz = 2\)
\(3x + 5y + 3z = 3\)
has no unique solution. Then the value of \(m\) is equal to
Answer: D
\(m = \) -2.
A square system lacks a unique solution exactly when its coefficient determinant is zero:
\[ \begin{aligned} \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & -m \\ 3 & 5 & 3 \end{vmatrix} &= 1(9 + 5m) - 1(6 + 3m) + 1(10 - 9) \\ &= 4 + 2m \end{aligned} \]
Setting \(4 + 2m = 0\) gives \(m = -2\). (Then \(x = 1, y = z = 0\) is one solution among infinitely many, so the system is consistent but not uniquely solvable.) The value 2 comes from a sign slip in the \(-m\) entry.
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