Consider the system of equations: $x + y + z = 6$, $x + 2y + 3z = 10$, and $x + 2y + k z = 35$. The value of $k$ for which the system has infinitely many solutions is:
Options:
(A) 3
(B) 2
(C) 1
(D) 0
Answer Key:
A
Step-by-Step Solution & Explanation
Writing the augmented matrix $[A|B]$:
$\begin{bmatrix} 1 & 1 & 1 & 6 \\ 1 & 2 & 3 & 10 \\ 1 & 2 & k & k+7 \end{bmatrix}$
Performing $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$:
$\begin{bmatrix} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & k - 3 & (k + 7) - 10 \end{bmatrix}$
For infinitely many solutions, $\operatorname{{rank}}(A) = \operatorname{{rank}}(A|B) < 3$, which requires:
$k - 3 = 0 \implies k = 3$
Checking the constant side: $(3 + 7) - 10 = 10 - 10 = 0$. Thus the third row becomes all zeros, yielding infinite solutions.