Consider the system of linear equations: $$\begin{aligned} x + 2y + 3z &= 4 \\ 2x + 4y + 6z &= k \end{aligned}$$ For what value of $k$ does this system possess infinitely many solutions?

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles By the Rouché-Capelli theorem: A linear system $A \vec{x} = \vec{b}$ is consistent if and only if the rank of the coefficient matrix equals the rank of the augmented matrix: $\text{Rank}(A) = \text{Rank}(A|b)$. For infinitely many solutions with fewer equations than unknowns, the equations must be linearly dependent. ### 2. Analytical Formulation & Calculation Notice that the coefficients of the second equation are exactly twice the first equation: $$2(x + 2y + 3z) = 2x + 4y + 6z$$ For consistency and linear dependence: $$k = 2 \times 4 = 8$$ If $k = 8$, $\text{Rank}(A) = \text{Rank}(A|b) = 1 < 3$ (number of unknowns), yielding an infinite 2-parameter family of solutions. If $k \neq 8$, $\text{Rank}(A) = 1 \neq \text{Rank}(A|b) = 2$, yielding no solution (inconsistent). ### 3. Distractor Forensics - Any value $k \neq 8$ yields parallel planes with zero solutions. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Linear dependence requires the constant vector $\vec{b}$ to be scaled by the identical scalar factor. - **Standard Reference:** Kreyszig, E., *Advanced Engineering Mathematics*, Chapter 7.

← Prev Question Next Question →