IB DP Mathematics • 2024 • 7 Marks

Number & Algebra: Systems of Linear Equations & Gaussian Elimination

Official examination question with verified M1/A1 mark scheme and step-by-step mathematical reasoning.

Problem Statement

Consider the system of linear equations: $\begin{cases} x + 2y - z = 4 \\ 2x - y + 3z = 3 \\ 3x + y + kz = 7 \end{cases}$ where $k \in \mathbb{R}$. (a) Find the value of $k$ for which the system does not have a unique solution. [4 Marks] (b) For this value of $k$, determine whether the system is consistent (infinitely many solutions) or inconsistent (no solutions). [3 Marks]

Verified Solution & Marking Scheme

Part (a): Evaluate the Coefficient Matrix Determinant
The system fails to possess a unique solution when $\det(A) = 0$: $\det(A) = \begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 3 & 1 & k \end{vmatrix}$ Expand along row 1: $= 1(-k - 3) - 2(2k - 9) - 1(2 - (-3))$ $= -k - 3 - 4k + 18 - 5 = -5k + 10$ Setting $\det(A) = 0 \implies -5k + 10 = 0 \implies k = 2$.
Part (b): Perform Gaussian Elimination for k = 2
Augmented matrix with $k = 2$: $\begin{pmatrix} 1 & 2 & -1 & 4 \\ 2 & -1 & 3 & 3 \\ 3 & 1 & 2 & 7 \end{pmatrix}$ Row operations: $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$: $\begin{pmatrix} 1 & 2 & -1 & 4 \\ 0 & -5 & 5 & -5 \\ 0 & -5 & 5 & -5 \end{pmatrix}$ Now $R_3 \to R_3 - R_2$: $\begin{pmatrix} 1 & 2 & -1 & 4 \\ 0 & -5 & 5 & -5 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ The row of zeros with a zero constant indicates that the equations are linearly dependent.
Conclusion for k = 2
Since the bottom row is $0x + 0y + 0z = 0$, the system is consistent with infinitely many solutions (representing three planes intersecting along a common line).
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