Exercise · ZW-GAUSSIAN-ELIMINATION-010 · ★★★ Advanced
For which values of \(k\) does the system \(\begin{cases}x+y=1\\2x+2y=k\end{cases}\) have no solution?
Choose one option.
Show answer
- B. \(k\neq 2\)
Explanation
\(R_2-2R_1\) gives \(0=k-2\). If \(k=2\) there are infinitely many solutions; if \(k\neq 2\) it is contradictory.
Review the concept: Gaussian elimination
Gaussian elimination solves a linear system by reducing its augmented matrix, with coefficients \(A\) and vector \(\mathbf b\), to an echelon form \(U\) using elementary row operations.
\[[A\mid \mathbf b]\ \longrightarrow\ [U\mid \mathbf c]\]
More practice problems: gaussian elimination
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- ZW-GAUSSIAN-ELIMINATION-007 A consistent system with 3 equations and 4 unknowns has 2 pivots in echelon form. How many free var…
- ZW-GAUSSIAN-ELIMINATION-008 Which of these matrices is in row echelon form?