Exercise · ZW-CRAMERS-RULE-007 · ★★☆ Intermediate
For which values of \(k\) does \(\begin{cases}kx+2y=3\\ 2x+ky=1\end{cases}\) have a unique solution?
Choose one option.
Show answer
- D. \(k\neq2\text{ and }k\neq-2\)
Explanation
\(\det A=k^2-4=(k-2)(k+2)\). There is a unique solution exactly when \(\det A\neq0\), that is, \(k\neq\pm2\).
Review the concept: Cramer's rule
Cramer's rule gives the unique solution of a square system, whose coefficient matrix \(A\) has nonzero determinant, as a quotient of determinants.
\[x_i=\frac{\det A_i}{\det A}\]
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