Exercise · ZW-CRAMERS-RULE-006 · ★★☆ Intermediate
For \(\begin{cases}2x+4y=6\\ x+2y=3\end{cases}\), what do you conclude when trying Cramer's rule ?
Choose one option.
A It gives \(x=y=0\), because \(\det A=0\) B It has a unique solution, since the constants are not zero C It cannot be applied because \(\det A=0\); another method is needed to study the system D It is inconsistent, because \(\det A=0\) always means no solution Show answer C. It cannot be applied because \(\det A=0\); another method is needed to study the system← Back to the question
Explanation \(\det A=4-4=0\), so Cramer does not apply. \(\det A=0\) alone does not settle it: here the first equation is twice the second and the system has infinitely many solutions.
Cramer's rule gives the unique solution of a square system, whose coefficient matrix \(A\) has nonzero determinant , as a quotient of determinants.
\[x_i=\frac{\det A_i}{\det A}\]
More practice problems: cramer's rule