Solve the system for x and y using Cramer's rule. Assume a and b are nonzero constants.
-$$\begin{array} { l }
a x + y = \frac { 1 } { a } \
b x + y = \frac { 1 } { b }
\end{array}$$
A) $\left\{ \left( - \frac { 1 } { a b } , \frac { a + b } { a b } \right) \right\}$

B) $\left\{ \left( \frac { 1 } { a b } , a + b \right) \right\}$

C) $\left\{ \left( - \frac { 1 } { a b } , \frac { b - a } { a b } \right) \right\}$

D) $\left\{ \left( \frac { 1 } { a b } , \frac { a - b } { a b } \right) \right\}$

Question

Solve the system for x and y using Cramer's rule. Assume a and b are nonzero constants.
-$$\begin{array} { l } 
a x + y = \frac { 1 } { a } \
b x + y = \frac { 1 } { b }
\end{array}$$
A) $\left\{ \left( - \frac { 1 } { a b } , \frac { a + b } { a b } \right) \right\}$

B) $\left\{ \left( \frac { 1 } { a b } , a + b \right) \right\}$

C) $\left\{ \left( - \frac { 1 } { a b } , \frac { b - a } { a b } \right) \right\}$

D) $\left\{ \left( \frac { 1 } { a b } , \frac { a - b } { a b } \right) \right\}$

Quizplus · Accepted Answer

Using Cramer's rule, the solution for $x$ and $y$ in a system of linear equations is given by the ratio of determinants. For $x$, the determinant of the coefficient matrix is $ab - b = ab(1 - \frac{1}{a})$, and for the modified matrix (replacing $x$'s column with the constants), it's $\frac{1}{a} + \frac{1}{b}$. Solving these gives $x = -\frac{1}{ab}$ and $y = \frac{a + b}{ab}$.

Solve the System for X and Y Using Cramer's Rule ax+y=1abx+y=1b\begin{array} { l } a x + y = \frac { 1 } { a } \\b x + y = \frac { 1 } { b }\end{array}ax+y=a1​bx+y=b1​​

Solve the System for X and Y Using Cramer's Rule $\begin{array} { l } a x + y = \frac { 1 } { a } \\b x + y = \frac { 1 } { b }\end{array}$