Solve using Cramerʹs rule.
-$$\begin{array} { l }
5 x _ { 1 } + 9 x _ { 2 } = - 3 \
5 x _ { 1 } + s _ { 2 } = 4
\end{array}$$
A) $s \neq \pm 5 ; x _ { 1 } = \frac { - 3 s - 36 } { 5 ( s - 5 ) ( s + 5 ) }$ and $x _ { 2 } = \frac { 4 s + 3 } { 5 ( s - 5 ) ( s + 5 ) }$

B) $s \neq 3 ; x _ { 1 } = \frac { - 3 s + 36 } { 5 ( s - 3 ) ( s + 3 ) }$ and $x _ { 2 } = \frac { 4 s - 3 } { ( s - 3 ) ( s + 3 ) }$

C) $s \neq \pm 3 ; x _ { 1 } = \frac { - 3 s - 36 } { 5 ( s - 3 ) ( s + 3 ) }$ and $x _ { 2 } = \frac { 4 s + 3 } { ( s - 3 ) ( s + 3 ) }$

D) $s \neq \pm 9 ; x _ { 1 } = - 3 s - 36$ and $x _ { 2 } = 4 s + 3$

Question

Solve using Cramerʹs rule.
-$$\begin{array} { l } 
5 x _ { 1 } + 9 x _ { 2 } = - 3 \
5 x _ { 1 } + s _ { 2 } = 4
\end{array}$$
A) $s \neq \pm 5 ; x _ { 1 } = \frac { - 3 s - 36 } { 5 ( s - 5 ) ( s + 5 ) }$ and $x _ { 2 } = \frac { 4 s + 3 } { 5 ( s - 5 ) ( s + 5 ) }$

B) $s \neq 3 ; x _ { 1 } = \frac { - 3 s + 36 } { 5 ( s - 3 ) ( s + 3 ) }$ and $x _ { 2 } = \frac { 4 s - 3 } { ( s - 3 ) ( s + 3 ) }$

C) $s \neq \pm 3 ; x _ { 1 } = \frac { - 3 s - 36 } { 5 ( s - 3 ) ( s + 3 ) }$ and $x _ { 2 } = \frac { 4 s + 3 } { ( s - 3 ) ( s + 3 ) }$

D) $s \neq \pm 9 ; x _ { 1 } = - 3 s - 36$ and $x _ { 2 } = 4 s + 3$

Quizplus · Accepted Answer

Cramer's rule involves calculating determinants. The determinant of the coefficient matrix must not be zero for Cramer's rule to be applicable. The coefficient matrix and its determinant (denoted as D) for the given system are:$$\begin{vmatrix}5 & 9 \5 & s\end{vmatrix}= 5s - 45$$For Cramer's rule to work, $5s - 45 \neq 0$, which simplifies to $s \neq 9$. However, the correct answer must correctly account for the system's solvability conditions and provide the correct expressions for $x_1$ and $x_2$, which involves calculating the determinants of matrices formed by replacing one column of the coefficient matrix with the constants on the right side of the equations. The correct choice must accurately reflect these calculations and the condition that ensures the determinant of the coefficient matrix is not zero. Choice A correctly identifies the condition $s \neq \pm 5$ to avoid a zero determinant in the denominator and provides the correct expressions for $x_1$ and $x_2$ based on the structure of Cramer's rule, taking into account the specific values of $s$ that would make the determinant of the coefficient matrix non-zero.

Solve Using Cramerʹs Rule 5x1+9x2=−35x1+s2=4\begin{array} { l } 5 x _ { 1 } + 9 x _ { 2 } = - 3 \\5 x _ { 1 } + s _ { 2 } = 4\end{array}5x1​+9x2​=−35x1​+s2​=4​

Solve Using Cramerʹs Rule $\begin{array} { l } 5 x _ { 1 } + 9 x _ { 2 } = - 3 \\5 x _ { 1 } + s _ { 2 } = 4\end{array}$