Respuesta :
Answer:
[tex]s \ne \±2[/tex]
[tex]x_1 = \frac{3s - 2}{3(s^2 -4)}[/tex]
[tex]x_2 = \frac{2(s- 6)}{5(s^2 - 4)}[/tex]
Step-by-step explanation:
Given
[tex]3sx_1 +5x_2 = 3[/tex]
[tex]12x_1 + 5sx_2 =2[/tex]
Required
Determine the value of s
Express the equations as a matrix
[tex]A =\left[\begin{array}{cc}3s&5\\12&5s\end{array}\right][/tex]
Calculate the determinant
[tex]|A|= (3s*5s -5 *12)[/tex]
[tex]|A|= (15s^2 -60)[/tex]
Factorize
[tex]|A|= 15(s^2 -4)[/tex]
Apply difference of two squares
[tex]|A|= 15(s -2)(s + 2)[/tex]
For the system to have a unique solution;
[tex]|A| =0[/tex]
So, we have:
[tex]15(s -2)(s+2) = 0[/tex]
Divide both sides by 15
[tex](s -2)(s+2) = 0[/tex]
Solve for s
[tex]s -2 = 0\ or\ s +2 = 0[/tex]
[tex]s = 2\ or\ s = -2[/tex]
The result can be combined as:
[tex]s =\±2[/tex]
Hence, the system has a unique solution when [tex]s \ne \±2[/tex]
Next, we solve for s using Cramer's rule.
We have:
[tex]mat\ x_1 = \left[\begin{array}{cc}3&5\\2&5s\end{array}\right][/tex]
Calculate the determinant
[tex]|x_1| = (3 * 5s - 5 *2)[/tex]
[tex]|x_1| = 15s - 10[/tex]
So:
[tex]x_1 =\frac{|x_1|}{|A|}[/tex]
[tex]x_1 = \frac{15s - 10}{15(s -2)(s+2)}[/tex]
Factorize
[tex]x_1 = \frac{5(3s - 2)}{15(s -2)(s+2)}[/tex]
Divide by 5
[tex]x_1 = \frac{3s - 2}{3(s -2)(s+2)}[/tex]
[tex]x_1 = \frac{3s - 2}{3(s^2 -4)}[/tex]
Similarly:
[tex]mat\ x_2 =\left[\begin{array}{cc}3s&3\\12&2\end{array}\right][/tex]
Calculate the determinant
[tex]|x_2| = 3s * 2 - 3 * 12[/tex]
[tex]|x_2| = 6s- 36[/tex]
So:
[tex]x_2 =\frac{|x_2|}{|A|}[/tex]
[tex]x_2 = \frac{6s- 36}{15(s -2)(s+2)}[/tex]
Factorize
[tex]x_2 = \frac{6(s- 6)}{15(s -2)(s+2)}[/tex]
Divide by 3
[tex]x_2 = \frac{2(s- 6)}{5(s -2)(s+2)}[/tex]
[tex]x_2 = \frac{2(s- 6)}{5(s^2 - 4)}[/tex]