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]

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