Using Cramer's Rule, what is the value of [tex]$x$[/tex] in the solution to the system of equations below?

[tex]\[
\begin{array}{l}
-x - 3y = -3 \\
-2x - 5y = -8
\end{array}
\][/tex]

A. [tex]\(\frac{\left|\begin{array}{cc}-3 & -3 \\ -8 & -5\end{array}\right|}{-1} = -2\)[/tex]

B. [tex]\(\frac{\left|\begin{array}{cc}-3 & -3 \\ -8 & -5\end{array}\right|}{-1} = 9\)[/tex]

C. [tex]\(\left|\begin{array}{cc}-1 & -3 \\ -2 & -8\end{array}\right|\)[/tex]



Answer :

Let's use Cramer's Rule to find the value of [tex]\( x \)[/tex] for the given system of equations:

[tex]\[ \begin{cases} -x - 3y = -3 \\ -2x - 5y = -8 \end{cases} \][/tex]

Cramer's Rule states that for a system of linear equations [tex]\( A \mathbf{x} = \mathbf{b} \)[/tex], the solution for [tex]\( x \)[/tex] can be found using determinants:
[tex]\[ x = \frac{\det(A_x)}{\det(A)} \][/tex]
where [tex]\( \det(A) \)[/tex] is the determinant of the coefficient matrix [tex]\( A \)[/tex], and [tex]\( \det(A_x) \)[/tex] is the determinant of the matrix [tex]\( A \)[/tex] with its [tex]\( x \)[/tex]-column replaced by the constants [tex]\( \mathbf{b} \)[/tex] from the right-hand side.

First, let's identify the coefficient matrix [tex]\( A \)[/tex] and the constants vector [tex]\( \mathbf{b} \)[/tex]:

[tex]\[ A = \begin{pmatrix} -1 & -3 \\ -2 & -5 \end{pmatrix} \][/tex]
[tex]\[ \mathbf{b} = \begin{pmatrix} -3 \\ -8 \end{pmatrix} \][/tex]

Next, we need to find the determinant of matrix [tex]\( A \)[/tex]:

[tex]\[ \det(A) = \begin{vmatrix} -1 & -3 \\ -2 & -5 \end{vmatrix} = (-1)(-5) - (-3)(-2) = 5 - 6 = -1 \][/tex]

Now, we form the matrix [tex]\( A_x \)[/tex] by replacing the first column of [tex]\( A \)[/tex] with the constants [tex]\( \mathbf{b} \)[/tex]:

[tex]\[ A_x = \begin{pmatrix} -3 & -3 \\ -8 & -5 \end{pmatrix} \][/tex]

Next, we calculate the determinant of [tex]\( A_x \)[/tex]:

[tex]\[ \det(A_x) = \begin{vmatrix} -3 & -3 \\ -8 & -5 \end{vmatrix} = (-3)(-5) - (-3)(-8) = 15 - 24 = -9 \][/tex]

Now, using Cramer's Rule, we can find [tex]\( x \)[/tex]:

[tex]\[ x = \frac{\det(A_x)}{\det(A)} = \frac{-9}{-1} = 9 \][/tex]

Therefore, the value of [tex]\( x \)[/tex] is:
[tex]\[ x = 9 \][/tex]