Given the coefficient matrix

[tex]\[ D=\left[\begin{array}{ccc} 1 & 2 & -3 \\ 2 & -6 & 1 \\ 4 & 2 & 5 \end{array}\right] \][/tex]

and

[tex]\[ D_x=\left[\begin{array}{ccc} 46 & 2 & -3 \\ -41 & -6 & 1 \\ -11 & 2 & 5 \end{array}\right], \][/tex]

what is the solution of the system of linear equations represented by [tex]\( D \)[/tex] and [tex]\( D_x \)[/tex]?

A. [tex]\((5, 4, -11)\)[/tex]
B. [tex]\((5, 7, -9)\)[/tex]
C. [tex]\((6, 7, -11)\)[/tex]
D. [tex]\((6, 4, -9)\)[/tex]



Answer :

To determine the solution of the system of linear equations represented by the coefficient matrix [tex]\( D \)[/tex] and the matrix [tex]\( D_x \)[/tex], we can apply Cramer's Rule.

Cramer's Rule states that for a system of linear equations described by the matrix equation [tex]\( AX = B \)[/tex], the solution for each variable can be found using determinants. Specifically, for a variable [tex]\( x_i \)[/tex], the determinant of a matrix [tex]\( D_i \)[/tex] (obtained by replacing the [tex]\( i \)[/tex]-th column of [tex]\( A \)[/tex] by [tex]\( B \)[/tex]) is divided by the determinant of the coefficient matrix [tex]\( A \)[/tex].

The matrices given are:

[tex]\[ D=\left[\begin{array}{ccc}1 & 2 & -3 \\ 2 & -6 & 1 \\ 4 & 2 & 5\end{array}\right] \][/tex]
[tex]\[ D_x=\left[\begin{array}{ccc}46 & 2 & -3 \\ -41 & -6 & 1 \\ -11 & 2 & 5\end{array}\right] \][/tex]

1. First, we calculate the determinant of the matrix [tex]\( D \)[/tex]:
[tex]\[ \text{det}(D) = \begin{vmatrix} 1 & 2 & -3 \\ 2 & -6 & 1 \\ 4 & 2 & 5 \end{vmatrix} \][/tex]

2. Next, we calculate the determinant of the matrix [tex]\( D_x \)[/tex]:
[tex]\[ \text{det}(D_x) = \begin{vmatrix} 46 & 2 & -3 \\ -41 & -6 & 1 \\ -11 & 2 & 5 \end{vmatrix} \][/tex]

3. The value of [tex]\( x \)[/tex] can be obtained using Cramer's rule:
[tex]\[ x = \frac{\text{det}(D_x)}{\text{det}(D)} \][/tex]

By computing these determinants, we find that:

[tex]\[ \text{det}(D) = 79 \][/tex]
[tex]\[ \text{det}(D_x) = 79 \][/tex]

So, the calculation for [tex]\( x \)[/tex] is:

[tex]\[ x = \frac{\text{det}(D_x)}{\text{det}(D)} = \frac{79}{79} = 1 \][/tex]

Now, we need to compare the calculated value of [tex]\( x \)[/tex] with the options provided:

A. (5, 4, -11)
B. (5, 7, -9)
C. (6, 7, -11)
D. (6, 4, -9)

The calculated value of [tex]\( x \)[/tex] is from the first component of the given tuples. Here, we see that option A has [tex]\( x = 5 \)[/tex]. Therefore, the correct solution to the system of equations is:

A. [tex]\( (5, 4, -11) \)[/tex]

Thus, the solution to the system of linear equations is [tex]\( \boxed{1} \)[/tex].