Sure! Let's solve the given system of linear equations using determinants and the inverse of the coefficient matrix.
The system of equations is:
[tex]\[
\begin{aligned}
3x + 2y + 2 &= 6, \\
x - 2y + 32 &= 2, \\
5x + 2y - 42 &= 3.
\end{aligned}
\][/tex]
First, we need to rewrite each equation in the standard form [tex]\(Ax = B\)[/tex], where [tex]\(A\)[/tex] is the matrix of coefficients, [tex]\(x\)[/tex] is the column matrix of variables, and [tex]\(B\)[/tex] is the column matrix of constants.
Rewriting the equations:
[tex]\[
\begin{aligned}
3x + 2y &= 4, \\
x - 2y &= -30, \\
5x + 2y &= 45.
\end{aligned}
\][/tex]
The coefficient matrix [tex]\(A\)[/tex] and constants matrix [tex]\(B\)[/tex] are:
[tex]\[
A = \begin{pmatrix}
3 & 2 \\
1 & -2 \\
5 & 2
\end{pmatrix}, \quad
B = \begin{pmatrix}
4 \\
-30 \\
45
\end{pmatrix}
\][/tex]
Since this is a 3x3 system, let's set up the augmented matrix and proceed using the inverse matrix method:
1. Find the determinant of [tex]\(A\)[/tex]:
The determinant of a matrix [tex]\(A\)[/tex] is denoted as [tex]\(|A|\)[/tex] and for a 3x3 matrix, it is calculated as follows:
[tex]\[
|A| = \begin{vmatrix}
3 & 2 & 0 \\
1 & -2 & 0 \\
5 & 2 & 0
\end{vmatrix}
\][/tex]
To find the determinant, let's expand along the third column (noting that the elements in the third column are all zeros). With all zeros in one column, the determinant of matrix [tex]\(A\)[/tex] is:
[tex]\[
|A| = 0
\][/tex]
A determinant of zero implies that the matrix [tex]\(A\)[/tex] is singular, which means it does not have an inverse. Therefore, the system of equations does not have a unique solution; it may either have no solution or an infinite number of solutions.
Since the determinant is zero, we face a special case. Let’s look into a deeper scrutiny:
2. Determine if the system has infinitely many solutions or no solution:
To do this, we can convert the system into row-echelon form and check for consistency.
[tex]\[
\begin{aligned}
3x + 2y &= 4 &\quad (1) \\
x - 2y &= -30 &\quad (2) \\
5x + 2y &= 45 &\quad (3)
\end{aligned}
\][/tex]
Let's eliminate [tex]\(y\)[/tex] from equations (1) and (3):
From equation (2): [tex]\( x = -30 + 2y \)[/tex].
Substitute [tex]\(x\)[/tex] in equations (1) and (3):
[tex]\[
\begin{aligned}
3(-30 + 2y) + 2y &= 4 \\
-90 + 6y + 2y &= 4 \\
8y &= 94 \\
y &= \frac{94}{8} = 11.75
\end{aligned}
\][/tex]
Substitute [tex]\(y = 11.75\)[/tex] back into [tex]\(x = -30 + 2y\)[/tex]:
[tex]\[
x = -30 + 2(11.75) = -30 + 23.5 = -6.5
\][/tex]
Let's check this solution in the third equation:
[tex]\[
5(-6.5) + 2(11.75) = -32.5 + 23.5 = -9
\][/tex]
This seems to yield a contradiction as the calculated value does not satisfy the third equation. Therefore, revising the steps show the system doesn't have a consistent solution, concluding the discussed system of equations does not provide a correct solution with the set values.
However, knowing that determinant of zero suggests parallel or overly determined system needing more breed calculation effort.
