To determine if the given system of equations has a unique solution, we can use the concepts of linear algebra, specifically by looking at the determinants. Here's a step-by-step explanation of how you can determine whether there is a unique solution:
1. Rewrite the Equations in Standard Form:
The given system of equations is:
[tex]\[
\begin{aligned}
2x + y &= 4 \quad \text{(Equation 1)} \\
2y &= 6 - 2x \quad \text{(Equation 2)}
\end{aligned}
\][/tex]
To standardize Equation 2 for comparison, we rewrite it in the form [tex]\(ax + by = c\)[/tex]:
Rearrange Equation 2:
[tex]\[
2y = 6 - 2x \implies 2y + 2x = 6 \implies 2x + 2y = 6 \quad \text{(Equation 2')}
\][/tex]
2. Write the Coefficient Matrix and the Constant Vector:
From Equation 1 and Equation 2', the system can be rewritten in matrix form as:
[tex]\[
\begin{aligned}
2x + y &= 4 \\
2x + 2y &= 6
\end{aligned}
\][/tex]
The coefficient matrix [tex]\(A\)[/tex] and the constant vector [tex]\(\mathbf{b}\)[/tex] are:
[tex]\[
A = \begin{pmatrix}
2 & 1 \\
2 & 2
\end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix}
4 \\
6
\end{pmatrix}
\][/tex]
3. Calculate the Determinant of the Coefficient Matrix:
The determinant of matrix [tex]\(A\)[/tex] will tell us whether the system has a unique solution. If the determinant is non-zero, the system has a unique solution. If it's zero, the system does not have a unique solution.
To find the determinant of matrix [tex]\(A\)[/tex]:
[tex]\[
\text{det}(A) = \begin{vmatrix}
2 & 1 \\
2 & 2
\end{vmatrix} = (2 \cdot 2) - (1 \cdot 2) = 4 - 2 = 2
\][/tex]
The determinant of the coefficient matrix [tex]\(A\)[/tex] is 2.
4. Interpret the Determinant:
Since the determinant is non-zero ([tex]\(\text{det}(A) = 2\)[/tex]), we can conclude that the system of equations has a unique solution.
Thus, the system [tex]\(\begin{aligned} 2x + y &= 4 \\ 2y &= 6 - 2x \end{aligned}\)[/tex] has one unique solution.
