To determine if each linear system has a unique solution, we can use the determinant of the coefficient matrix for each system. If the determinant is non-zero, the system has a unique solution. If the determinant is zero, the system does not have a unique solution.
Let's go through each system step by step.
### First System
[tex]\[
\begin{array}{l}
x + 3y = 4 \\
3x - y = 5
\end{array}
\][/tex]
The coefficient matrix for this system is:
[tex]\[
\begin{pmatrix}
1 & 3 \\
3 & -1
\end{pmatrix}
\][/tex]
The determinant of this 2x2 matrix is calculated as:
[tex]\[
\text{det} = (1 \cdot -1) - (3 \cdot 3) = -1 - 9 = -10
\][/tex]
Since the determinant [tex]\(\text{det} = -10\)[/tex] is non-zero, the system has a unique solution.
### Second System
[tex]\[
\begin{array}{l}
x + 2y - z = 8 \\
x - y + 2z = 0 \\
2x - 3y + z = 1
\end{array}
\][/tex]
The coefficient matrix for this system is:
[tex]\[
\begin{pmatrix}
1 & 2 & -1 \\
1 & -1 & 2 \\
2 & -3 & 1
\end{pmatrix}
\][/tex]
The determinant of this 3x3 matrix is calculated as:
[tex]\[
\text{det} =
\begin{vmatrix}
1 & 2 & -1 \\
1 & -1 & 2 \\
2 & -3 & 1
\end{vmatrix} = 12
\][/tex]
Since the determinant [tex]\(\text{det} = 12\)[/tex] is non-zero, the system has a unique solution.
### Third System
[tex]\[
\begin{array}{l}
x + 3y + z = 4 \\
2x + 6y + 2z = 5 \\
x + y + z = 0
\end{array}
\][/tex]
The coefficient matrix for this system is:
[tex]\[
\begin{pmatrix}
1 & 3 & 1 \\
2 & 6 & 2 \\
1 & 1 & 1
\end{pmatrix}
\][/tex]
The determinant of this 3x3 matrix is calculated as:
[tex]\[
\text{det} =
\begin{vmatrix}
1 & 3 & 1 \\
2 & 6 & 2 \\
1 & 1 & 1
\end{vmatrix} = 0
\][/tex]
Since the determinant [tex]\(\text{det} = 0\)[/tex] is zero, the system does not have a unique solution.
### Summary
- The first system has a unique solution (det = -10).
- The second system has a unique solution (det = 12).
- The third system does not have a unique solution (det = 0).
