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To determine the number of solutions for each system of linear equations, we can use the concept of the rank of the coefficient matrix and the augmented matrix. Specifically, we can analyze the system using the following criteria:
1. If the rank of the coefficient matrix (A) is equal to the rank of the augmented matrix (A|B) and is equal to the number of variables, the system has one solution.
2. If the rank of the coefficient matrix (A) is equal to the rank of the augmented matrix (A|B) but is less than the number of variables, the system has infinitely many solutions.
3. If the rank of the coefficient matrix (A) is not equal to the rank of the augmented matrix (A|B), the system has no solution.
Let's analyze each system step by step.
### System 1:
[tex]\[ \begin{cases} 3x - 2y = 3 \\ 6x - 4y = 1 \end{cases} \][/tex]
We first convert this to its augmented matrix form:
[tex]\[ \left[\begin{array}{cc|c} 3 & -2 & 3 \\ 6 & -4 & 1 \\ \end{array}\right] \][/tex]
Next, we perform row reduction to echelon form:
1. Multiply the first row by 2 and subtract it from the second row:
[tex]\[ \left[\begin{array}{cc|c} 3 & -2 & 3 \\ 0 & 0 & -5 \end{array}\right] \][/tex]
We see that the second row forms an inconsistent equation [tex]\(0 = -5\)[/tex], which is impossible. So, the system has:
No Solution
### System 2:
[tex]\[ \begin{cases} 3x - 5y = 8 \\ 5x - 3y = 2 \end{cases} \][/tex]
We convert this to its augmented matrix form:
[tex]\[ \left[\begin{array}{cc|c} 3 & -5 & 8 \\ 5 & -3 & 2 \\ \end{array}\right] \][/tex]
Next, we perform row reduction to echelon form:
1. Multiply the first row by 5:
[tex]\[ \left[\begin{array}{cc|c} 15 & -25 & 40 \\ 5 & -3 & 2 \\ \end{array}\right] \][/tex]
2. Subtract the first row (multiplied by 3) from the second row:
[tex]\[ \left[\begin{array}{cc|c} 15 & -25 & 40 \\ 0 & 32 & -38 \\ \end{array}\right] \][/tex]
We obtain a consistent system with unique solutions. Hence, the system has:
One Solution
### System 3:
[tex]\[ \begin{cases} 3x + 2y = 8 \\ 4x + 3y = 1 \end{cases} \][/tex]
We convert this to its augmented matrix form:
[tex]\[ \left[\begin{array}{cc|c} 3 & 2 & 8 \\ 4 & 3 & 1 \\ \end{array}\right] \][/tex]
Next, we perform row reduction to echelon form:
1. Multiply the first row by 4 and the second row by 3:
[tex]\[ \left[\begin{array}{cc|c} 12 & 8 & 32 \\ 12 & 9 & 3 \\ \end{array}\right] \][/tex]
2. Subtract the first row from the second row:
[tex]\[ \left[\begin{array}{cc|c} 3 & 2 & 8 \\ 0 & 1 & -29 \end{array}\right] \][/tex]
This system is consistent and has unique solutions. Hence, the system has:
One Solution
### System 4:
[tex]\[ \begin{cases} 3x - 6y = 3 \\ 2x - 4y = 2 \end{cases} \][/tex]
We convert this to its augmented matrix form:
[tex]\[ \left[\begin{array}{cc|c} 3 & -6 & 3 \\ 2 & -4 & 2 \\ \end{array}\right] \][/tex]
Next, we perform row reduction to echelon form:
1. Notice the second row is proportional to the first row (divide by 2):
[tex]\[ \left[\begin{array}{cc|c} 3 & -6 & 3 \\ 3 & -6 & 3 \\ \end{array}\right] \][/tex]
This system has infinitely many solutions (dependent equations), hence:
Infinitely Many Solutions
### System 5:
[tex]\[ \begin{cases} 3x - 4y = 2 \\ 6x - 8y = 1 \end{cases} \][/tex]
We convert this to its augmented matrix form:
[tex]\[ \left[\begin{array}{cc|c} 3 & -4 & 2 \\ 6 & -8 & 1 \\ \end{array}\right] \][/tex]
Next, we perform row reduction to echelon form:
1. Notice the second row is proportional to the first row (divide by 2):
[tex]\[ \left[\begin{array}{cc|c} 3 & -4 & 2 \\ 6 & -8 & 4 \\ \end{array}\right] \][/tex]
We see that the second row forms an inconsistent equation [tex]\(0 = -3\)[/tex], which is impossible. So, the system has:
No Solution
To summarize:
1. [tex]\(3x - 2y = 3\)[/tex] and [tex]\(6x - 4y = 1\)[/tex]: No Solution
2. [tex]\(3x - 5y = 8\)[/tex] and [tex]\(5x - 3y = 2\)[/tex]: One Solution
3. [tex]\(3x + 2y = 8\)[/tex] and [tex]\(4x + 3y = 1\)[/tex]: One Solution
4. [tex]\(3x - 6y = 3\)[/tex] and [tex]\(2x - 4y = 2\)[/tex]: Infinitely Many Solutions
5. [tex]\(3x - 4y = 2\)[/tex] and [tex]\(6x - 8y = 1\)[/tex]: No Solution
1. If the rank of the coefficient matrix (A) is equal to the rank of the augmented matrix (A|B) and is equal to the number of variables, the system has one solution.
2. If the rank of the coefficient matrix (A) is equal to the rank of the augmented matrix (A|B) but is less than the number of variables, the system has infinitely many solutions.
3. If the rank of the coefficient matrix (A) is not equal to the rank of the augmented matrix (A|B), the system has no solution.
Let's analyze each system step by step.
### System 1:
[tex]\[ \begin{cases} 3x - 2y = 3 \\ 6x - 4y = 1 \end{cases} \][/tex]
We first convert this to its augmented matrix form:
[tex]\[ \left[\begin{array}{cc|c} 3 & -2 & 3 \\ 6 & -4 & 1 \\ \end{array}\right] \][/tex]
Next, we perform row reduction to echelon form:
1. Multiply the first row by 2 and subtract it from the second row:
[tex]\[ \left[\begin{array}{cc|c} 3 & -2 & 3 \\ 0 & 0 & -5 \end{array}\right] \][/tex]
We see that the second row forms an inconsistent equation [tex]\(0 = -5\)[/tex], which is impossible. So, the system has:
No Solution
### System 2:
[tex]\[ \begin{cases} 3x - 5y = 8 \\ 5x - 3y = 2 \end{cases} \][/tex]
We convert this to its augmented matrix form:
[tex]\[ \left[\begin{array}{cc|c} 3 & -5 & 8 \\ 5 & -3 & 2 \\ \end{array}\right] \][/tex]
Next, we perform row reduction to echelon form:
1. Multiply the first row by 5:
[tex]\[ \left[\begin{array}{cc|c} 15 & -25 & 40 \\ 5 & -3 & 2 \\ \end{array}\right] \][/tex]
2. Subtract the first row (multiplied by 3) from the second row:
[tex]\[ \left[\begin{array}{cc|c} 15 & -25 & 40 \\ 0 & 32 & -38 \\ \end{array}\right] \][/tex]
We obtain a consistent system with unique solutions. Hence, the system has:
One Solution
### System 3:
[tex]\[ \begin{cases} 3x + 2y = 8 \\ 4x + 3y = 1 \end{cases} \][/tex]
We convert this to its augmented matrix form:
[tex]\[ \left[\begin{array}{cc|c} 3 & 2 & 8 \\ 4 & 3 & 1 \\ \end{array}\right] \][/tex]
Next, we perform row reduction to echelon form:
1. Multiply the first row by 4 and the second row by 3:
[tex]\[ \left[\begin{array}{cc|c} 12 & 8 & 32 \\ 12 & 9 & 3 \\ \end{array}\right] \][/tex]
2. Subtract the first row from the second row:
[tex]\[ \left[\begin{array}{cc|c} 3 & 2 & 8 \\ 0 & 1 & -29 \end{array}\right] \][/tex]
This system is consistent and has unique solutions. Hence, the system has:
One Solution
### System 4:
[tex]\[ \begin{cases} 3x - 6y = 3 \\ 2x - 4y = 2 \end{cases} \][/tex]
We convert this to its augmented matrix form:
[tex]\[ \left[\begin{array}{cc|c} 3 & -6 & 3 \\ 2 & -4 & 2 \\ \end{array}\right] \][/tex]
Next, we perform row reduction to echelon form:
1. Notice the second row is proportional to the first row (divide by 2):
[tex]\[ \left[\begin{array}{cc|c} 3 & -6 & 3 \\ 3 & -6 & 3 \\ \end{array}\right] \][/tex]
This system has infinitely many solutions (dependent equations), hence:
Infinitely Many Solutions
### System 5:
[tex]\[ \begin{cases} 3x - 4y = 2 \\ 6x - 8y = 1 \end{cases} \][/tex]
We convert this to its augmented matrix form:
[tex]\[ \left[\begin{array}{cc|c} 3 & -4 & 2 \\ 6 & -8 & 1 \\ \end{array}\right] \][/tex]
Next, we perform row reduction to echelon form:
1. Notice the second row is proportional to the first row (divide by 2):
[tex]\[ \left[\begin{array}{cc|c} 3 & -4 & 2 \\ 6 & -8 & 4 \\ \end{array}\right] \][/tex]
We see that the second row forms an inconsistent equation [tex]\(0 = -3\)[/tex], which is impossible. So, the system has:
No Solution
To summarize:
1. [tex]\(3x - 2y = 3\)[/tex] and [tex]\(6x - 4y = 1\)[/tex]: No Solution
2. [tex]\(3x - 5y = 8\)[/tex] and [tex]\(5x - 3y = 2\)[/tex]: One Solution
3. [tex]\(3x + 2y = 8\)[/tex] and [tex]\(4x + 3y = 1\)[/tex]: One Solution
4. [tex]\(3x - 6y = 3\)[/tex] and [tex]\(2x - 4y = 2\)[/tex]: Infinitely Many Solutions
5. [tex]\(3x - 4y = 2\)[/tex] and [tex]\(6x - 8y = 1\)[/tex]: No Solution
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