Answer :
Alright, let's solve this system of linear equations using Gaussian elimination to determine if the system has one solution, infinitely many solutions, or no solution at all.
We start with the system of linear equations:
[tex]\[ \begin{array}{rcl} 5x + 18y - 15z &=& 18 \\ 3x + 12y - 11z &=& 8 \\ x + 3y - 2z &=& 5 \end{array} \][/tex]
We convert this system into an augmented matrix:
[tex]\[ \left[\begin{array}{ccc|c} 5 & 18 & -15 & 18 \\ 3 & 12 & -11 & 8 \\ 1 & 3 & -2 & 5 \end{array}\right] \][/tex]
Next, we apply Gaussian elimination to transform this matrix into reduced row-echelon form (RREF). After performing the row operations, the augmented matrix in RREF looks like this:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 3 & 12 \\ 0 & 1 & -\frac{5}{3} & -\frac{7}{3} \\ 0 & 0 & 0 & 0 \end{array}\right] \][/tex]
Let's interpret the resulting matrix:
1. The first row represents:
[tex]\[ x + 3z = 12 \][/tex]
2. The second row represents:
[tex]\[ y - \frac{5}{3}z = -\frac{7}{3} \][/tex]
3. The third row is all zeros, which indicates that it's a free row and does not provide any new information about the variables.
Since we have one row (third row) full of zeros, it implies that the system has infinitely many solutions.
To express these solutions, we solve the equations derived from the first and second rows:
From the first row:
[tex]\[ x + 3z = 12 \implies x = 12 - 3z \][/tex]
From the second row:
[tex]\[ y - \frac{5}{3}z = -\frac{7}{3} \implies y = -\frac{7}{3} + \frac{5}{3}z \][/tex]
Now we can describe the solution set in terms of the free variable [tex]\(z\)[/tex]:
[tex]\[ (x, y, z) = \left(12 - 3z, -\frac{7}{3} + \frac{5}{3}z, z\right) \][/tex]
So the solution set can be expressed as:
[tex]\[ \{ (12 - 3z, -\frac{7}{3} + \frac{5}{3}z, z) \mid z \text{ is any real number}\} \][/tex]
Therefore, the correct choice is:
B. There are infinitely many solutions. The solution set is [tex]\(\{(12 - 3z, -\frac{7}{3} + \frac{5}{3}z, z)\}\)[/tex], where [tex]\(z\)[/tex] is any real number.
We start with the system of linear equations:
[tex]\[ \begin{array}{rcl} 5x + 18y - 15z &=& 18 \\ 3x + 12y - 11z &=& 8 \\ x + 3y - 2z &=& 5 \end{array} \][/tex]
We convert this system into an augmented matrix:
[tex]\[ \left[\begin{array}{ccc|c} 5 & 18 & -15 & 18 \\ 3 & 12 & -11 & 8 \\ 1 & 3 & -2 & 5 \end{array}\right] \][/tex]
Next, we apply Gaussian elimination to transform this matrix into reduced row-echelon form (RREF). After performing the row operations, the augmented matrix in RREF looks like this:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 3 & 12 \\ 0 & 1 & -\frac{5}{3} & -\frac{7}{3} \\ 0 & 0 & 0 & 0 \end{array}\right] \][/tex]
Let's interpret the resulting matrix:
1. The first row represents:
[tex]\[ x + 3z = 12 \][/tex]
2. The second row represents:
[tex]\[ y - \frac{5}{3}z = -\frac{7}{3} \][/tex]
3. The third row is all zeros, which indicates that it's a free row and does not provide any new information about the variables.
Since we have one row (third row) full of zeros, it implies that the system has infinitely many solutions.
To express these solutions, we solve the equations derived from the first and second rows:
From the first row:
[tex]\[ x + 3z = 12 \implies x = 12 - 3z \][/tex]
From the second row:
[tex]\[ y - \frac{5}{3}z = -\frac{7}{3} \implies y = -\frac{7}{3} + \frac{5}{3}z \][/tex]
Now we can describe the solution set in terms of the free variable [tex]\(z\)[/tex]:
[tex]\[ (x, y, z) = \left(12 - 3z, -\frac{7}{3} + \frac{5}{3}z, z\right) \][/tex]
So the solution set can be expressed as:
[tex]\[ \{ (12 - 3z, -\frac{7}{3} + \frac{5}{3}z, z) \mid z \text{ is any real number}\} \][/tex]
Therefore, the correct choice is:
B. There are infinitely many solutions. The solution set is [tex]\(\{(12 - 3z, -\frac{7}{3} + \frac{5}{3}z, z)\}\)[/tex], where [tex]\(z\)[/tex] is any real number.
