Answer :
To solve the system of equations using matrices and row operations, we begin by representing the system as an augmented matrix:
[tex]\[ \begin{pmatrix} 1 & 6 & 3 & | & 3 \\ 3 & -3 & 9 & | & 9 \\ 4 & 3 & 12 & | & 12 \\ \end{pmatrix} \][/tex]
We will use Gaussian elimination to reduce this matrix to row echelon form and determine the nature of the solutions.
### Step 1: Normalizing the first row
We make the leading coefficient of the first row (row 1) equal to 1. This is already the case, so we do not need to do anything to row 1.
### Step 2: Eliminating the first column entries below the leading 1
Subtract 3 times the first row from the second row:
[tex]\[ R2 = R2 - 3R1 \][/tex]
[tex]\[ R2 = [3, -3, 9, |, 9] - 3 [1, 6, 3, |, 3] = [3 - 3, -3 - 18, 9 - 9, |, 9 - 9] = [0, -21, 0, |, 0] \][/tex]
Subtract 4 times the first row from the third row:
[tex]\[ R3 = R3 - 4R1 \][/tex]
[tex]\[ R3 = [4, 3, 12, |, 12] - 4 [1, 6, 3, |, 3] = [4 - 4, 3 - 24, 12 - 12, |, 12 - 12] = [0, -21, 0, |, 0] \][/tex]
The augmented matrix becomes:
[tex]\[ \begin{pmatrix} 1 & 6 & 3 & | & 3 \\ 0 & -21 & 0 & | & 0 \\ 0 & -21 & 0 & | & 0 \\ \end{pmatrix} \][/tex]
### Step 3: Normalizing the second row
Divide the second row by -21:
[tex]\[ R2 = \frac{R2}{-21} = [0, -21, 0, |, 0] \div -21 = [0, 1, 0, |, 0] \][/tex]
Normalize the augmented matrix:
[tex]\[ \begin{pmatrix} 1 & 6 & 3 & | & 3 \\ 0 & 1 & 0 & | & 0 \\ 0 & -21 & 0 & | & 0 \\ \end{pmatrix} \][/tex]
### Step 4: Eliminating the second column in the third row
Subtract the second row from the third row:
[tex]\[ R3 = R3 - R2 = [0, -21, 0, |, 0] - [0, 1, 0, |, 0] = [0, 0, 0, |, 0] \][/tex]
Our final matrix becomes:
[tex]\[ \begin{pmatrix} 1 & 6 & 3 & | & 3 \\ 0 & 1 & 0 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ \end{pmatrix} \][/tex]
### Interpreting the rows
From the reduced matrix, we have the equivalent system:
[tex]\[ \left\{ \begin{aligned} x + 6y + 3z &= 3 \\ y &= 0 \\ 0 &= 0 \\ \end{aligned} \right. \][/tex]
From the second equation, [tex]\( y = 0 \)[/tex].
Substituting [tex]\( y = 0 \)[/tex] into the first equation:
[tex]\[ x + 6(0) + 3z = 3 \implies x + 3z = 3 \implies x = 3 - 3z \][/tex]
Thus, [tex]\( x = 3 - 3z \)[/tex], [tex]\( y = 0 \)[/tex], and [tex]\( z \)[/tex] can be any real number.
Given this analysis, the system has infinitely many solutions, which can be written as:
[tex]\[ \{(x, y, z) \mid x = 3 - 3z, y = 0, z \text{ is any real number} \} \][/tex]
Select the appropriate choice:
B. There are infinitely many solutions. The solution can be written as [tex]\(\{(x, y, z) \mid x=3-3z, y=0, z\)[/tex] is any real number \}.
[tex]\[ \begin{pmatrix} 1 & 6 & 3 & | & 3 \\ 3 & -3 & 9 & | & 9 \\ 4 & 3 & 12 & | & 12 \\ \end{pmatrix} \][/tex]
We will use Gaussian elimination to reduce this matrix to row echelon form and determine the nature of the solutions.
### Step 1: Normalizing the first row
We make the leading coefficient of the first row (row 1) equal to 1. This is already the case, so we do not need to do anything to row 1.
### Step 2: Eliminating the first column entries below the leading 1
Subtract 3 times the first row from the second row:
[tex]\[ R2 = R2 - 3R1 \][/tex]
[tex]\[ R2 = [3, -3, 9, |, 9] - 3 [1, 6, 3, |, 3] = [3 - 3, -3 - 18, 9 - 9, |, 9 - 9] = [0, -21, 0, |, 0] \][/tex]
Subtract 4 times the first row from the third row:
[tex]\[ R3 = R3 - 4R1 \][/tex]
[tex]\[ R3 = [4, 3, 12, |, 12] - 4 [1, 6, 3, |, 3] = [4 - 4, 3 - 24, 12 - 12, |, 12 - 12] = [0, -21, 0, |, 0] \][/tex]
The augmented matrix becomes:
[tex]\[ \begin{pmatrix} 1 & 6 & 3 & | & 3 \\ 0 & -21 & 0 & | & 0 \\ 0 & -21 & 0 & | & 0 \\ \end{pmatrix} \][/tex]
### Step 3: Normalizing the second row
Divide the second row by -21:
[tex]\[ R2 = \frac{R2}{-21} = [0, -21, 0, |, 0] \div -21 = [0, 1, 0, |, 0] \][/tex]
Normalize the augmented matrix:
[tex]\[ \begin{pmatrix} 1 & 6 & 3 & | & 3 \\ 0 & 1 & 0 & | & 0 \\ 0 & -21 & 0 & | & 0 \\ \end{pmatrix} \][/tex]
### Step 4: Eliminating the second column in the third row
Subtract the second row from the third row:
[tex]\[ R3 = R3 - R2 = [0, -21, 0, |, 0] - [0, 1, 0, |, 0] = [0, 0, 0, |, 0] \][/tex]
Our final matrix becomes:
[tex]\[ \begin{pmatrix} 1 & 6 & 3 & | & 3 \\ 0 & 1 & 0 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ \end{pmatrix} \][/tex]
### Interpreting the rows
From the reduced matrix, we have the equivalent system:
[tex]\[ \left\{ \begin{aligned} x + 6y + 3z &= 3 \\ y &= 0 \\ 0 &= 0 \\ \end{aligned} \right. \][/tex]
From the second equation, [tex]\( y = 0 \)[/tex].
Substituting [tex]\( y = 0 \)[/tex] into the first equation:
[tex]\[ x + 6(0) + 3z = 3 \implies x + 3z = 3 \implies x = 3 - 3z \][/tex]
Thus, [tex]\( x = 3 - 3z \)[/tex], [tex]\( y = 0 \)[/tex], and [tex]\( z \)[/tex] can be any real number.
Given this analysis, the system has infinitely many solutions, which can be written as:
[tex]\[ \{(x, y, z) \mid x = 3 - 3z, y = 0, z \text{ is any real number} \} \][/tex]
Select the appropriate choice:
B. There are infinitely many solutions. The solution can be written as [tex]\(\{(x, y, z) \mid x=3-3z, y=0, z\)[/tex] is any real number \}.
