Answer :
Sure, let's go through the steps to solve this system of linear equations.
We are given the following system of equations:
[tex]\[ \left\{ \begin{aligned} x - y + 2z &= -2 \\ 5x + 2y + 9z &= -11 \\ x + 6y + z &= -3 \end{aligned} \right. \][/tex]
Step 1: Represent the system in matrix form
We can write the system in the form of [tex]\( A\mathbf{x} = \mathbf{b} \)[/tex], where [tex]\( A \)[/tex] is the coefficient matrix, [tex]\( \mathbf{x} \)[/tex] is the column vector of variables, and [tex]\( \mathbf{b} \)[/tex] is the column vector of constants.
[tex]\[ A = \begin{pmatrix} 1 & -1 & 2 \\ 5 & 2 & 9 \\ 1 & 6 & 1 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} -2 \\ -11 \\ -3 \end{pmatrix} \][/tex]
Step 2: Determine the rank of the matrix
To determine if the system has a solution, we need to check the rank of the coefficient matrix [tex]\( A \)[/tex] and the rank of the augmented matrix [tex]\( [A | \mathbf{b}] \)[/tex].
The rank of the coefficient matrix [tex]\( A \)[/tex] and the augmented matrix are both found to be 2.
Since the rank of [tex]\( A \)[/tex] (2) is equal to the rank of the augmented matrix [tex]\( [A | \mathbf{b}] \)[/tex] (2), and the rank is less than the number of variables (3), this indicates that the system has infinitely many solutions.
Step 3: Parameterize the solution
We will express [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in terms of [tex]\( z \)[/tex].
1. Let's take the first two equations:
[tex]\[ x - y + 2z = -2 \tag{1} \][/tex]
[tex]\[ 5x + 2y + 9z = -11 \tag{2} \][/tex]
Solve them step-by-step. From equation (1), we can express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] and [tex]\( z \)[/tex]:
[tex]\[ y = x + 2z + 2 \tag{3} \][/tex]
2. Substitute equation (3) into equation (2):
[tex]\[ 5x + 2(x + 2z + 2) + 9z = -11 \][/tex]
Simplify the equation:
[tex]\[ 5x + 2x + 4z + 4 + 9z = -11 \][/tex]
[tex]\[ 7x + 13z = -15 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x = -\frac{13}{7}z - \frac{15}{7} \tag{4} \][/tex]
3. Substitute equation (4) into equation (3) to get [tex]\( y \)[/tex]:
[tex]\[ y = -\frac{13}{7}z - \frac{15}{7} + 2z + 2 \][/tex]
Simplify:
[tex]\[ y = -\frac{13}{7}z + \frac{14}{7}z - \frac{15}{7} + \frac{14}{7} \][/tex]
[tex]\[ y = \frac{1}{7}z - \frac{1}{7} \tag{5} \][/tex]
So, the solutions are parameterized as:
[tex]\[ x = -\frac{13}{7}z - \frac{15}{7} \][/tex]
[tex]\[ y = \frac{1}{7}z - \frac{1}{7} \][/tex]
Since the system has infinitely many solutions, the correct choice is:
Choice A. There are infinitely many solutions. If [tex]\( z \)[/tex] is allowed to be any real number, then
[tex]\[ x = -\frac{13}{7}z - \frac{15}{7} \][/tex]
and
[tex]\[ y = \frac{1}{7}z - \frac{1}{7} \][/tex]
So, the final answer is:
A. There are infinitely many solutions. If [tex]\( z \)[/tex] is allowed to be any real number, then [tex]\( x = -\frac{13}{7}z - \frac{15}{7} \)[/tex] and [tex]\( y = \frac{1}{7}z - \frac{1}{7} \)[/tex].
We are given the following system of equations:
[tex]\[ \left\{ \begin{aligned} x - y + 2z &= -2 \\ 5x + 2y + 9z &= -11 \\ x + 6y + z &= -3 \end{aligned} \right. \][/tex]
Step 1: Represent the system in matrix form
We can write the system in the form of [tex]\( A\mathbf{x} = \mathbf{b} \)[/tex], where [tex]\( A \)[/tex] is the coefficient matrix, [tex]\( \mathbf{x} \)[/tex] is the column vector of variables, and [tex]\( \mathbf{b} \)[/tex] is the column vector of constants.
[tex]\[ A = \begin{pmatrix} 1 & -1 & 2 \\ 5 & 2 & 9 \\ 1 & 6 & 1 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} -2 \\ -11 \\ -3 \end{pmatrix} \][/tex]
Step 2: Determine the rank of the matrix
To determine if the system has a solution, we need to check the rank of the coefficient matrix [tex]\( A \)[/tex] and the rank of the augmented matrix [tex]\( [A | \mathbf{b}] \)[/tex].
The rank of the coefficient matrix [tex]\( A \)[/tex] and the augmented matrix are both found to be 2.
Since the rank of [tex]\( A \)[/tex] (2) is equal to the rank of the augmented matrix [tex]\( [A | \mathbf{b}] \)[/tex] (2), and the rank is less than the number of variables (3), this indicates that the system has infinitely many solutions.
Step 3: Parameterize the solution
We will express [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in terms of [tex]\( z \)[/tex].
1. Let's take the first two equations:
[tex]\[ x - y + 2z = -2 \tag{1} \][/tex]
[tex]\[ 5x + 2y + 9z = -11 \tag{2} \][/tex]
Solve them step-by-step. From equation (1), we can express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] and [tex]\( z \)[/tex]:
[tex]\[ y = x + 2z + 2 \tag{3} \][/tex]
2. Substitute equation (3) into equation (2):
[tex]\[ 5x + 2(x + 2z + 2) + 9z = -11 \][/tex]
Simplify the equation:
[tex]\[ 5x + 2x + 4z + 4 + 9z = -11 \][/tex]
[tex]\[ 7x + 13z = -15 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x = -\frac{13}{7}z - \frac{15}{7} \tag{4} \][/tex]
3. Substitute equation (4) into equation (3) to get [tex]\( y \)[/tex]:
[tex]\[ y = -\frac{13}{7}z - \frac{15}{7} + 2z + 2 \][/tex]
Simplify:
[tex]\[ y = -\frac{13}{7}z + \frac{14}{7}z - \frac{15}{7} + \frac{14}{7} \][/tex]
[tex]\[ y = \frac{1}{7}z - \frac{1}{7} \tag{5} \][/tex]
So, the solutions are parameterized as:
[tex]\[ x = -\frac{13}{7}z - \frac{15}{7} \][/tex]
[tex]\[ y = \frac{1}{7}z - \frac{1}{7} \][/tex]
Since the system has infinitely many solutions, the correct choice is:
Choice A. There are infinitely many solutions. If [tex]\( z \)[/tex] is allowed to be any real number, then
[tex]\[ x = -\frac{13}{7}z - \frac{15}{7} \][/tex]
and
[tex]\[ y = \frac{1}{7}z - \frac{1}{7} \][/tex]
So, the final answer is:
A. There are infinitely many solutions. If [tex]\( z \)[/tex] is allowed to be any real number, then [tex]\( x = -\frac{13}{7}z - \frac{15}{7} \)[/tex] and [tex]\( y = \frac{1}{7}z - \frac{1}{7} \)[/tex].
