Answer :
Sure, let’s solve the system of linear equations using Cramer's Rule. The system of equations given is:
[tex]\[ \begin{cases} -x + 2y - 3z = 1 \\ 2x + z = 0 \\ 3x - 4y + 4z = 2 \end{cases} \][/tex]
First, let's write the system in matrix form [tex]\( AX = B \)[/tex], where [tex]\( A \)[/tex] is the coefficient matrix, [tex]\( X \)[/tex] is the column vector of variables, and [tex]\( B \)[/tex] is the column vector of constants.
[tex]\[ A = \begin{pmatrix} -1 & 2 & -3 \\ 2 & 0 & 1 \\ 3 & -4 & 4 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} \][/tex]
### Step 1: Calculate Determinant of [tex]\( A \)[/tex]
The determinant of [tex]\( A \)[/tex] is denoted as [tex]\( \text{det}(A) \)[/tex].
[tex]\[ \text{det}(A) = 10.000000000000002 \][/tex]
Since [tex]\(\text{det}(A) \neq 0\)[/tex], the system has a unique solution.
### Step 2: Calculate Determinant of [tex]\( A_1 \)[/tex]
Matrix [tex]\( A_1 \)[/tex] is formed by replacing the first column of [tex]\( A \)[/tex] with the column vector [tex]\( B \)[/tex]:
[tex]\[ A_1 = \begin{pmatrix} 1 & 2 & -3 \\ 0 & 0 & 1 \\ 2 & -4 & 4 \end{pmatrix} \][/tex]
The determinant of [tex]\( A_1 \)[/tex] is [tex]\(\text{det}(A_1) = 7.999999999999998\)[/tex].
### Step 3: Calculate Determinant of [tex]\( A_2 \)[/tex]
Matrix [tex]\( A_2 \)[/tex] is formed by replacing the second column of [tex]\( A \)[/tex] with the column vector [tex]\( B \)[/tex]:
[tex]\[ A_2 = \begin{pmatrix} -1 & 1 & -3 \\ 2 & 0 & 1 \\ 3 & 2 & 4 \end{pmatrix} \][/tex]
The determinant of [tex]\( A_2 \)[/tex] is [tex]\(\text{det}(A_2) = -15.000000000000007\)[/tex].
### Step 4: Calculate Determinant of [tex]\( A_3 \)[/tex]
Matrix [tex]\( A_3 \)[/tex] is formed by replacing the third column of [tex]\( A \)[/tex] with the column vector [tex]\( B \)[/tex]:
[tex]\[ A_3 = \begin{pmatrix} -1 & 2 & 1 \\ 2 & 0 & 0 \\ 3 & -4 & 2 \end{pmatrix} \][/tex]
The determinant of [tex]\( A_3 \)[/tex] is [tex]\(\text{det}(A_3) = -16.000000000000007\)[/tex].
### Step 5: Solve for [tex]\( x, y, \)[/tex] and [tex]\( z \)[/tex] using Cramer's Rule
Cramer's Rule states that the solution for the variables [tex]\( x, y, z \)[/tex] can be found using:
[tex]\[ x = \frac{\text{det}(A_1)}{\text{det}(A)} = \frac{7.999999999999998}{10.000000000000002} = 0.7999999999999997 \][/tex]
[tex]\[ y = \frac{\text{det}(A_2)}{\text{det}(A)} = \frac{-15.000000000000007}{10.000000000000002} = -1.5000000000000004 \][/tex]
[tex]\[ z = \frac{\text{det}(A_3)}{\text{det}(A)} = \frac{-16.000000000000007}{10.000000000000002} = -1.6000000000000005 \][/tex]
### Final Solution
The solution to the system of equations is:
[tex]\[ \boxed{x = 0.7999999999999997, \, y = -1.5000000000000004, \, z = -1.6000000000000005} \][/tex]
[tex]\[ \begin{cases} -x + 2y - 3z = 1 \\ 2x + z = 0 \\ 3x - 4y + 4z = 2 \end{cases} \][/tex]
First, let's write the system in matrix form [tex]\( AX = B \)[/tex], where [tex]\( A \)[/tex] is the coefficient matrix, [tex]\( X \)[/tex] is the column vector of variables, and [tex]\( B \)[/tex] is the column vector of constants.
[tex]\[ A = \begin{pmatrix} -1 & 2 & -3 \\ 2 & 0 & 1 \\ 3 & -4 & 4 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} \][/tex]
### Step 1: Calculate Determinant of [tex]\( A \)[/tex]
The determinant of [tex]\( A \)[/tex] is denoted as [tex]\( \text{det}(A) \)[/tex].
[tex]\[ \text{det}(A) = 10.000000000000002 \][/tex]
Since [tex]\(\text{det}(A) \neq 0\)[/tex], the system has a unique solution.
### Step 2: Calculate Determinant of [tex]\( A_1 \)[/tex]
Matrix [tex]\( A_1 \)[/tex] is formed by replacing the first column of [tex]\( A \)[/tex] with the column vector [tex]\( B \)[/tex]:
[tex]\[ A_1 = \begin{pmatrix} 1 & 2 & -3 \\ 0 & 0 & 1 \\ 2 & -4 & 4 \end{pmatrix} \][/tex]
The determinant of [tex]\( A_1 \)[/tex] is [tex]\(\text{det}(A_1) = 7.999999999999998\)[/tex].
### Step 3: Calculate Determinant of [tex]\( A_2 \)[/tex]
Matrix [tex]\( A_2 \)[/tex] is formed by replacing the second column of [tex]\( A \)[/tex] with the column vector [tex]\( B \)[/tex]:
[tex]\[ A_2 = \begin{pmatrix} -1 & 1 & -3 \\ 2 & 0 & 1 \\ 3 & 2 & 4 \end{pmatrix} \][/tex]
The determinant of [tex]\( A_2 \)[/tex] is [tex]\(\text{det}(A_2) = -15.000000000000007\)[/tex].
### Step 4: Calculate Determinant of [tex]\( A_3 \)[/tex]
Matrix [tex]\( A_3 \)[/tex] is formed by replacing the third column of [tex]\( A \)[/tex] with the column vector [tex]\( B \)[/tex]:
[tex]\[ A_3 = \begin{pmatrix} -1 & 2 & 1 \\ 2 & 0 & 0 \\ 3 & -4 & 2 \end{pmatrix} \][/tex]
The determinant of [tex]\( A_3 \)[/tex] is [tex]\(\text{det}(A_3) = -16.000000000000007\)[/tex].
### Step 5: Solve for [tex]\( x, y, \)[/tex] and [tex]\( z \)[/tex] using Cramer's Rule
Cramer's Rule states that the solution for the variables [tex]\( x, y, z \)[/tex] can be found using:
[tex]\[ x = \frac{\text{det}(A_1)}{\text{det}(A)} = \frac{7.999999999999998}{10.000000000000002} = 0.7999999999999997 \][/tex]
[tex]\[ y = \frac{\text{det}(A_2)}{\text{det}(A)} = \frac{-15.000000000000007}{10.000000000000002} = -1.5000000000000004 \][/tex]
[tex]\[ z = \frac{\text{det}(A_3)}{\text{det}(A)} = \frac{-16.000000000000007}{10.000000000000002} = -1.6000000000000005 \][/tex]
### Final Solution
The solution to the system of equations is:
[tex]\[ \boxed{x = 0.7999999999999997, \, y = -1.5000000000000004, \, z = -1.6000000000000005} \][/tex]
