Answer :
Let's use Cramer's rule to solve each of the given systems of linear equations step-by-step.
### Part (a)
The system of linear equations is:
[tex]\[ \begin{cases} 2x - 3y - z = 4 \\ x - 2y - z = 1 \\ x - y + 2z = 9 \end{cases} \][/tex]
We can represent this system in matrix form as [tex]\(AX = B\)[/tex], where:
[tex]\[ A = \begin{pmatrix} 2 & -3 & -1 \\ 1 & -2 & -1 \\ 1 & -1 & 2 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 4 \\ 1 \\ 9 \end{pmatrix} \][/tex]
To find [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] using Cramer's rule, we first find the determinant of matrix [tex]\(A\)[/tex], denoted as [tex]\(\det(A)\)[/tex].
The determinants of matrices [tex]\(A_x\)[/tex], [tex]\(A_y\)[/tex], and [tex]\(A_z\)[/tex] are found by replacing the corresponding columns of [tex]\(A\)[/tex] with [tex]\(B\)[/tex]:
1. [tex]\(\det(A_x)\)[/tex]: Replace the first column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_x = \begin{pmatrix} 4 & -3 & -1 \\ 1 & -2 & -1 \\ 9 & -1 & 2 \end{pmatrix} \][/tex]
2. [tex]\(\det(A_y)\)[/tex]: Replace the second column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_y = \begin{pmatrix} 2 & 4 & -1 \\ 1 & 1 & -1 \\ 1 & 9 & 2 \end{pmatrix} \][/tex]
3. [tex]\(\det(A_z)\)[/tex]: Replace the third column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_z = \begin{pmatrix} 2 & -3 & 4 \\ 1 & -2 & 1 \\ 1 & -1 & 9 \end{pmatrix} \][/tex]
Next, the solutions for [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] can be found by:
[tex]\[ x = \frac{\det(A_x)}{\det(A)}, \quad y = \frac{\det(A_y)}{\det(A)}, \quad z = \frac{\det(A_z)}{\det(A)} \][/tex]
By calculating the determinants, we get:
[tex]\[ x \approx 2, \quad y \approx -1, \quad z \approx 3 \][/tex]
### Part (b)
The system of linear equations is:
[tex]\[ \begin{cases} x + y - z = 1 \\ y + z = 2 \\ x - y = 0 \end{cases} \][/tex]
We can represent this system in matrix form as [tex]\(AX = B\)[/tex], where:
[tex]\[ A = \begin{pmatrix} 1 & 1 & -1 \\ 0 & 1 & 1 \\ 1 & -1 & 0 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} \][/tex]
Finding the determinants for [tex]\(A_x\)[/tex], [tex]\(A_y\)[/tex], and [tex]\(A_z\)[/tex]:
1. [tex]\(\det(A_x)\)[/tex]: Replace the first column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_x = \begin{pmatrix} 1 & 1 & -1 \\ 2 & 1 & 1 \\ 0 & -1 & 0 \end{pmatrix} \][/tex]
2. [tex]\(\det(A_y)\)[/tex]: Replace the second column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_y = \begin{pmatrix} 1 & 1 & -1 \\ 0 & 2 & 1 \\ 1 & 0 & 0 \end{pmatrix} \][/tex]
3. [tex]\(\det(A_z)\)[/tex]: Replace the third column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_z = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 1 & -1 & 0 \end{pmatrix} \][/tex]
By calculating the determinants, we find solutions:
[tex]\[ x = 1, \quad y = 1, \quad z = 1 \][/tex]
### Part (c)
The system of linear equations is:
[tex]\[ \begin{cases} 2x + 3y = 4 \\ 4y + 3x = -5 \\ 4x + z = 5 \end{cases} \][/tex]
We can represent this system in matrix form as [tex]\(AX = B\)[/tex], where:
[tex]\[ A = \begin{pmatrix} 2 & 3 & 0 \\ 3 & 4 & 0 \\ 4 & 0 & 1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 4 \\ -5 \\ 5 \end{pmatrix} \][/tex]
Finding the determinants for [tex]\(A_x\)[/tex], [tex]\(A_y\)[/tex], and [tex]\(A_z\)[/tex]:
1. [tex]\(\det(A_x)\)[/tex]: Replace the first column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_x = \begin{pmatrix} 4 & 3 & 0 \\ -5 & 4 & 0 \\ 5 & 0 & 1 \end{pmatrix} \][/tex]
2. [tex]\(\det(A_y)\)[/tex]: Replace the second column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_y = \begin{pmatrix} 2 & 4 & 0 \\ 3 & -5 & 0 \\ 4 & 5 & 1 \end{pmatrix} \][/tex]
3. [tex]\(\det(A_z)\)[/tex]: Replace the third column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_z = \begin{pmatrix} 2 & 3 & 4 \\ 3 & 4 & -5 \\ 4 & 0 & 5 \end{pmatrix} \][/tex]
By calculating the determinants, we find solutions:
[tex]\[ x \approx -31, \quad y \approx 22, \quad z \approx 129 \][/tex]
Thus, the solutions to the systems of linear equations are:
- Part (a): [tex]\(x \approx 2\)[/tex], [tex]\(y \approx -1\)[/tex], [tex]\(z \approx 3\)[/tex]
- Part (b): [tex]\(x = 1\)[/tex], [tex]\(y = 1\)[/tex], [tex]\(z = 1\)[/tex]
- Part (c): [tex]\(x \approx -31\)[/tex], [tex]\(y \approx 22\)[/tex], [tex]\(z \approx 129\)[/tex]
### Part (a)
The system of linear equations is:
[tex]\[ \begin{cases} 2x - 3y - z = 4 \\ x - 2y - z = 1 \\ x - y + 2z = 9 \end{cases} \][/tex]
We can represent this system in matrix form as [tex]\(AX = B\)[/tex], where:
[tex]\[ A = \begin{pmatrix} 2 & -3 & -1 \\ 1 & -2 & -1 \\ 1 & -1 & 2 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 4 \\ 1 \\ 9 \end{pmatrix} \][/tex]
To find [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] using Cramer's rule, we first find the determinant of matrix [tex]\(A\)[/tex], denoted as [tex]\(\det(A)\)[/tex].
The determinants of matrices [tex]\(A_x\)[/tex], [tex]\(A_y\)[/tex], and [tex]\(A_z\)[/tex] are found by replacing the corresponding columns of [tex]\(A\)[/tex] with [tex]\(B\)[/tex]:
1. [tex]\(\det(A_x)\)[/tex]: Replace the first column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_x = \begin{pmatrix} 4 & -3 & -1 \\ 1 & -2 & -1 \\ 9 & -1 & 2 \end{pmatrix} \][/tex]
2. [tex]\(\det(A_y)\)[/tex]: Replace the second column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_y = \begin{pmatrix} 2 & 4 & -1 \\ 1 & 1 & -1 \\ 1 & 9 & 2 \end{pmatrix} \][/tex]
3. [tex]\(\det(A_z)\)[/tex]: Replace the third column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_z = \begin{pmatrix} 2 & -3 & 4 \\ 1 & -2 & 1 \\ 1 & -1 & 9 \end{pmatrix} \][/tex]
Next, the solutions for [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] can be found by:
[tex]\[ x = \frac{\det(A_x)}{\det(A)}, \quad y = \frac{\det(A_y)}{\det(A)}, \quad z = \frac{\det(A_z)}{\det(A)} \][/tex]
By calculating the determinants, we get:
[tex]\[ x \approx 2, \quad y \approx -1, \quad z \approx 3 \][/tex]
### Part (b)
The system of linear equations is:
[tex]\[ \begin{cases} x + y - z = 1 \\ y + z = 2 \\ x - y = 0 \end{cases} \][/tex]
We can represent this system in matrix form as [tex]\(AX = B\)[/tex], where:
[tex]\[ A = \begin{pmatrix} 1 & 1 & -1 \\ 0 & 1 & 1 \\ 1 & -1 & 0 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} \][/tex]
Finding the determinants for [tex]\(A_x\)[/tex], [tex]\(A_y\)[/tex], and [tex]\(A_z\)[/tex]:
1. [tex]\(\det(A_x)\)[/tex]: Replace the first column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_x = \begin{pmatrix} 1 & 1 & -1 \\ 2 & 1 & 1 \\ 0 & -1 & 0 \end{pmatrix} \][/tex]
2. [tex]\(\det(A_y)\)[/tex]: Replace the second column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_y = \begin{pmatrix} 1 & 1 & -1 \\ 0 & 2 & 1 \\ 1 & 0 & 0 \end{pmatrix} \][/tex]
3. [tex]\(\det(A_z)\)[/tex]: Replace the third column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_z = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 1 & -1 & 0 \end{pmatrix} \][/tex]
By calculating the determinants, we find solutions:
[tex]\[ x = 1, \quad y = 1, \quad z = 1 \][/tex]
### Part (c)
The system of linear equations is:
[tex]\[ \begin{cases} 2x + 3y = 4 \\ 4y + 3x = -5 \\ 4x + z = 5 \end{cases} \][/tex]
We can represent this system in matrix form as [tex]\(AX = B\)[/tex], where:
[tex]\[ A = \begin{pmatrix} 2 & 3 & 0 \\ 3 & 4 & 0 \\ 4 & 0 & 1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 4 \\ -5 \\ 5 \end{pmatrix} \][/tex]
Finding the determinants for [tex]\(A_x\)[/tex], [tex]\(A_y\)[/tex], and [tex]\(A_z\)[/tex]:
1. [tex]\(\det(A_x)\)[/tex]: Replace the first column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_x = \begin{pmatrix} 4 & 3 & 0 \\ -5 & 4 & 0 \\ 5 & 0 & 1 \end{pmatrix} \][/tex]
2. [tex]\(\det(A_y)\)[/tex]: Replace the second column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_y = \begin{pmatrix} 2 & 4 & 0 \\ 3 & -5 & 0 \\ 4 & 5 & 1 \end{pmatrix} \][/tex]
3. [tex]\(\det(A_z)\)[/tex]: Replace the third column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
[tex]\[ A_z = \begin{pmatrix} 2 & 3 & 4 \\ 3 & 4 & -5 \\ 4 & 0 & 5 \end{pmatrix} \][/tex]
By calculating the determinants, we find solutions:
[tex]\[ x \approx -31, \quad y \approx 22, \quad z \approx 129 \][/tex]
Thus, the solutions to the systems of linear equations are:
- Part (a): [tex]\(x \approx 2\)[/tex], [tex]\(y \approx -1\)[/tex], [tex]\(z \approx 3\)[/tex]
- Part (b): [tex]\(x = 1\)[/tex], [tex]\(y = 1\)[/tex], [tex]\(z = 1\)[/tex]
- Part (c): [tex]\(x \approx -31\)[/tex], [tex]\(y \approx 22\)[/tex], [tex]\(z \approx 129\)[/tex]
