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### 21. Solve the system of linear equations using Cramer's rule
Given system of equations:
1. [tex]\(-x + 4y - z = 1\)[/tex]
2. [tex]\(2x - y + z = 0\)[/tex]
3. [tex]\(x + y + z = 1\)[/tex]
Step 1: Form the coefficient and constant matrices
The coefficient matrix [tex]\(A\)[/tex] and constant vector [tex]\(\mathbf{B}\)[/tex] are:
[tex]\[ A = \begin{pmatrix} -1 & 4 & -1 \\ 2 & -1 & 1 \\ 1 & 1 & 1 \end{pmatrix} \][/tex]
[tex]\[ \mathbf{B} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \][/tex]
Step 2: Calculate the determinant of the coefficient matrix [tex]\( \det(A) \)[/tex]
The determinant of [tex]\(A\)[/tex]:
[tex]\[ \det(A) = -5 \][/tex]
Step 3: Construct matrices for each variable and calculate their determinants
To find the determinant for each variable, modify the coefficient matrix by replacing the corresponding column with the constant vector [tex]\(\mathbf{B}\)[/tex].
- For [tex]\(x\)[/tex], replace the 1st column with [tex]\(\mathbf{B}\)[/tex]:
[tex]\[ A_x = \begin{pmatrix} 1 & 4 & -1 \\ 0 & -1 & 1 \\ 1 & 1 & 1 \end{pmatrix} \][/tex]
[tex]\[ \det(A_x) = 1 \][/tex]
- For [tex]\(y\)[/tex], replace the 2nd column with [tex]\(\mathbf{B}\)[/tex]:
[tex]\[ A_y = \begin{pmatrix} -1 & 1 & -1 \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{pmatrix} \][/tex]
[tex]\[ \det(A_y) = -2 \][/tex]
- For [tex]\(z\)[/tex], replace the 3rd column with [tex]\(\mathbf{B}\)[/tex]:
[tex]\[ A_z = \begin{pmatrix} -1 & 4 & 1 \\ 2 & -1 & 0 \\ 1 & 1 & 1 \end{pmatrix} \][/tex]
[tex]\[ \det(A_z) = -4 \][/tex]
Step 4: Solve for variables [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex]
Using Cramer's rule:
[tex]\[ x = \frac{\det(A_x)}{\det(A)} = \frac{1}{-5} = -0.2 \][/tex]
[tex]\[ y = \frac{\det(A_y)}{\det(A)} = \frac{-2}{-5} = 0.4 \][/tex]
[tex]\[ z = \frac{\det(A_z)}{\det(A)} = \frac{-4}{-5} = 0.8 \][/tex]
So, the solutions are:
[tex]\[ x = -0.2, \quad y = 0.4, \quad z = 0.8 \][/tex]
### 22. Find [tex]\(\cos \theta\)[/tex]
Given vectors:
[tex]\(\vec{A} = \mathbf{i} + \mathbf{j}\)[/tex] and [tex]\(\vec{B} = -2\mathbf{i} + 3\mathbf{j}\)[/tex].
Step 1: Dot product of [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex]
[tex]\[ \vec{A} \cdot \vec{B} = (1)(-2) + (1)(3) = -2 + 3 = 1 \][/tex]
Step 2: Magnitudes of [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex]
[tex]\[ |\vec{A}| = \sqrt{1^2 + 1^2} = \sqrt{2} \][/tex]
[tex]\[ |\vec{B}| = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \][/tex]
Step 3: Calculate [tex]\(\cos \theta\)[/tex]
[tex]\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} = \frac{1}{\sqrt{2} \cdot \sqrt{13}} = \frac{1}{\sqrt{26}} \][/tex]
### 23. Find the area of the parallelogram
Given vectors:
[tex]\(\vec{U} = 2\mathbf{i} + 3\mathbf{j} + \mathbf{k}\)[/tex] and [tex]\(\vec{V} = 4\mathbf{i} + \mathbf{j} + 2\mathbf{k}\)[/tex].
Step 1: Compute the cross-product [tex]\(\vec{U} \times \vec{V}\)[/tex]
[tex]\[ \vec{U} \times \vec{V} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 1 \\ 4 & 1 & 2 \end{vmatrix} \][/tex]
[tex]\[ = \mathbf{i}(3\cdot 2 - 1\cdot 1) - \mathbf{j}(2\cdot 2 - 1\cdot 4) + \mathbf{k}(2\cdot 1 - 3\cdot 4) \][/tex]
[tex]\[ = \mathbf{i}(6 - 1) - \mathbf{j}(4 - 4) + \mathbf{k}(2 - 12) \][/tex]
[tex]\[ = 5\mathbf{i} - 0\mathbf{j} - 10\mathbf{k} \][/tex]
[tex]\[ = 5\mathbf{i} - 10\mathbf{k} \][/tex]
Step 2: Magnitude of the cross product
[tex]\[ |\vec{U} \times \vec{V}| = \sqrt{5^2 + (-10)^2} = \sqrt{25 + 100} = \sqrt{125} = 5\sqrt{5} \][/tex]
So, the area of the parallelogram is:
[tex]\[ 5\sqrt{5} \][/tex]
### 24. Find the value of [tex]\(x\)[/tex]
Given the determinant equation:
[tex]\[ \left|\begin{array}{cc} x & x+4 \\ x+2 & 6 \end{array}\right| = 6x + 1 \][/tex]
Step 1: Calculate the determinant
[tex]\[ |A| = x \cdot 6 - (x+4)(x+2) = 6x - (x^2 + 2x + 4x + 8) = 6x - (x^2 + 6x + 8) \][/tex]
[tex]\[ = 6x - x^2 - 6x - 8 = -x^2 - 8 \][/tex]
So,
[tex]\[ -x^2 - 8 = 6x + 1 \][/tex]
Step 2: Solve for [tex]\(x\)[/tex]
[tex]\[ -x^2 - 8 = 6x + 1 \][/tex]
[tex]\[ -x^2 - 6x - 8 = 1 \][/tex]
[tex]\[ -x^2 - 6x - 9 = 0 \][/tex]
[tex]\[ x^2 + 6x + 9 = 0 \][/tex]
[tex]\((x + 3)^2 = 0\)[/tex]
[tex]\[ x + 3 = 0 \implies x = -3 \][/tex]
So, the value of [tex]\(x\)[/tex] is:
[tex]\[ x = -3 \][/tex]
Here you go, I have solved each part of the question systematically.
### 21. Solve the system of linear equations using Cramer's rule
Given system of equations:
1. [tex]\(-x + 4y - z = 1\)[/tex]
2. [tex]\(2x - y + z = 0\)[/tex]
3. [tex]\(x + y + z = 1\)[/tex]
Step 1: Form the coefficient and constant matrices
The coefficient matrix [tex]\(A\)[/tex] and constant vector [tex]\(\mathbf{B}\)[/tex] are:
[tex]\[ A = \begin{pmatrix} -1 & 4 & -1 \\ 2 & -1 & 1 \\ 1 & 1 & 1 \end{pmatrix} \][/tex]
[tex]\[ \mathbf{B} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \][/tex]
Step 2: Calculate the determinant of the coefficient matrix [tex]\( \det(A) \)[/tex]
The determinant of [tex]\(A\)[/tex]:
[tex]\[ \det(A) = -5 \][/tex]
Step 3: Construct matrices for each variable and calculate their determinants
To find the determinant for each variable, modify the coefficient matrix by replacing the corresponding column with the constant vector [tex]\(\mathbf{B}\)[/tex].
- For [tex]\(x\)[/tex], replace the 1st column with [tex]\(\mathbf{B}\)[/tex]:
[tex]\[ A_x = \begin{pmatrix} 1 & 4 & -1 \\ 0 & -1 & 1 \\ 1 & 1 & 1 \end{pmatrix} \][/tex]
[tex]\[ \det(A_x) = 1 \][/tex]
- For [tex]\(y\)[/tex], replace the 2nd column with [tex]\(\mathbf{B}\)[/tex]:
[tex]\[ A_y = \begin{pmatrix} -1 & 1 & -1 \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{pmatrix} \][/tex]
[tex]\[ \det(A_y) = -2 \][/tex]
- For [tex]\(z\)[/tex], replace the 3rd column with [tex]\(\mathbf{B}\)[/tex]:
[tex]\[ A_z = \begin{pmatrix} -1 & 4 & 1 \\ 2 & -1 & 0 \\ 1 & 1 & 1 \end{pmatrix} \][/tex]
[tex]\[ \det(A_z) = -4 \][/tex]
Step 4: Solve for variables [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex]
Using Cramer's rule:
[tex]\[ x = \frac{\det(A_x)}{\det(A)} = \frac{1}{-5} = -0.2 \][/tex]
[tex]\[ y = \frac{\det(A_y)}{\det(A)} = \frac{-2}{-5} = 0.4 \][/tex]
[tex]\[ z = \frac{\det(A_z)}{\det(A)} = \frac{-4}{-5} = 0.8 \][/tex]
So, the solutions are:
[tex]\[ x = -0.2, \quad y = 0.4, \quad z = 0.8 \][/tex]
### 22. Find [tex]\(\cos \theta\)[/tex]
Given vectors:
[tex]\(\vec{A} = \mathbf{i} + \mathbf{j}\)[/tex] and [tex]\(\vec{B} = -2\mathbf{i} + 3\mathbf{j}\)[/tex].
Step 1: Dot product of [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex]
[tex]\[ \vec{A} \cdot \vec{B} = (1)(-2) + (1)(3) = -2 + 3 = 1 \][/tex]
Step 2: Magnitudes of [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex]
[tex]\[ |\vec{A}| = \sqrt{1^2 + 1^2} = \sqrt{2} \][/tex]
[tex]\[ |\vec{B}| = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \][/tex]
Step 3: Calculate [tex]\(\cos \theta\)[/tex]
[tex]\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} = \frac{1}{\sqrt{2} \cdot \sqrt{13}} = \frac{1}{\sqrt{26}} \][/tex]
### 23. Find the area of the parallelogram
Given vectors:
[tex]\(\vec{U} = 2\mathbf{i} + 3\mathbf{j} + \mathbf{k}\)[/tex] and [tex]\(\vec{V} = 4\mathbf{i} + \mathbf{j} + 2\mathbf{k}\)[/tex].
Step 1: Compute the cross-product [tex]\(\vec{U} \times \vec{V}\)[/tex]
[tex]\[ \vec{U} \times \vec{V} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 1 \\ 4 & 1 & 2 \end{vmatrix} \][/tex]
[tex]\[ = \mathbf{i}(3\cdot 2 - 1\cdot 1) - \mathbf{j}(2\cdot 2 - 1\cdot 4) + \mathbf{k}(2\cdot 1 - 3\cdot 4) \][/tex]
[tex]\[ = \mathbf{i}(6 - 1) - \mathbf{j}(4 - 4) + \mathbf{k}(2 - 12) \][/tex]
[tex]\[ = 5\mathbf{i} - 0\mathbf{j} - 10\mathbf{k} \][/tex]
[tex]\[ = 5\mathbf{i} - 10\mathbf{k} \][/tex]
Step 2: Magnitude of the cross product
[tex]\[ |\vec{U} \times \vec{V}| = \sqrt{5^2 + (-10)^2} = \sqrt{25 + 100} = \sqrt{125} = 5\sqrt{5} \][/tex]
So, the area of the parallelogram is:
[tex]\[ 5\sqrt{5} \][/tex]
### 24. Find the value of [tex]\(x\)[/tex]
Given the determinant equation:
[tex]\[ \left|\begin{array}{cc} x & x+4 \\ x+2 & 6 \end{array}\right| = 6x + 1 \][/tex]
Step 1: Calculate the determinant
[tex]\[ |A| = x \cdot 6 - (x+4)(x+2) = 6x - (x^2 + 2x + 4x + 8) = 6x - (x^2 + 6x + 8) \][/tex]
[tex]\[ = 6x - x^2 - 6x - 8 = -x^2 - 8 \][/tex]
So,
[tex]\[ -x^2 - 8 = 6x + 1 \][/tex]
Step 2: Solve for [tex]\(x\)[/tex]
[tex]\[ -x^2 - 8 = 6x + 1 \][/tex]
[tex]\[ -x^2 - 6x - 8 = 1 \][/tex]
[tex]\[ -x^2 - 6x - 9 = 0 \][/tex]
[tex]\[ x^2 + 6x + 9 = 0 \][/tex]
[tex]\((x + 3)^2 = 0\)[/tex]
[tex]\[ x + 3 = 0 \implies x = -3 \][/tex]
So, the value of [tex]\(x\)[/tex] is:
[tex]\[ x = -3 \][/tex]
Here you go, I have solved each part of the question systematically.
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