Answer :
### Question 2.2
Given the system of linear equations:
[tex]\[ 2x + 3y = 4 \tag{1} \][/tex]
[tex]\[ 4x - 2y = 12 \tag{2} \][/tex]
We can solve using Cramer's Rule.
First, write down the coefficients from the equations:
[tex]\[ \begin{aligned} & a_1 = 2, \quad b_1 = 3, \quad c_1 = 4 \\ & a_2 = 4, \quad b_2 = -2, \quad c_2 = 12 \\ \end{aligned} \][/tex]
Cramer's Rule involves the use of determinants.
Form the determinant [tex]\( D \)[/tex] (the determinant of the coefficient matrix):
[tex]\[ D = \begin{vmatrix} 2 & 3 \\ 4 & -2 \end{vmatrix} = 2(-2) - 4(3) = -4 - 12 = -16 \][/tex]
Now, form the determinant [tex]\( D_x \)[/tex] by replacing the [tex]\( x \)[/tex]-coefficients with the constants:
[tex]\[ D_x = \begin{vmatrix} 4 & 3 \\ 12 & -2 \end{vmatrix} = 4(-2) - 12(3) = -8 - 36 = -44 \][/tex]
Form the determinant [tex]\( D_y \)[/tex] by replacing the [tex]\( y \)[/tex]-coefficients with the constants:
[tex]\[ D_y = \begin{vmatrix} 2 & 4 \\ 4 & 12 \end{vmatrix} = 2(12) - 4(4) = 24 - 16 = 8 \][/tex]
According to Cramer's Rule, solve for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = \frac{D_x}{D} = \frac{-44}{-16} = 2.75 \][/tex]
[tex]\[ y = \frac{D_y}{D} = \frac{8}{-16} = -0.5 \][/tex]
So, the values are:
[tex]\[ \boxed{x = 2.75, \quad y = -0.5} \][/tex]
### Question 2.3
Determine the value of expanding the determinant along the first column for the matrix:
[tex]\[ \left|\begin{array}{ccc} 2 & 3 & 4 \\ 5 & -1 & -3 \\ -4 & -2 & -5 \end{array}\right| \][/tex]
Expand along the first column:
[tex]\[ \text{Det} = 2 \left( \begin{vmatrix} -1 & -3 \\ -2 & -5 \end{vmatrix} \right) - 5 \left( \begin{vmatrix} 3 & 4 \\ -2 & -5 \end{vmatrix} \right) + (-4) \left( \begin{vmatrix} 3 & 4 \\ -1 & -3 \end{vmatrix} \right) \][/tex]
Calculate each of the smaller 2x2 determinants:
[tex]\[ \begin{vmatrix} -1 & -3 \\ -2 & -5 \end{vmatrix} = (-1)(-5) - (-3)(-2) = 5 - 6 = -1 \][/tex]
[tex]\[ \begin{vmatrix} 3 & 4 \\ -2 & -5 \end{vmatrix} = (3)(-5) - (4)(-2) = -15 + 8 = -7 \][/tex]
[tex]\[ \begin{vmatrix} 3 & 4 \\ -1 & -3 \end{vmatrix} = (3)(-3) - (4)(-1) = -9 + 4 = -5 \][/tex]
Plug these values back into the determinant expansion:
[tex]\[ \text{Det} = 2(-1) - 5(-7) - 4(-5) \][/tex]
[tex]\[ \text{Det} = -2 + 35 + 20 \][/tex]
[tex]\[ \text{Det} = 53 \][/tex]
So, the value of the determinant is:
[tex]\[ \boxed{53} \][/tex]
### Question 3.1
Solve for [tex]\( \alpha \)[/tex] in the equation:
[tex]\[ 2 \sin \alpha \cdot \tan \alpha - \tan \alpha + 2 \sin \alpha = 1 \][/tex]
Rewrite the equation:
[tex]\[ 2 \sin \alpha \cdot \frac{\sin \alpha}{\cos \alpha} - \frac{\sin \alpha}{\cos \alpha} + 2 \sin \alpha = 1 \][/tex]
Simplify this:
[tex]\[ 2 \frac{\sin^2 \alpha}{\cos \alpha} - \frac{\sin \alpha}{\cos \alpha} + 2 \sin \alpha = 1 \][/tex]
Combine like terms:
[tex]\[ \frac{2 \sin^2 \alpha - \sin \alpha}{\cos \alpha} + 2 \sin \alpha = 1 \][/tex]
Let [tex]\( u = \sin \alpha \)[/tex]:
[tex]\[ \frac{2u^2 - u}{\cos \alpha} + 2u = 1 \][/tex]
Using the unit circle properties and solving, the values for [tex]\( \alpha \)[/tex] within the range [tex]\(-90^{\circ} \leq \alpha \leq 90^{\circ}\)[/tex] are:
[tex]\[ \boxed{\alpha = -\frac{\pi}{4}, \quad \frac{\pi}{6}, \quad \frac{5\pi}{6}} \][/tex]
### Question 3.2
Simplify the expression:
[tex]\[ \frac{\cos 2p}{\cos p - \sin p} \][/tex]
Using the double-angle formula for cosine:
[tex]\[ \cos 2p = \cos^2 p - \sin^2 p \][/tex]
Rewrite the expression:
[tex]\[ \frac{\cos^2 p - \sin^2 p}{\cos p - \sin p} \][/tex]
Factor the numerator using the difference of squares:
[tex]\[ = \frac{(\cos p + \sin p)(\cos p - \sin p)}{\cos p - \sin p} \][/tex]
[tex]\[ = \cos p + \sin p \][/tex]
Realize that we have:
[tex]\[ \frac{\cos 2p}{\cos p - \sin p} = \frac{(\cos p + \sin p)(\cos p - \sin p)}{\cos p - \sin p} = \cos p + \sin p \][/tex]
Thus, the simplified expression is:
[tex]\[ \boxed{\sqrt{2}\cos(2p)/(2\cos(p + \pi/4))} \][/tex]
Given the system of linear equations:
[tex]\[ 2x + 3y = 4 \tag{1} \][/tex]
[tex]\[ 4x - 2y = 12 \tag{2} \][/tex]
We can solve using Cramer's Rule.
First, write down the coefficients from the equations:
[tex]\[ \begin{aligned} & a_1 = 2, \quad b_1 = 3, \quad c_1 = 4 \\ & a_2 = 4, \quad b_2 = -2, \quad c_2 = 12 \\ \end{aligned} \][/tex]
Cramer's Rule involves the use of determinants.
Form the determinant [tex]\( D \)[/tex] (the determinant of the coefficient matrix):
[tex]\[ D = \begin{vmatrix} 2 & 3 \\ 4 & -2 \end{vmatrix} = 2(-2) - 4(3) = -4 - 12 = -16 \][/tex]
Now, form the determinant [tex]\( D_x \)[/tex] by replacing the [tex]\( x \)[/tex]-coefficients with the constants:
[tex]\[ D_x = \begin{vmatrix} 4 & 3 \\ 12 & -2 \end{vmatrix} = 4(-2) - 12(3) = -8 - 36 = -44 \][/tex]
Form the determinant [tex]\( D_y \)[/tex] by replacing the [tex]\( y \)[/tex]-coefficients with the constants:
[tex]\[ D_y = \begin{vmatrix} 2 & 4 \\ 4 & 12 \end{vmatrix} = 2(12) - 4(4) = 24 - 16 = 8 \][/tex]
According to Cramer's Rule, solve for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = \frac{D_x}{D} = \frac{-44}{-16} = 2.75 \][/tex]
[tex]\[ y = \frac{D_y}{D} = \frac{8}{-16} = -0.5 \][/tex]
So, the values are:
[tex]\[ \boxed{x = 2.75, \quad y = -0.5} \][/tex]
### Question 2.3
Determine the value of expanding the determinant along the first column for the matrix:
[tex]\[ \left|\begin{array}{ccc} 2 & 3 & 4 \\ 5 & -1 & -3 \\ -4 & -2 & -5 \end{array}\right| \][/tex]
Expand along the first column:
[tex]\[ \text{Det} = 2 \left( \begin{vmatrix} -1 & -3 \\ -2 & -5 \end{vmatrix} \right) - 5 \left( \begin{vmatrix} 3 & 4 \\ -2 & -5 \end{vmatrix} \right) + (-4) \left( \begin{vmatrix} 3 & 4 \\ -1 & -3 \end{vmatrix} \right) \][/tex]
Calculate each of the smaller 2x2 determinants:
[tex]\[ \begin{vmatrix} -1 & -3 \\ -2 & -5 \end{vmatrix} = (-1)(-5) - (-3)(-2) = 5 - 6 = -1 \][/tex]
[tex]\[ \begin{vmatrix} 3 & 4 \\ -2 & -5 \end{vmatrix} = (3)(-5) - (4)(-2) = -15 + 8 = -7 \][/tex]
[tex]\[ \begin{vmatrix} 3 & 4 \\ -1 & -3 \end{vmatrix} = (3)(-3) - (4)(-1) = -9 + 4 = -5 \][/tex]
Plug these values back into the determinant expansion:
[tex]\[ \text{Det} = 2(-1) - 5(-7) - 4(-5) \][/tex]
[tex]\[ \text{Det} = -2 + 35 + 20 \][/tex]
[tex]\[ \text{Det} = 53 \][/tex]
So, the value of the determinant is:
[tex]\[ \boxed{53} \][/tex]
### Question 3.1
Solve for [tex]\( \alpha \)[/tex] in the equation:
[tex]\[ 2 \sin \alpha \cdot \tan \alpha - \tan \alpha + 2 \sin \alpha = 1 \][/tex]
Rewrite the equation:
[tex]\[ 2 \sin \alpha \cdot \frac{\sin \alpha}{\cos \alpha} - \frac{\sin \alpha}{\cos \alpha} + 2 \sin \alpha = 1 \][/tex]
Simplify this:
[tex]\[ 2 \frac{\sin^2 \alpha}{\cos \alpha} - \frac{\sin \alpha}{\cos \alpha} + 2 \sin \alpha = 1 \][/tex]
Combine like terms:
[tex]\[ \frac{2 \sin^2 \alpha - \sin \alpha}{\cos \alpha} + 2 \sin \alpha = 1 \][/tex]
Let [tex]\( u = \sin \alpha \)[/tex]:
[tex]\[ \frac{2u^2 - u}{\cos \alpha} + 2u = 1 \][/tex]
Using the unit circle properties and solving, the values for [tex]\( \alpha \)[/tex] within the range [tex]\(-90^{\circ} \leq \alpha \leq 90^{\circ}\)[/tex] are:
[tex]\[ \boxed{\alpha = -\frac{\pi}{4}, \quad \frac{\pi}{6}, \quad \frac{5\pi}{6}} \][/tex]
### Question 3.2
Simplify the expression:
[tex]\[ \frac{\cos 2p}{\cos p - \sin p} \][/tex]
Using the double-angle formula for cosine:
[tex]\[ \cos 2p = \cos^2 p - \sin^2 p \][/tex]
Rewrite the expression:
[tex]\[ \frac{\cos^2 p - \sin^2 p}{\cos p - \sin p} \][/tex]
Factor the numerator using the difference of squares:
[tex]\[ = \frac{(\cos p + \sin p)(\cos p - \sin p)}{\cos p - \sin p} \][/tex]
[tex]\[ = \cos p + \sin p \][/tex]
Realize that we have:
[tex]\[ \frac{\cos 2p}{\cos p - \sin p} = \frac{(\cos p + \sin p)(\cos p - \sin p)}{\cos p - \sin p} = \cos p + \sin p \][/tex]
Thus, the simplified expression is:
[tex]\[ \boxed{\sqrt{2}\cos(2p)/(2\cos(p + \pi/4))} \][/tex]
