Answer :
Sure, let's solve each part step-by-step.
Problem 5: Solve by Matrix method:
Given the system of equations:
[tex]\[ 2x + y = 3 \][/tex]
[tex]\[ 3x + 2y = 2 \][/tex]
1. Form the coefficient matrix \( A \) and the constant matrix \( B \):
[tex]\[ A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \][/tex]
2. Solve the system \( AX = B \) for \( X \):
[tex]\[ X = A^{-1}B \][/tex]
Using matrix methods, the solution for \( x \) and \( y \) is:
[tex]\[ x = 4, \quad y = -5 \][/tex]
So, the solution to the system is \(( x, y ) = ( 4, -5 )\).
Problem 6:
Given, a square \( PQRS \) has a vertex \( P(2,3) \) and the equation of the diagonal \( PR \) is:
[tex]\[ x - 3y = 2 \][/tex]
1. Determine the slope of diagonal \( PR \):
Rewrite the equation in slope-intercept form:
[tex]\[ y = \frac{1}{3}x - \frac{2}{3} \][/tex]
The slope \( m \) of \( PR \) is \( \frac{1}{3} \).
2. Find the slope of the sides of the square:
The sides of the square are perpendicular to the diagonal \( PR \). The slope of a line perpendicular to \( PR \) is the negative reciprocal of \( \frac{1}{3} \), which is \( -3 \).
3. Equations of the sides through \( P(2,3) \):
- Equation of Side 1:
One side passes through \( P(2,3) \) with the slope \( -3 \).
[tex]\[ y - 3 = -3(x - 2) \][/tex]
Simplify the equation:
[tex]\[ y - 3 = -3x + 6 \implies 3x + y = 9 \][/tex]
Thus, the equation of Side 1 is:
[tex]\[ 3x + y = 9 \][/tex]
- Equation of Side 2:
Another side through \( P \) is perpendicular to the side whose equation we just found, so its slope is the negative reciprocal of \( -3 \), which is \( \frac{1}{3} \).
[tex]\[ y - 3 = \frac{1}{3}(x - 2) \][/tex]
Simplify the equation:
[tex]\[ y - 3 = \frac{1}{3}x - \frac{2}{3} \implies 3y - 9 = x - 2 \implies x - 3y = -7 \][/tex]
Thus, the equation of Side 2 is:
[tex]\[ x - 3y = -7 \][/tex]
So, the equations of the two sides of the square through \( P(2,3) \) are:
[tex]\[ 3x + y = 9 \][/tex]
[tex]\[ x - 3y = -7 \][/tex]
Problem 5: Solve by Matrix method:
Given the system of equations:
[tex]\[ 2x + y = 3 \][/tex]
[tex]\[ 3x + 2y = 2 \][/tex]
1. Form the coefficient matrix \( A \) and the constant matrix \( B \):
[tex]\[ A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \][/tex]
2. Solve the system \( AX = B \) for \( X \):
[tex]\[ X = A^{-1}B \][/tex]
Using matrix methods, the solution for \( x \) and \( y \) is:
[tex]\[ x = 4, \quad y = -5 \][/tex]
So, the solution to the system is \(( x, y ) = ( 4, -5 )\).
Problem 6:
Given, a square \( PQRS \) has a vertex \( P(2,3) \) and the equation of the diagonal \( PR \) is:
[tex]\[ x - 3y = 2 \][/tex]
1. Determine the slope of diagonal \( PR \):
Rewrite the equation in slope-intercept form:
[tex]\[ y = \frac{1}{3}x - \frac{2}{3} \][/tex]
The slope \( m \) of \( PR \) is \( \frac{1}{3} \).
2. Find the slope of the sides of the square:
The sides of the square are perpendicular to the diagonal \( PR \). The slope of a line perpendicular to \( PR \) is the negative reciprocal of \( \frac{1}{3} \), which is \( -3 \).
3. Equations of the sides through \( P(2,3) \):
- Equation of Side 1:
One side passes through \( P(2,3) \) with the slope \( -3 \).
[tex]\[ y - 3 = -3(x - 2) \][/tex]
Simplify the equation:
[tex]\[ y - 3 = -3x + 6 \implies 3x + y = 9 \][/tex]
Thus, the equation of Side 1 is:
[tex]\[ 3x + y = 9 \][/tex]
- Equation of Side 2:
Another side through \( P \) is perpendicular to the side whose equation we just found, so its slope is the negative reciprocal of \( -3 \), which is \( \frac{1}{3} \).
[tex]\[ y - 3 = \frac{1}{3}(x - 2) \][/tex]
Simplify the equation:
[tex]\[ y - 3 = \frac{1}{3}x - \frac{2}{3} \implies 3y - 9 = x - 2 \implies x - 3y = -7 \][/tex]
Thus, the equation of Side 2 is:
[tex]\[ x - 3y = -7 \][/tex]
So, the equations of the two sides of the square through \( P(2,3) \) are:
[tex]\[ 3x + y = 9 \][/tex]
[tex]\[ x - 3y = -7 \][/tex]
