Question Statement We are given a system of linear equations. The task is to express each system in matrix form and determine if the lines represented by the equations are concurrent. Specifically, we need to analyze the following cases:
Case (a)
x + 3 y β 2 = 0 x + 3y - 2 = 0 x + 3 y β 2 = 0 2 x β y + 4 = 0 2x - y + 4 = 0 2 x β y + 4 = 0 x β 11 y + 14 = 0 x - 11y + 14 = 0 x β 11 y + 14 = 0 Case (b)
2 x + 3 y + 4 = 0 2x + 3y + 4 = 0 2 x + 3 y + 4 = 0 x β 2 y β 3 = 0 x - 2y - 3 = 0 x β 2 y β 3 = 0 3 x + y β 8 = 0 3x + y - 8 = 0 3 x + y β 8 = 0 Case (c)
3 x β 4 y β 2 = 0 3x - 4y - 2 = 0 3 x β 4 y β 2 = 0 x + 2 y β 4 = 0 x + 2y - 4 = 0 x + 2 y β 4 = 0 3 x β 2 y + 5 = 0 3x - 2y + 5 = 0 3 x β 2 y + 5 = 0 Background and Explanation To solve this problem, we need to represent each system of linear equations in matrix form. A system of equations can be expressed as:
A x = b Ax = b A x = b Where:
A A A x x x x x x y y y b b b
Once we express the system in matrix form, we will use the determinant of the coefficient matrix A A A β£ A β£ = 0 |A| = 0 β£ A β£ = 0 concurrent , meaning they intersect at a single point. If β£ A β£ β 0 |A| \neq 0 β£ A β£ ξ = 0 not concurrent , meaning they do not all meet at a single point.
Solution Case (a)
The system of equations is:
x + 3 y β 2 = 02 x β y + 4 = 0 x β 11 y + 14 = 0 \begin{aligned}
x + 3y - 2 &= 0
2x - y + 4 &= 0
x - 11y + 14 &= 0
\end{aligned} x + 3 y β 2 β = 02 x β y + 4 β = 0 x β 11 y + 14 β = 0 β This system can be written in matrix form as:
[ 1 3 β 22 β 1 41 β 11 14 ] [ x y z ] = [ 000 ] \left[\begin{array}{ccc}
1 & 3 & -2
2 & -1 & 4
1 & -11 & 14
\end{array}\right]
\left[\begin{array}{c}
x
y
z
\end{array}\right]
=
\left[\begin{array}{c}
0
0
0
\end{array}\right] [ 1 β 3 β β 22 β β 1 β 41 β β 11 β 14 β ] [ x yz β ] = [ 000 β ] Determinant of the Coefficient Matrix:
We calculate the determinant of the coefficient matrix A A A
β£ A β£ = β£ 1 3 β 22 β 1 41 β 11 14 β£ |A| = \begin{vmatrix} 1 & 3 & -2 2 & -1 & 4 1 & -11 & 14 \end{vmatrix} β£ A β£ = β 1 β 3 β β 22 β β 1 β 41 β β 11 β 14 β β Expanding this determinant:
β£ A β£ = 1 Γ ( β£ β 1 4 β 11 14 β£ ) β 3 Γ ( β£ 2 41 14 β£ ) + ( β 2 ) Γ ( β£ 2 β 11 β 11 β£ ) |A| = 1 \times \left( \begin{vmatrix} -1 & 4 -11 & 14 \end{vmatrix} \right)
- 3 \times \left( \begin{vmatrix} 2 & 4 1 & 14 \end{vmatrix} \right)
+ (-2) \times \left( \begin{vmatrix} 2 & -1 1 & -11 \end{vmatrix} \right) β£ A β£ = 1 Γ ( β β 1 β 4 β 11 β 14 β β ) β 3 Γ ( β 2 β 41 β 14 β β ) + ( β 2 ) Γ ( β 2 β β 11 β β 11 β β ) Calculating each 2x2 determinant:
β£ β 1 4 β 11 14 β£ = ( β 1 ) ( 14 ) β ( 4 ) ( β 11 ) = β 14 + 44 = 30 \begin{vmatrix} -1 & 4 -11 & 14 \end{vmatrix} = (-1)(14) - (4)(-11) = -14 + 44 = 30 β β 1 β 4 β 11 β 14 β β = ( β 1 ) ( 14 ) β ( 4 ) ( β 11 ) = β 14 + 44 = 30 β£ 2 41 14 β£ = ( 2 ) ( 14 ) β ( 4 ) ( 1 ) = 28 β 4 = 24 \begin{vmatrix} 2 & 4 1 & 14 \end{vmatrix} = (2)(14) - (4)(1) = 28 - 4 = 24 β 2 β 41 β 14 β β = ( 2 ) ( 14 ) β ( 4 ) ( 1 ) = 28 β 4 = 24 β£ 2 β 11 β 11 β£ = ( 2 ) ( β 11 ) β ( β 1 ) ( 1 ) = β 22 + 1 = β 21 \begin{vmatrix} 2 & -1 1 & -11 \end{vmatrix} = (2)(-11) - (-1)(1) = -22 + 1 = -21 β 2 β β 11 β β 11 β β = ( 2 ) ( β 11 ) β ( β 1 ) ( 1 ) = β 22 + 1 = β 21 Substituting these values back:
β£ A β£ = 1 ( 30 ) β 3 ( 24 ) β 2 ( β 21 ) = 30 β 72 + 42 = 0 |A| = 1(30) - 3(24) - 2(-21) = 30 - 72 + 42 = 0 β£ A β£ = 1 ( 30 ) β 3 ( 24 ) β 2 ( β 21 ) = 30 β 72 + 42 = 0 Since β£ A β£ = 0 |A| = 0 β£ A β£ = 0 concurrent .
Case (b)
The system of equations is:
2 x + 3 y + 4 = 0 x β 2 y β 3 = 03 x + y β 8 = 0 \begin{aligned}
2x + 3y + 4 &= 0
x - 2y - 3 &= 0
3x + y - 8 &= 0
\end{aligned} 2 x + 3 y + 4 β = 0 x β 2 y β 3 β = 03 x + y β 8 β = 0 β This system can be written in matrix form as:
[ 2 3 41 β 2 β 33 1 β 8 ] [ x y z ] = [ 000 ] \left[\begin{array}{ccc}
2 & 3 & 4
1 & -2 & -3
3 & 1 & -8
\end{array}\right]
\left[\begin{array}{c}
x
y
z
\end{array}\right]
=
\left[\begin{array}{c}
0
0
0
\end{array}\right] [ 2 β 3 β 41 β β 2 β β 33 β 1 β β 8 β ] [ x yz β ] = [ 000 β ] Determinant of the Coefficient Matrix:
We calculate the determinant of the coefficient matrix A A A
β£ A β£ = β£ 2 3 41 β 2 β 33 1 β 8 β£ |A| = \begin{vmatrix} 2 & 3 & 4 1 & -2 & -3 3 & 1 & -8 \end{vmatrix} β£ A β£ = β 2 β 3 β 41 β β 2 β β 33 β 1 β β 8 β β Expanding this determinant:
β£ A β£ = 2 Γ β£ β 2 β 31 β 8 β£ β 3 Γ β£ 1 β 33 β 8 β£ + 4 Γ β£ 1 β 23 1 β£ |A| = 2 \times \begin{vmatrix} -2 & -3 1 & -8 \end{vmatrix}
- 3 \times \begin{vmatrix} 1 & -3 3 & -8 \end{vmatrix}
+ 4 \times \begin{vmatrix} 1 & -2 3 & 1 \end{vmatrix} β£ A β£ = 2 Γ β β 2 β β 31 β β 8 β β β 3 Γ β 1 β β 33 β β 8 β β + 4 Γ β 1 β β 23 β 1 β β Calculating each 2x2 determinant:
β£ β 2 β 31 β 8 β£ = ( β 2 ) ( β 8 ) β ( β 3 ) ( 1 ) = 16 + 3 = 19 \begin{vmatrix} -2 & -3 1 & -8 \end{vmatrix} = (-2)(-8) - (-3)(1) = 16 + 3 = 19 β β 2 β β 31 β β 8 β β = ( β 2 ) ( β 8 ) β ( β 3 ) ( 1 ) = 16 + 3 = 19 β£ 1 β 33 β 8 β£ = ( 1 ) ( β 8 ) β ( β 3 ) ( 3 ) = β 8 + 9 = 1 \begin{vmatrix} 1 & -3 3 & -8 \end{vmatrix} = (1)(-8) - (-3)(3) = -8 + 9 = 1 β 1 β β 33 β β 8 β β = ( 1 ) ( β 8 ) β ( β 3 ) ( 3 ) = β 8 + 9 = 1 β£ 1 β 23 1 β£ = ( 1 ) ( 1 ) β ( β 2 ) ( 3 ) = 1 + 6 = 7 \begin{vmatrix} 1 & -2 3 & 1 \end{vmatrix} = (1)(1) - (-2)(3) = 1 + 6 = 7 β 1 β β 23 β 1 β β = ( 1 ) ( 1 ) β ( β 2 ) ( 3 ) = 1 + 6 = 7 Substituting these values back:
β£ A β£ = 2 ( 19 ) β 3 ( 1 ) + 4 ( 7 ) = 38 β 3 + 28 = 63 |A| = 2(19) - 3(1) + 4(7) = 38 - 3 + 28 = 63 β£ A β£ = 2 ( 19 ) β 3 ( 1 ) + 4 ( 7 ) = 38 β 3 + 28 = 63 Since β£ A β£ = 7 β 0 |A| = 7 \neq 0 β£ A β£ = 7 ξ = 0 not concurrent .
Case (c)
The system of equations is:
3 x β 4 y β 2 = 0 x + 2 y β 4 = 03 x β 2 y + 5 = 0 \begin{aligned}
3x - 4y - 2 &= 0
x + 2y - 4 &= 0
3x - 2y + 5 &= 0
\end{aligned} 3 x β 4 y β 2 β = 0 x + 2 y β 4 β = 03 x β 2 y + 5 β = 0 β This system can be written in matrix form as:
[ 3 β 4 β 21 2 β 43 β 2 5 ] [ x y z ] = [ 000 ] \left[\begin{array}{ccc}
3 & -4 & -2
1 & 2 & -4
3 & -2 & 5
\end{array}\right]
\left[\begin{array}{c}
x
y
z
\end{array}\right]
=
\left[\begin{array}{c}
0
0
0
\end{array}\right] [ 3 β β 4 β β 21 β 2 β β 43 β β 2 β 5 β ] [ x yz β ] = [ 000 β ] Determinant of the Coefficient Matrix:
We calculate the determinant of the coefficient matrix A A A
β£ A β£ = β£ 3 β 4 β 21 2 β 43 β 2 5 β£ |A| = \begin{vmatrix} 3 & -4 & -2 1 & 2 & -4 3 & -2 & 5 \end{vmatrix} β£ A β£ = β 3 β β 4 β β 21 β 2 β β 43 β β 2 β 5 β β Expanding this determinant:
β£ A β£ = 3 Γ β£ 2 β 4 β 2 5 β£ β ( β 4 ) Γ β£ 1 β 43 5 β£ + ( β 2 ) Γ β£ 1 23 β 2 β£ |A| = 3 \times \begin{vmatrix} 2 & -4 -2 & 5 \end{vmatrix}
- (-4) \times \begin{vmatrix} 1 & -4 3 & 5 \end{vmatrix}
+ (-2) \times \begin{vmatrix} 1 & 2 3 & -2 \end{vmatrix} β£ A β£ = 3 Γ β 2 β β 4 β 2 β 5 β β β ( β 4 ) Γ β 1 β β 43 β 5 β β + ( β 2 ) Γ β 1 β 23 β β 2 β β Calculating each 2x2 determinant:
β£ 2 β 4 β 2 5 β£ = ( 2 ) ( 5 ) β ( β 4 ) ( β 2 ) = 10 β 8 = 2 \begin{vmatrix} 2 & -4 -2 & 5 \end{vmatrix} = (2)(5) - (-4)(-2) = 10 - 8 = 2 β 2 β β 4 β 2 β 5 β β = ( 2 ) ( 5 ) β ( β 4 ) ( β 2 ) = 10 β 8 = 2 β£ 1 β 43 5 β£ = ( 1 ) ( 5 ) β ( β 4 ) ( 3 ) = 5 + 12 = 17 \begin{vmatrix} 1 & -4 3 & 5 \end{vmatrix} = (1)(5) - (-4)(3) = 5 + 12 = 17 β 1 β β 43 β 5 β β = ( 1 ) ( 5 ) β ( β 4 ) ( 3 ) = 5 + 12 = 17 β£ 1 23 β 2 β£ = ( 1 ) ( β 2 ) β ( 2 ) ( 3 ) = β 2 β 6 = β 8 \begin{vmatrix} 1 & 2 3 & -2 \end{vmatrix} = (1)(-2) - (2)(3) = -2 - 6 = -8 β 1 β 23 β β 2 β β = ( 1 ) ( β 2 ) β ( 2 ) ( 3 ) = β 2 β 6 = β 8 Substituting these values back:
β£ A β£ = 3 ( 2 ) β ( β 4 ) ( 17 ) + ( β 2 ) ( β 8 ) = 6 + 68 + 16 = 90 |A| = 3(2) - (-4)(17) + (-2)(-8) = 6 + 68 + 16 = 90 β£ A β£ = 3 ( 2 ) β ( β 4 ) ( 17 ) + ( β 2 ) ( β 8 ) = 6 + 68 + 16 = 90 Since β£ A β£ = 90 β 0 |A| = 90 \neq 0 β£ A β£ = 90 ξ = 0 not concurrent .
Conclusion
Case (a) : The lines are concurrent.Case (b) : The lines are not concurrent.Case (c) : The lines are not concurrent. 
