We want to solve the given system of equations using matrices to find the point of intersection of the lines. To do so, we will start by rewriting Equation (I) so that the variable terms are in the same order in both equations.
2y-4x=3 & (I) 4x-3y=- 6 & (II) ⇔ - 4x+2y=3 4x-3y=- 6
We need to write the system as a matrix equation.
- 4x+2y=3 4x-3y=- 6
⇒
[
c
- 4x+2y
4x-3y
]
=
[
c
3
- 6
]
Now, we will write the left-hand side of the above matrix equation as the product of the coefficient matrix and the variable matrix, while the right-hand side is the constant matrix.
[
c
- 4x+2y
4x-3y
]
=
[
c
3
- 6
]
⇕
[
cc
- 4 & 2
4 & - 3
]
*
[
c
x
y
]
=
[
c
3
- 6
]
We will solve the matrix equation by multiplying both sides by the inverse of the coefficient matrix to isolate the x and y variables. When multiplying a matrix by its inverse, we obtain the identity matrix.
A^(- 1)=1/ad-bc [ cc d & - b - c & a ] where ad-bc ≠0
The expression ad-bc is known as the determinant of a 2* 2 matrix. Because it is in the denominator of a fraction, if the determinant is zero the matrix cannot have an inverse. Consider our coefficient matrix.
[
cc
- 4 & 2 4 & - 3
]
Let's calculate its determinant.
Since the determinant is not zero, the matrix has an inverse. We can now apply the formula for the inverse. Note that we usually refer to the determinant using the notation ad-bc=det(A).
For simplicity, instead of multiplying the matrix by the scalar, we will leave the expression as it is.
Multiplying Matrices
In our exercise, we had to multiply a 2* 2 matrix by a 2* 1 matrix. This is possible because the number of columns of the first matrix is equal to the number of rows of the second one. Moreover, the product has as many rows as the first matrix and as many columns as the second one. Therefore, the product is a 2* 1 matrix.
[
cc
a & b
c & d
]
*
[
c
x
y
]
=
[
c
ax+by
cx+dy
]
2 * 2 2 * 1 2 * 1
Let's multiply the matrices of the exercise.
