Solve the system of equations by first expressing it in matrix form as and then evaluating.
a). 3x-2y=5
4x-y =-10
b). 3x -2y =-2
4x -y = 3
To solve a system of linear equations by first expressing it in matrix form, we can use the equation AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.
Let's first solve the system of equations for part (a):
3x - 2y = 5 (Equation 1)
4x - y = -10 (Equation 2)
To express the system in matrix form, we need to rearrange the equations:
3x - 2y = 5 => 3x - 2y - 5 = 0 (Equation 1)
4x - y = -10 => 4x - y + 10 = 0 (Equation 2)
Now, we can write the coefficient matrix A, the variable matrix X, and the constant matrix B:
A = |3 -2| X = |x| B = |5|
|4 -1| |y| |-10|
To evaluate the matrix equation AX = B, we can use matrix algebra.
First, we will find the inverse of A:
Calculate the determinant of A: |A| = (3*(-1)) - (-2*4) = -3 + 8 = 5
Since the determinant is non-zero, A is invertible.
Next, find the inverse of A: A^(-1) = (1/|A|) * adj(A)
Where adj(A) is the adjugate of A, which is obtained by swapping the elements on the main diagonal and changing the sign of the elements outside the main diagonal.
A^(-1) = (1/5) * |-1 2| = |-1/5 2/5|
|4 -3| |-4/5 3/5|
Now, multiply both sides of the equation AX = B by A^(-1) to solve for X:
A^(-1) * AX = A^(-1) * B
X = A^(-1) * B
Multiplying the matrices together, we have:
X = |-1/5 2/5| * |5| = |1|
|-4/5 3/5| |-10| |-2|
So, the solution to the system of equations is x = 1 and y = -2.
Now, let's solve the system of equations for part (b):
3x - 2y = -2 (Equation 1)
4x - y = 3 (Equation 2)
To express the system in matrix form, rearrange the equations:
3x - 2y = -2 => 3x - 2y + 2 = 0 (Equation 1)
4x - y = 3 => 4x - y - 3 = 0 (Equation 2)
The coefficient matrix A, the variable matrix X, and the constant matrix B are:
A = |3 -2| X = |x| B = |-2|
|4 -1| |y| | 3|
To evaluate the matrix equation AX = B, follow the same steps as before.
First, calculate the determinant of A: |A| = (3*(-1)) - (-2*4) = -3 + 8 = 5
Since the determinant is non-zero, A is invertible.
Next, find the inverse of A: A^(-1) = (1/|A|) * adj(A)
Calculate the inverse of A^(-1) = (1/5) * |-1 2| = |-1/5 2/5|
| 4 -3| |-4/5 3/5|
Multiply both sides of the equation AX = B by A^(-1) to solve for X:
A^(-1) * AX = A^(-1) * B
X = A^(-1) * B
Multiplying the matrices together:
X = |-1/5 2/5| * |-2| = |1|
|-4/5 3/5| | 3| |-1|
Therefore, the solution to the system of equations is x = 1 and y = -1.