Solve the system of equations [[4x_{1} + x_{2} + x_{3} = 4], [x_{1} + 4x_{2} - 2x_{3} = 4], [3x_{1} + 2x_{2} - 4x_{3} = 6]] , by using LU
Decomposition method with I_{33} = I_{22} = I_{33} = 1
To solve the system of equations using LU decomposition, we need to find the LU decomposition of the coefficient matrix, A.
The coefficient matrix, A, is given by:
A = [[4, 1, 1], [1, 4, -2], [3, 2, -4]]
We can find the LU decomposition of A using the following steps:
1. Set L = I (identity matrix) and U = A.
L = [[1, 0, 0], [0, 1, 0], [0, 0, 1]]
U = [[4, 1, 1], [1, 4, -2], [3, 2, -4]]
2. Perform Gaussian elimination to eliminate the elements below the main diagonal of U.
- Multiply the first row of U by 1/4.
U = [[1, 1/4, 1/4], [1, 4, -2], [3, 2, -4]]
- Subtract the first row of U multiplied by 1 from the second row of U.
U = [[1, 1/4, 1/4], [0, 15/4, -9/4], [3, 2, -4]]
- Subtract the first row of U multiplied by 3 from the third row of U.
U = [[1, 1/4, 1/4], [0, 15/4, -9/4], [0, 5/4, -19/4]]
3. The resulting U matrix is the upper triangular matrix.
U = [[1, 1/4, 1/4], [0, 15/4, -9/4], [0, 5/4, -19/4]]
4. The L matrix can be found by looking at the multipliers used during the Gaussian elimination steps.
Since we multiplied the first row of U by 1/4 to eliminate the elements below the main diagonal, the first element of L is 1/4.
L = [[1/4, 0, 0], [0, 1, 0], [0, 0, 1]]
Since we subtracted the first row of U multiplied by 1 to eliminate the elements below the main diagonal in the second row, the second element of L is 1.
L = [[1/4, 0, 0], [1, 1, 0], [0, 0, 1]]
Since we subtracted the first row of U multiplied by 3 to eliminate the elements below the main diagonal in the third row, the third element of L is 3.
L = [[1/4, 0, 0], [1, 1, 0], [3, 0, 1]]
5. The LU decomposition of A is given by:
L = [[1/4, 0, 0], [1, 1, 0], [3, 0, 1]]
U = [[1, 1/4, 1/4], [0, 15/4, -9/4], [0, 5/4, -19/4]]
Now that we have the LU decomposition of the coefficient matrix, we can solve the system of equations.
Let's denote the unknown vector as X = [x1, x2, x3].
From the LU decomposition, we have:
AX = LU X
LUX = b
(UX) = Y
Let's solve for Y first.
From the equation (U*X) = Y, we can see that:
y1 = 4
(15/4)*y1 + (-9/4)*y2 = 4
(5/4)*y2 + (-19/4)*y3 = 6
Substituting the value of y1 from the first equation, we get:
(15/4)*4 + (-9/4)*y2 = 4
15 + (-9/4)*y2 = 4
(-9/4)*y2 = -11
y2 = (4/9)*11
Substituting the value of y2 into the third equation, we get:
(5/4)*(4/9)*11 + (-19/4)*y3 = 6
(5/9)*55 + (-19/4)*y3 = 6
(-19/4)*y3 = 6 - (5/9)*55
(-19/4)*y3 = 6 - 61.11
(-19/4)*y3 = -55.11
y3 = (-4/19)*55.11
Therefore, we have:
y1 = 4
y2 = (4/9)*11
y3 = (-4/19)*55.11
Now, let's solve for X.
From the equation UX = Y, we can see that:
x3 = y3 / (-19/4)
x3 = (-4/19)*55.11 / (-19/4)
x3 = 55.11
Substituting the value of x3 into the second equation, we get:
(15/4)*x2 + (-9/4)*x3 = y2
(15/4)*x2 + (-9/4)*55.11 = (4/9)*11
(15/4)*x2 = (4/9)*11 + (9/4)*55.11
(15/4)*x2 = (4/9)*11 + (9/4)*55.11
(15/4)*x2 = (4/9)*11 + (9/4)*55.11
(15/4)*x2 = 88.78
x2 = (4/15)*88.78
Similarly, substituting the value of x3 into the first equation, we get:
x1 + (1/4)*x2 + (1/4)*x3 = y1
x1 + (1/4)*(4/15)*88.78 + (1/4)*55.11 = 4
x1 + (1/15)*88.78 + (1/4)*55.11 = 4
x1 + 5.9187 + 13.7775 = 4
x1 = 4 - 5.9187 - 13.7775
x1 = -15.6962
Therefore, we have:
x1 = -15.6962
x2 = (4/15)*88.78
x3 = 55.11
Hence, the solution to the system of equations is:
x1 = -15.6962
x2 = 23.6728
x3 = 55.11