. Solve the system of equations \{\begin{matrix}4x_{1}+x_{2}+x_{3}=4\\ x_{1}+4x_{2}-2x_{3}=4\\ 3x_{1}+2x_{2}-4x_{3}=6\end{matrix}\} by using LU Decomposition method with I_{11}=l_{22}=l_{33}=1
To solve the system of equations using LU decomposition, we need to first decompose the coefficient matrix [A] into lower (L) and upper (U) triangular matrices.
Given that I_{11} = l_{22} = l_{33} = 1, we can write the decomposition as:
[A] = [L][U]
where:
[L] = | 1 0 0 |
| l21 1 0 |
| l31 l32 1 |
[U] = | u11 u12 u13 |
| 0 u22 u23 |
| 0 0 u33 |
Now, let's perform LU decomposition.
From the given coefficients, we have:
u11 = 4
u12 = 1
u13 = 1
u22 = 4 - l21
u23 = -2 - l32
u33 = 6 - l31 - 2l32
We can now substitute these values into the equation [L][U] = [A] and solve for the unknowns l21, l31, and l32.
For the first row of [A], we have:
4 = 1 * u11 + 0 * u12 + 0 * u13
= 1 * 4 + 0 * 1 + 0 * 1
= 4
Therefore, we have u11 = 4.
For the second row of [A], we have:
1 = l21 * u11 + 1 * u12 + 0 * u13
= l21 * 4 + 1 * 1 + 0 * 1
= 4l21 + 1
From this, we can solve for l21:
l21 = (1 - 1) / 4 = 0
Using this value, we can find u22:
u22 = 4 - l21
= 4 - 0
= 4
Now, let's move on to the third row of [A]:
6 = l31 * u11 + l32 * u12 + 1 * u13
= l31 * 4 + l32 * 1 + 1 * 1
= 4l31 + l32 + 1
From this, we can solve for l31:
l31 = (6 - l32 - 1) / 4 = (5 - l32) / 4
Plugging this value into the equation for u33, we have:
u33 = 6 - l31 - 2l32
= 6 - [(5 - l32) / 4] - 2l32
= 24/4 - 5/4 + l32/4 - 8l32/4
= (-1 + 7l32) / 4
Now, we have the LU decomposition:
[L] = | 1 0 0 |
| 0 1 0 |
| (5 - l32)/4 l32 1 |
[U] = | 4 1 1 |
| 0 4 -2 - 0l32 |
| 0 0 (-1 + 7l32) / 4 |
Now, we can solve the system of equations using LU decomposition.
First, solve the equation [L][Y] = [B], where [Y] is a vector containing the solutions for the variables x1, x2, and x3, and [B] is a vector containing the constant terms on the right-hand side of the equations.
For this system of equations:
[B] = | 4 |
| 4 |
| 6 |
We have:
x1 = 4 / 1
x2 = (4 - l32 * 4) / 4
x3 = (6 - (5 - l32)/4 - 2 * l32 * 4) / [( -1 + 7l32) / 4]
Simplifying, we get:
x1 = 4
x2 = 1 - l32
x3 = (1 - l32) / (1 - 7l32)
This is the solution to the system of equations using LU decomposition with the given values of I_{11}, l_{22}, and l_{33}.