Solve the system of equations x1 + x2 + x3 = 1, 3x1 + x2 – 3x3 = 5 and x1 – 2x2 – 5x3 = 10 by LU decomposition method.

Let's go through the steps of performing the LU decomposition manually.

Given the system of equations:

$$\begin{aligned} x_1 + x_2 + x_3 &= 1 \\ 3x_1 + x_2 - 3x_3 &= 5 \\ x_1 - 2x_2 - 5x_3 &= 10 \end{aligned}$$

The coefficient matrix $A$ is:

$$A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 1 & -3 \\ 1 & -2 & -5 \end{bmatrix}$$

We want to decompose $A$ into $L$ (lower triangular) and $U$ (upper triangular) such that:

$$A = LU$$

Where $L$ and $U$ are:

$$L = \begin{bmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{bmatrix} \quad \text{and} \quad U = \begin{bmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{bmatrix}$$

Step 1: Initialize $U$ as $A$ and $L$ as the identity matrix

Start by initializing $U = A$ and $L = I$:

$$U = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 1 & -3 \\ 1 & -2 & -5 \end{bmatrix} , \quad L = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

Step 2: Eliminate elements below the diagonal in the first column

We want to eliminate the elements in the second and third rows, first column, of $U$ by performing elementary row operations and storing the corresponding multipliers in $L$.

• Eliminate $U_{21}$:

To eliminate the element $3$ (in position $U_{21}$), we perform the operation:

$$R_2 = R_2 - 3R_1$$

This results in the new $U$ matrix:

$$U = \begin{bmatrix} 1 & 1 & 1 \\ 0 & -2 & -6 \\ 1 & -2 & -5 \end{bmatrix}$$

The multiplier used is $l_{21} = 3$, so update the $L$ matrix:

$$L = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

• Eliminate $U_{31}$:

To eliminate the element $1$ (in position $U_{31}$), we perform the operation:

$$R_3 = R_3 - 1R_1$$

This results in the new $U$ matrix:

$$U = \begin{bmatrix} 1 & 1 & 1 \\ 0 & -2 & -6 \\ 0 & -3 & -6 \end{bmatrix}$$

The multiplier used is $l_{31} = 1$, so update the $L$ matrix:

$$L = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$$

Step 3: Eliminate elements below the diagonal in the second column

Next, we move to the second column of $U$. We want to eliminate the element in the third row, second column, by performing an elementary row operation and storing the corresponding multiplier in $L$.

• Eliminate $U_{32}$:

To eliminate the element $-3$ (in position $U_{32}$), we perform the operation:

$$R_3 = R_3 - \frac{3}{2}R_2$$

This results in the new $U$ matrix:

$$U = \begin{bmatrix} 1 & 1 & 1 \\ 0 & -2 & -6 \\ 0 & 0 & 3 \end{bmatrix}$$

The multiplier used is $l_{32} = \frac{3}{2}$, so update the $L$ matrix:

$$L = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 1 & \frac{3}{2} & 1 \end{bmatrix}$$

Step 4: Final $L$ and $U$ matrices

Thus, the LU decomposition of matrix $A$ gives:

$$L = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 1 & \frac{3}{2} & 1 \end{bmatrix} \quad \text{and} \quad U = \begin{bmatrix} 1 & 1 & 1 \\ 0 & -2 & -6 \\ 0 & 0 & 3 \end{bmatrix}$$

Step 5: Solve for $y$ in $Ly = b$

Now, to solve the system $Ax = b$, we first solve $Ly = b$ using forward substitution.

The system is:

$$L y = \begin{bmatrix} 1 \\ 5 \\ 10 \end{bmatrix}$$

This gives us the following equations:

$$\begin{aligned} y_1 &= 1 \\ 3y_1 + y_2 &= 5 \\ y_1 + \frac{3}{2}y_2 + y_3 &= 10 \end{aligned}$$

From the first equation:

$$y_1 = 1$$

From the second equation:

$$3(1) + y_2 = 5 \implies y_2 = 2$$

From the third equation:

$$1 + \frac{3}{2}(2) + y_3 = 10 \implies 1 + 3 + y_3 = 10 \implies y_3 = 6$$

Thus, $y = \begin{bmatrix} 1 \\ 2 \\ 6 \end{bmatrix}$.

Step 6: Solve for $x$ in $Ux = y$

Next, we solve $Ux = y$ using backward substitution.

The system is:

$$\begin{bmatrix} 1 & 1 & 1 \\ 0 & -2 & -6 \\ 0 & 0 & 3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 6 \end{bmatrix}$$

This gives us the following equations:

$$\begin{aligned} x_1 + x_2 + x_3 &= 1 \\ -2x_2 - 6x_3 &= 2 \\ 3x_3 &= 6 \end{aligned}$$

From the third equation:

$$x_3 = 2$$

From the second equation:

$$-2x_2 - 6(2) = 2 \implies -2x_2 - 12 = 2 \implies -2x_2 = 14 \implies x_2 = -7$$

From the first equation:

$$x_1 + (-7) + 2 = 1 \implies x_1 - 5 = 1 \implies x_1 = 6$$

Thus, the solution is:

$$x = \begin{bmatrix} 6 \\ -7 \\ 2 \end{bmatrix}$$

Final Answer:

The solution to the system is:

$$x_1 = 6, \quad x_2 = -7, \quad x_3 = 2$$