Answer
\[
\mathbf{x} = \begin{bmatrix}
-5 \\
1 \\
3
\end{bmatrix}
\]
Work Step by Step
Given
\[
A = \begin{bmatrix}
2 & -2 & 4 \\
1 & -3 & 1 \\
3 & 7 & 5
\end{bmatrix}, \quad
\mathbf{b} = \begin{bmatrix}
0 \\
-5 \\
7
\end{bmatrix}
\]
\[
A = LU \text{ where } L = \begin{bmatrix}
1 & 0 & 0 \\
1/2 & 1 & 0 \\
3/2 & -5 & 1
\end{bmatrix}, \quad
U = \begin{bmatrix}
2 & -2 & 4 \\
0 & -2 & -1 \\
0 & 0 & -6
\end{bmatrix}
\]
Step 1: Solve $L\mathbf{y} = \mathbf{b}$ for $\mathbf{y}$
We need to solve:
\[
\begin{bmatrix}
1 & 0 & 0 \\
1/2 & 1 & 0 \\
3/2 & -5 & 1
\end{bmatrix}
\begin{bmatrix}
y_1 \\
y_2 \\
y_3
\end{bmatrix}
=
\begin{bmatrix}
0 \\
-5 \\
7
\end{bmatrix}
\]
Forward substitution (top to bottom):
Row 1:
\begin{align*}
1 \cdot y_1 &= 0 \\
y_1 &= 0
\end{align*}
Row 2:
\begin{align*}
\frac{1}{2} \cdot y_1 + 1 \cdot y_2 &= -5 \\
\frac{1}{2} \cdot 0 + y_2 &= -5 \\
y_2 &= -5
\end{align*}
Row 3:
\begin{align*}
\frac{3}{2} \cdot y_1 + (-5) \cdot y_2 + 1 \cdot y_3 &= 7 \\
\frac{3}{2} \cdot 0 + (-5) \cdot (-5) + y_3 &= 7 \\
0 + 25 + y_3 &= 7 \\
y_3 &= -18
\end{align*}
So: $\mathbf{y} = \begin{bmatrix} 0 \\ -5 \\ -18 \end{bmatrix}$
Step 2: Solve $U\mathbf{x} = \mathbf{y}$ for $\mathbf{x}$
We need to solve:
\[
\begin{bmatrix}
2 & -2 & 4 \\
0 & -2 & -1 \\
0 & 0 & -6
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}
=
\begin{bmatrix}
0 \\
-5 \\
-18
\end{bmatrix}
\]
Backward substitution (bottom to top):
Row 3:
\begin{align*}
-6 \cdot x_3 &= -18 \\
x_3 &= 3
\end{align*}
Row 2:
\begin{align*}
-2 \cdot x_2 + (-1) \cdot x_3 &= -5 \\
-2 \cdot x_2 + (-1) \cdot 3 &= -5 \\
-2 \cdot x_2 - 3 &= -5 \\
-2 \cdot x_2 &= -2 \\
x_2 &= 1
\end{align*}
Row 1:
\begin{align*}
2 \cdot x_1 + (-2) \cdot x_2 + 4 \cdot x_3 &= 0 \\
2 \cdot x_1 + (-2) \cdot 1 + 4 \cdot 3 &= 0 \\
2 \cdot x_1 - 2 + 12 &= 0 \\
2 \cdot x_1 + 10 &= 0 \\
2 \cdot x_1 &= -10 \\
x_1 &= -5
\end{align*}
So the final answer is:
\[
\mathbf{x} = \begin{bmatrix}
-5 \\
1 \\
3
\end{bmatrix}
\]
