Another LU decomposition by hand!
This problem is just like this one - it's just another version so you can have more practice!
Compute the LU Decomposition of
$$
A = \left(
\begin{array}{ccc}
1 & 2 & -2 \\
-3 & -7 & 5 \\
-1 & -3 & 2
\end{array}
\right)
$$
and use this to solve the system $A\vec{x} = \langle 0, 0, 1 \rangle$.
Comments
To obtain the matrix $U$ I used the row operations:
$
3R_1+R_2\\
R_1+R_3 \\
-R_2+R_3 \\
$
on $A$ with the result
$$
U=
\
\left[
\begin{array}{ccc}
1 & 2 & -2 \\
0 & -1 & -1\\
0 & 0 & 1\\
\end{array}
\right]
\
$$
As usual $L$ is made from the coefficients on the scaled row from the rows operations, but with the opposite sign. $L$ also has 1s on its diagonal. The result is:
$$
L=
\
\left[
\begin{array}{ccc}
1 & 0 & 0 \\
-3 & 1 & 0\\
-1 & 1 & 1\\
\end{array}
\right]
\
$$
Now to solve $\langle 0,0, 1\rangle$ I used no operations for L as forward row operations have no effect on the vector. Next augmenting it on $U$ and using the operations:
$
R_3+R_2 \\
2R_3+R_1 \\
2R_2+R_1 \\
-R_2 \\
$
resulted in:
$$
\
\left[
\begin{array}{ccc|c}
1 & 0 & 0 & 4\\
0 & 1 & 0 & -1\\
0 & 0 & 1 & 1\\
\end{array}
\right]
\
$$
Interesting - Note that I gave this a "Good answer" vote, rather than a "THE Answer" vote. Generally, forward/back substitution is more efficient than a full reduction to reduced row echelon form like this.
$$A = \begin{pmatrix}
1 & 2 & -2\\
-3 & -7 & 5\\
-1 & -3 & 2
\end{pmatrix}\\$$
Use row operations to get into upper triangular form:
$$(1)R_1 + R_3 \\
\big{(} -(1) = L_{31} \big{)}$$
$$\begin{pmatrix}
1 & 2 & -2 \\
-3 & -7 & 5 \\
0 & -1 & 0
\end{pmatrix}\\$$
$$(3)R_1 + R_2 \\
\big{(} -(3) = L_{21} \big{)}$$
$$\begin{pmatrix}
1 & 2 & -2 \\
0 & -1 & -1 \\
0 & -1 & 0
\end{pmatrix}\\$$
$$(-1)R_2 + R_3 \\
\big{(} -(-1) = L_{32} \big{)}$$
$$\boxed{U = \begin{pmatrix}
1 & 2 & -2 \\
0 & -1 & -1 \\
0 & 0 & 1
\end{pmatrix}}\\$$
Construct L matrix,
$$\boxed{L = \begin{pmatrix}
1 & 0 & 0 \\
-3 & 1 & 0 \\
-1 & 1 & 1
\end{pmatrix}}\\$$
Use L and U to solve for $\vec{x}$ :
$$A \vec{x} = LU \vec{x} = \begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix}\\$$
$$\vec{c} = U \vec{x} \\$$
$$L \vec{c} = \begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix} \\$$
$$\begin{pmatrix}
1 & 0 & 0 \\
-3 & 1 & 0 \\
-1 & 1 & 1
\end{pmatrix}
\begin{pmatrix}
a \\
b \\
c
\end{pmatrix} =
\begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix}\\$$
$$\vec{c} = \begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix}\\$$
$$U \vec{x} = \vec{c}$$
$$\begin{pmatrix}
1 & 2 & -2 \\
0 & -1 & -1 \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
z
\end{pmatrix} =
\begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix}\\$$
$$\boxed{\vec{x} = \begin{pmatrix}
4 \\
-1 \\
1
\end{pmatrix}}$$