LU Decomposition by hand
Compute the LU Decomposition of
$$
A = \left(
\begin{array}{ccc}
1 & 4 & 1 \\
5 & 17 & 3 \\
1 & -2 & -6
\end{array}
\right)
$$
and use this to solve the system $A\vec{x} = \langle 0, 0, 1 \rangle$.
Note: While you can certainly use the computer to check your work, you should try it by hand as this is exactly the kind thing we might see on a quiz.
Comments
Step 1.) Solve for P, by hand:
I went ahead and did some row swaps, trying to fit the mold of the partial pivot method:
$$
PA = \left(\begin{array}{ccc}
5 & 17 & 3 \\
1 & 4 & 1 \\
1 & -2 & -6 \\
\end{array}\right)$$
Than python told me my P matrix was wrong and I had to think about why that was. It seams that 4 > -2 was the best way to go, but I didn't take into account that |-2 -17|>|4-17|, assuming that this was the correct choice, I followed suit with what PA of A should be
$$PA = \left(\begin{array}{ccc}
5 & 17 & 3 \\
1 & -2 & -6 \\
1 & 4 & 1\\
\end{array}\right)\rightarrow
P=\left(\begin{array}{ccc}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
\end{array}\right)$$
Step 2.) Solve for U and L
I found this shorthand method online, that makes writing this down a lot easier. http://www.ohiouniversityfaculty.com/youngt/IntNumMeth/lecture12.pdf
The short version of it is: On your way to get the U matrix, catalog your row operations in each matrix. This is done, by everytime you create a 0 to form a new pivot, put the multiplier you used to add the rows in (parenthese) of that zero. When you go to make your L matrix, take all the (parentheses) and flip the signs, along with the identity matrix. I'll demonstrate the steps here:
$$ PA = \left(\begin{array}{ccc}
5 & 17 & 3 \\
1 & -2 & -6 \\
1 & 4 & 1 \\
\end{array}\right) \sim
\left(\begin{array}{ccc}
5 & 17 & 3 \\
(-1/5) & -5.4 & -6.6 \\
1 & 4 & 1\\
\end{array}\right)
\sim$$
$$ \left(\begin{array}{ccc}
5 & 17 & 3 \\
(-1/5) & -5.4 & -6.6 \\
(-1/5) & 0.6 & 0.4\\
\end{array}\right) \sim
\left(\begin{array}{ccc}
5 & 17 & 3 \\
(-1/5) & -5.4 & -6.6 \\
(-1/5) & (1/9) & 1/3\\
\end{array}\right)$$
This gives us both the U and L matrix.
Where as:
$$ U = \left(\begin{array}{ccc}
5 & 17 & 3 \\
0 & -5.4 & -6.6 \\
0 & 0 & -1/3\\
\end{array}\right)$$
and
$$ L = \left(\begin{array}{ccc}
1 & 0 & 0 \\
1/5 & 1 & 0 \\
1/5 & -1/9 & 1\\
\end{array}\right)$$
Step 3.)
$$ A\vec{x} = \left(\begin{array{c}
0, 0, 1\\
\end{array}\right$$
I went by the way of the book and class notes for this one, which seems like the easiest notation for writing it out by hand.
Where as:
$$LU\vec{x} = \left(\begin{array}{c}
0, 0, 1\\
\end{array}\right)\rightarrow $$
$$L\vec{c} = \left(\begin{array}{c}
0, 0, 1 \\
\end{array}\right)$$
$$ =\left(\begin{array}{ccc}
1 & 0 & 0 \\
1/5 & 1 & 0 \\
1/5 & -1/9 & 1\\
\end{array}\right)
\begin{bmatrix}
a\\
b\\
c\\
\end{bmatrix}
= \begin{bmatrix}
0\\
0\\
1\\
\end{bmatrix}$$
$$ \rightarrow
a =0, b =0, c = 1$$
From here: Ux = c
$$U = \left(\begin{array}{ccc}
5 & 17 & 3 \\
0 & -5.4 & -6.6 \\
0 & 0 & -1/3\\
\end{array}\right)
\begin{bmatrix}
x\\
y\\
z\\
\end{bmatrix}$$
$$ \rightarrow 5x+17y+3z = 0$$
$$ -5.4y+66.z = 0$$
$$z = -3$$
$$ \rightarrow x = \frac{-32}{3}, y=\frac{11}{3}, and z = -3$$
I did some elementary row operations,
$$-5R_{1}+R_{2}\\
-R_{1}+R_{3}\\
-2R_{2}+R_{3},$$
and found the $L$ and $U$ matrices,
$$L = \begin{pmatrix}
1&0&0\\
5&1&0\\
1&2&1\end{pmatrix}$$
$$U = \begin{pmatrix}
1&4&1\\
0&-3&-2\\
0&0&-3\end{pmatrix}.$$
I then solved $L\vec{c} = \vec{b}$ and $U\vec{x} = \vec{c}$ through back substitution and found that
$$\vec{b} = \begin{pmatrix}
-\frac{5}{9}\\
\frac{2}{9}\\
-\frac{1}{3}\end{pmatrix}.$$
@frank I think your third elementary row operation should be $-2R_2 + R_3$.
Yes, that looks much better. Thanks!