LU Decomposition from Quiz 1
How do I do the LU Decomposition problem from Quiz 1? This asked us to find the LU Decomposition of
$$A = \left(
\begin{array}{ccc}
-3 & 4 & -2 \\
-6 & 7 & -5 \\
-12 & 17 & -8
\end{array}
\right)$$
and then use that to find the solution of
$$A\left(
\begin{array}{c}
x \\ y \\ z
\end{array}
\right) = \left(
\begin{array}{c}
0 \\ 0 \\ 1
\end{array}
\right).$$
Comments
$$A = \begin{pmatrix}
-3 & 4 & -2\\
-6 & 7 & -5\\
-12 & 17 & -8
\end{pmatrix}\\$$
Use row operations to get into upper triangular form:
$$(-4)R_1 + R_3 \\
\big{(} -(-4) = L_{31} \big{)}$$
$$\begin{pmatrix}
-3 & 4 & -2 \\
-6 & 7 & -5 \\
0 & 1 & 0
\end{pmatrix}\\$$
$$(-2)R_1 + R_2 \\
\big{(} -(-2) = L_{21} \big{)}$$
$$\begin{pmatrix}
-3 & 4 & -2 \\
0 & -1 & -1 \\
0 & 1 & 0
\end{pmatrix}\\$$
$$(1)R_2 + R_3 \\
\big{(} -(1) = L_{32} \big{)}$$
$$\boxed{U = \begin{pmatrix}
-3 & 4 & -2 \\
0 & -1 & -1 \\
0 & 0 & -1
\end{pmatrix}}\\$$
Construct L matrix,
$$\boxed{L = \begin{pmatrix}
1 & 0 & 0 \\
2 & 1 & 0 \\
4 & -1 & 1
\end{pmatrix}}\\$$
Use L and U to solve for $\vec{x}$ :
$$A \vec{x} = LU \vec{x} = \begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix}\\$$
$$\vec{c} = U \vec{x} \\$$
$$L \vec{c} = \begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix} \\$$
$$\begin{pmatrix}
1 & 0 & 0 \\
2 & 1 & 0 \\
4 & -1 & 1
\end{pmatrix}
\begin{pmatrix}
a \\
b \\
c
\end{pmatrix} =
\begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix}\\$$
$$\vec{c} = \begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix}\\$$
$$U \vec{x} = \vec{c}$$
$$\begin{pmatrix}
-3 & 4 & -2 \\
0 & -1 & -1 \\
0 & 0 & -1
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
z
\end{pmatrix} =
\begin{pmatrix}
0 \\
0 \\
1
\end{pmatrix}\\$$
$$\boxed{\vec{x} = \begin{pmatrix}
2 \\
1 \\
-1
\end{pmatrix}}$$
@joshua That looks great - thanks!