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
$$A\left(
\begin{array}{c}
x \\ y \\ z
\end{array}
\right) = \left(
\begin{array}{c}
0 \\ 0 \\ 1
\end{array}
\right).$$

mark

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 $\vec{x}$ :

    $$A \vec{x} = LU \vec{x} = \begin{pmatrix}
    0 \\
    0 \\
    1
    \end{pmatrix}\\$$

    $$\vec{c} = U \vec{x} \\$$

    $$L \vec{c} = \begin{pmatrix}
    0 \\
    0 \\
    1
    \end{pmatrix} \\$$

    $$\begin{pmatrix}
    1 & 0 & 0 \\
    2 & 1 & 0 \\
    4 & -1 & 1
    \end{pmatrix}
    \begin{pmatrix}
    a \\
    b \\
    c
    \end{pmatrix} =
    \begin{pmatrix}
    0 \\
    0 \\
    1
    \end{pmatrix}\\$$

    $$\vec{c} = \begin{pmatrix}
    0 \\
    0 \\
    1
    \end{pmatrix}\\$$

    $$U \vec{x} = \vec{c}$$

    $$\begin{pmatrix}
    -3 & 4 & -2 \\
    0 & -1 & -1 \\
    0 & 0 & -1
    \end{pmatrix}
    \begin{pmatrix}
    x \\
    y \\
    z
    \end{pmatrix} =
    \begin{pmatrix}
    0 \\
    0 \\
    1
    \end{pmatrix}\\$$

    $$\boxed{\vec{x} = \begin{pmatrix}
    2 \\
    1 \\
    -1
    \end{pmatrix}}$$

    audreymark
  • @joshua That looks great - thanks!

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