Another LU decomposition by hand!

edited February 2 in Problems

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

  • edited February 4

    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:
    $$
    \
    \left[
    \begin{array}{ccc|c}
    1 & 0 & 0 & 4\\
    0 & 1 & 0 & -1\\
    0 & 0 & 1 & 1\\
    \end{array}
    \right]
    \
    $$

    mark
  • 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.

  • edited February 9

    $$A = \begin{pmatrix}
    1 & 2 & -2\\
    -3 & -7 & 5\\
    -1 & -3 & 2
    \end{pmatrix}\\$$

    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 $\vec{x}$ :

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

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

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

    $$\begin{pmatrix}
    1 & 0 & 0 \\
    -3 & 1 & 0 \\
    -1 & 1 & 1
    \end{pmatrix}
    \begin{pmatrix}
    a \\
    b \\
    c
    \end{pmatrix} =
    \begin{pmatrix}
    0 \\
    0 \\
    1
    \end{pmatrix}\\$$

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

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

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

    $$\boxed{\vec{x} = \begin{pmatrix}
    4 \\
    -1 \\
    1
    \end{pmatrix}}$$

    mark
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