# 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 $A\vec{x} = \langle 0, 0, 1 \rangle$.

• 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] \$$

• 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 $\vec{x}$ :

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

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

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

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

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

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

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

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