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

• $$A = \begin{pmatrix} -3 & 4 & -2\\ -6 & 7 & -5\\ -12 & 17 & -8 \end{pmatrix}\\$$

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!