# LU Decomposition by hand

edited January 22 in Problems

Compute the LU Decomposition of
$$A = \left( \begin{array}{ccc} 1 & 4 & 1 \\ 5 & 17 & 3 \\ 1 & -2 & -6 \end{array} \right)$$
and use this to solve the system $A\vec{x} = \langle 0, 0, 1 \rangle$.

Note: While you can certainly use the computer to check your work, you should try it by hand as this is exactly the kind thing we might see on a quiz.

• edited February 3

Step 1.) Solve for P, by hand:
I went ahead and did some row swaps, trying to fit the mold of the partial pivot method:

$$PA = \left(\begin{array}{ccc} 5 & 17 & 3 \\ 1 & 4 & 1 \\ 1 & -2 & -6 \\ \end{array}\right)$$
Than python told me my P matrix was wrong and I had to think about why that was. It seams that 4 > -2 was the best way to go, but I didn't take into account that |-2 -17|>|4-17|, assuming that this was the correct choice, I followed suit with what PA of A should be

$$PA = \left(\begin{array}{ccc} 5 & 17 & 3 \\ 1 & -2 & -6 \\ 1 & 4 & 1\\ \end{array}\right)\rightarrow P=\left(\begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array}\right)$$

Step 2.) Solve for U and L
I found this shorthand method online, that makes writing this down a lot easier. http://www.ohiouniversityfaculty.com/youngt/IntNumMeth/lecture12.pdf
The short version of it is: On your way to get the U matrix, catalog your row operations in each matrix. This is done, by everytime you create a 0 to form a new pivot, put the multiplier you used to add the rows in (parenthese) of that zero. When you go to make your L matrix, take all the (parentheses) and flip the signs, along with the identity matrix. I'll demonstrate the steps here:

$$PA = \left(\begin{array}{ccc} 5 & 17 & 3 \\ 1 & -2 & -6 \\ 1 & 4 & 1 \\ \end{array}\right) \sim \left(\begin{array}{ccc} 5 & 17 & 3 \\ (-1/5) & -5.4 & -6.6 \\ 1 & 4 & 1\\ \end{array}\right) \sim$$

$$\left(\begin{array}{ccc} 5 & 17 & 3 \\ (-1/5) & -5.4 & -6.6 \\ (-1/5) & 0.6 & 0.4\\ \end{array}\right) \sim \left(\begin{array}{ccc} 5 & 17 & 3 \\ (-1/5) & -5.4 & -6.6 \\ (-1/5) & (1/9) & 1/3\\ \end{array}\right)$$

This gives us both the U and L matrix.

Where as:

$$U = \left(\begin{array}{ccc} 5 & 17 & 3 \\ 0 & -5.4 & -6.6 \\ 0 & 0 & -1/3\\ \end{array}\right)$$

and

$$L = \left(\begin{array}{ccc} 1 & 0 & 0 \\ 1/5 & 1 & 0 \\ 1/5 & -1/9 & 1\\ \end{array}\right)$$

Step 3.)
$$A\vec{x} = \left(\begin{array{c} 0, 0, 1\\ \end{array}\right$$

I went by the way of the book and class notes for this one, which seems like the easiest notation for writing it out by hand.

Where as:

$$LU\vec{x} = \left(\begin{array}{c} 0, 0, 1\\ \end{array}\right)\rightarrow$$

$$L\vec{c} = \left(\begin{array}{c} 0, 0, 1 \\ \end{array}\right)$$

$$=\left(\begin{array}{ccc} 1 & 0 & 0 \\ 1/5 & 1 & 0 \\ 1/5 & -1/9 & 1\\ \end{array}\right) \begin{bmatrix} a\\ b\\ c\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 1\\ \end{bmatrix}$$

$$\rightarrow a =0, b =0, c = 1$$

From here: Ux = c

$$U = \left(\begin{array}{ccc} 5 & 17 & 3 \\ 0 & -5.4 & -6.6 \\ 0 & 0 & -1/3\\ \end{array}\right) \begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}$$

$$\rightarrow 5x+17y+3z = 0$$
$$-5.4y+66.z = 0$$
$$z = -3$$

$$\rightarrow x = \frac{-32}{3}, y=\frac{11}{3}, and z = -3$$

• edited February 5

I did some elementary row operations,

$$-5R_{1}+R_{2}\\ -R_{1}+R_{3}\\ -2R_{2}+R_{3},$$

and found the $L$ and $U$ matrices,

$$L = \begin{pmatrix} 1&0&0\\ 5&1&0\\ 1&2&1\end{pmatrix}$$

$$U = \begin{pmatrix} 1&4&1\\ 0&-3&-2\\ 0&0&-3\end{pmatrix}.$$

I then solved $L\vec{c} = \vec{b}$ and $U\vec{x} = \vec{c}$ through back substitution and found that

$$\vec{b} = \begin{pmatrix} -\frac{5}{9}\\ \frac{2}{9}\\ -\frac{1}{3}\end{pmatrix}.$$

• @frank I think your third elementary row operation should be $-2R_2 + R_3$.

• Yes, that looks much better. Thanks!