I have:
$A=\begin{bmatrix} 2 & -1 & 2 & 3 & 4 \\ 4 & -2 & 7 & 7 & 6 \\ 2 & -1 & 20 & 9 & -8 \end{bmatrix}$
and I'm asked to LU-decomposition A, then solve $Ax=0$.
What I did:
With these steps:
$1)$ $R2=R2-2\times R1$
$2)$ $R3=R3-R1$
I got:
$U=\begin{bmatrix} 2 & -1 & 2 & 3 & 4 \\ 0 & 0 & 3 & 1 & -2 \\ 0 & 0 & 18 & 6 & -12 \end{bmatrix}$
$L=\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$
And because:
$Ax=b\hspace{12mm}L(Ux)=b$
$Ly=b\hspace{12mm}Ux=y$
I got:
$Ly=0$
$ \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $
Using this step:
$1)$ $R2=R2-2\times R1$
I got $y=0$
Then solving:
$Ux=y$
$ \begin{bmatrix} 2 & -1 & 2 & 3 & 4 \\ 0 & 0 & 3 & 1 & -2 \\ 0 & 0 & 18 & 6 & -12 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $
And using this steps:
$1)$ $R1=R1+2\times R2$
$2)$ $R3=\frac{R3}{6}$
$3)$ $R3=R3-R2$
I got:
$ \begin{cases} x_2=2x_1+8x_3+5x_4 \\ x_4=-3x_3+2x_5 \\ x_1\,,\,x_3\,,\,x_5\quad free \end{cases} $
And I chose free variables to be $x_1=1$ , $x_3=1$ , $x_5=1$ so I got:
$ \begin{cases} x_1=1 \\ x_2=5 \\ x_3=1 \\ x_4=-1 \\ x_5=1 \end{cases} $
Just wondering if I have done it right? And if not, then how should I do it?
Thanks in advance.
First of all, LU-decomposition is useful for a number of reasons, let me give you one. Suppose you have to solve $Ax = b_1, Ax = b_2, \ldots, Ax = b_n$, you can do this simultaneously by row reducing the augmented matrix $$\begin{pmatrix} A & | & b_1 & b_2 & \ldots &b_n \end{pmatrix}$$ and although this is very practical for small matrices $A$ it becomes very time costly to do this for large matrices. A better way is by finding the LU decomposition of $A$ and solving the following systems of equations: $$Ly = b \quad \text{ and } Ux = y.$$ The reason this works is because solving equations of systems given by a triangular matrix is extremely easy, using what is called forward (in case of a lower triangular matrix) or backward (in case of an upper triangular matrix) substitution. Indeed, suppose we have a lower triangular $3 \times 3$-matrix $L$ and the following system: $$\begin{pmatrix} u_{11} & 0 & 0 \\ u_{21} & u_{22} & 0 \\ u_{31} & u_{32} & u_{33} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2\\ x_3 \end{pmatrix} = \begin{pmatrix} b_1\\b_2\\b_3 \end{pmatrix}$$ then we have from the first row that $x_1 = \frac{b_1}{u_{11}}$ (in case $u_{11}$ is non zero) and plugging in this value for $x_1$ in the second row, we find a unique value for $x_2$ (in case $u_{22}$ is nonzero) and plugging in the values of $x_1, x_2$ in the third row gives us $x_3$. Hence the name 'forward substitution'.
Now for your question: You have found two correct LU decompositions, but I like the one in your question better than the one in your answer since the one in your answer uses an extra (unnecessary) step. So let me work with your first candidate for LU decompostion. I will take over your (correct) result, in order to make this answer easier to follow.
$$L=\begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} \quad \text{ and } \quad U=\begin{pmatrix} 2 & -1 & 2 & 3 & 4 \\ 0 & 0 & 3 & 1 & -2 \\ 0 & 0 & 18 & 6 & -12 \end{pmatrix}.$$ Your idea of using this LU decomposition of $A$ to solve $Ax = \vec{0}$ was correct: first we solve $Ly = \vec{0}$ and then we solve $Ux = y$. So let's take a closer look to the first step. We have the following matrix equation: $$\begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2\\ y_3 \end{pmatrix} = \begin{pmatrix} 0\\0\\0\end{pmatrix}.$$ Using the forward substitution I have described, we find from the first row that $y_1 = 0$. Plugging this value in row two, we find $2y_2 = 0$, hence $Y_2 = 0$ and by the same principle, we have from row $3$ that $Y_3 = 0$. So we find that $(y_1, y_2, y_3)^T = (0,0,0)^T$ (which should not be a surprise, since the $L$ matrix is invertible in this case, making the solution unique and the zero vector is obviously a solution).
Now for your second step. We want to solve the following matrix equation: $$\begin{pmatrix} 2 & -1 & 2 & 3 & 4 \\ 0 & 0 & 3 & 1 & -2 \\ 0 & 0 & 18 & 6 & -12 \end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{pmatrix} = y = \begin{pmatrix}0\\0\\0\end{pmatrix}.$$
You already saw in your answer to your own question that the free variables are $x_2, x_4, x_5$. The reason we have free variables is because we have three equations (and from the $U$ matrix, we see that the last row is equal to the second row, so only $2$ equations) for $5$ variables.
In order to solve this problem and to find all possible solutions, we can do the following: First we solve the system of equations setting $x_4 = 0 = x_5$ (the resulting system of equations has three equations and three variables). We also solve the system of equations where we set $x_2 = x_5 = 0$ and we also solve this system of equations for $x_2 = x_4 = 0$. (These are three separate steps.)
$\color{red}{\text{Solution 1: $x_4 = 0 = x_5$}}$ Using backward substitution, we find that (from the last row of $U$) $x_3 = 0$. From the second equation, we don't get new information. The first equation gives us $2x_1 = x_2$. Since $x_2$ is a free variable, we can set $x_2 = 2$. The resulting solution is $x = (1,2,0,0,0)^T$. Note how easy this solution follows from the upper triangular structure of $U$: we just substituted solutions.
$ \color{red}{\text{Solution 2: $x_2 = 0 = x_5$}}$ Using these values, we find from the last row that $18x_3 = -6x_4$. Picking $x_4 = 3$, we find that $x_3 = -1$. The second row does not give new information. From the first row, we find $2x_1 = -2x_3 - 3x_4$, hence we have that $x_1 = \frac{-7}{2}$. This gives the solution $x = (\frac{-7}{2}, 0, -1, 3, 0)^T$. If you do not like fractions, you could have taken $x_4 = 6$, which leads to the solution $(-7, 0, -2, 6, 0)^T$.
$ \color{red}{\text{Solution 3: $x_2 = 0 = x_4$}}$ Once again, using backward substitution, we find from the last equation that $18x_3 = -12x_5$ and since $x_5$ is a free variable, we can give it the value $x_5 = 6$, making $x_3 = -4$. The second row does not give any new information. The first row give us $2x_1 = -2x_3 -4x_5$. Using our solution for $x_3$ and the value of $x_5$, we find that $x_1 = \frac{1}{2}(-2 \cdot (-4) - 4 \cdot 6) = -8$. Hence we find the solution $x = (-8, 0, 4, 0, 6)^T$.
Now all possible solutions of the given system are linear combinations of these three solutions (so your solution in your answer should also be). Indeed, if $Ax_1 = 0 = Ax_2$, then we have that $A(\lambda_1x_1 + \lambda_2x_2) = \lambda_1Ax_1 + \lambda_2Ax_2 = 0+0 = 0$, where the $\lambda_i$ are real numbers.
This is the general way to proceed. What I did with giving some of the free variables the value $0$ (leading to my 'three' solutions) is actually what you did to obtain one possible solution. The reason I took zero is because this gives easier computations. Note that setting all three free variables equal to zero gives the trivial zero solution. The computations I have made should be correct, but if you would find some computational error, let me know :)