1. ## Coefficient Matrix Question

Here is the question I am having difficulty with:

If the echelon form of the coefficient matrix of a homogeneous system of linear equations is:

1 2 3 0 2 1
0 0 1 1 2 0
0 0 0 1 2 1
0 0 0 0 0 0
0 0 0 0 0 0

Find a set of basic solutions of this system

Any tips on where to start?

2. You have 3 equations and 5 unknows, so the dimension of the solution has 5-3 = 2 degrees of freedom (This is true for every system of linear equations. Number of unkowns minus number of linearly indepedent equations = degrees of freedom or dimension of the solution.) Think for example $x_1=\lambda$ and $x_2=\mu$ and solve it as a function of $\lambda$ and $\mu$

3. Originally Posted by dr4g0n9
Here is the question I am having difficulty with:

If the echelon form of the coefficient matrix of a homogeneous system of linear equations is:

1 2 3 0 2 1
0 0 1 1 2 0
0 0 0 1 2 1
0 0 0 0 0 0
0 0 0 0 0 0

Find a set of basic solutions of this system

Any tips on where to start?
Just set up the equations corresponding to that matrix: I am going to call the unknown numbers x, y, z, u, and v. The bottom non-zero row gives the equation u+ 2v= 1 so u= 1- 2v. The row just above that is z+ u+ 2v= 0 so z= -u- 2v and since u= 1- 2v, z= -(1- 2v)- 2v= -1+ 2v- 2v= -1. Finally, the top row gives x+ 2y+ 3z+ 2v= 1. Since z= -1, that becomes x+ 2y- 3+ 2v= 1 or x= 4-2y- 2v. Thus, we have all unknown numbers written in terms of y and v.

Another way to do this is to continue the row-reduction getting "0"s above the diagonal. Subtract three times the second row from the first row to get $\begin{bmatrix} 1 & 2 & 0 & -3 & -4 & 1 \\0 & 0 & 1 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0\end{bmatrix}$

Now subtract the third row from the second to get $\begin{bmatrix} 1 & 2 & 0 & -3 & -4 & 1 \\0 & 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 & 0\end{bmatrix}$.

Finally, subtract 3 times the third row from the first row to get $1 & 2 0 & 0 & 2 & 4 \\ 0 & 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0\end{bmatrix}$

Now the rows correspond to equations x+ 2y+ 2v= 4 so x= 4- 2y- 2v, z= -1, and u+ 2v= 1 so u= 1- 2v as before.

4. You can also just sort of look by inspection, you want to look from bottom up. You are looking for a 6x1 vector that satisfies this matrix equation being 0.

The bottom two rows give you no information. Row three you need to find a linear combination so that $a1 + b2 + c1=0$. (-1,1,-1) seems like a good choice, so make those your last 3 entries in your vector.

Now take that and look at the next row and see what you need to do to make it continue to work.
$1a + 1(-1) + 2(1) =0 \Rightarrow a=-1$

So now your vector looks like <*,*, -1, -1, 1, -1> and you just have 1 equation left to satisfy.

$1a + 2b + 3(-1) + 0(-1) + 2(1) + 1(-1) = 0 \Rightarrow a + 2b -2 =0 \Rightarrow a + 2b = 2$

<-2,2, -1, -1, 1, -1> works.

Oh I just thought you needed a solution, didn't realize you needed the whole set of solutions, should have read more carefully.