Finding solutions of a system of linear equations

Question

I was asked the following:

$$\begin{align}2x_1 - 4x_2 + 8x_4 &= 10\\ 3x_1 + 2x_2 &= 3\\ 3x_1 + 2x_3 + 6x_4 &= 11\\ 8x_1 + 6x_2 + 2x_3 + 14x_4 &= 24\end{align}$$ Find the solution set $W$. Is $W$ a vector space?

I know I could express one variable from one of the equations and substitute into the remaining ones. For example, from the second equation I get $x_1=1-\frac23x_2$.

After substituting this into the remaining equations I have $$ \begin{align*} -\frac{16}3x_2+8x_4&=8\\ -2x_2+2x_3+6x_4&=7\\ \frac23x_2+2x_3+14x_4&=16\\ \end{align*} $$ I could continue by expressing another variable and plug into other equations. But this approach seems to be rather cumbersome and if I have more than three equations I am prone to make mistakes. Is there a more elegant way to do this?

I also do not know what to say about the part of the assignment asking about vector space.

Answer 1

If you set up the augmented matrix for the system and perform RREF (Gaussian Elimination), you end up with:

$$ \left[\begin{array}{cccc|c} 1 & 0 & 0 & -2 & -1\\ 0 & 1 & 0 & 3 & 3\\ 0 & 0 & 1 & 6 & 7\\ 0 & 0 & 0 & 0 & 0 \end{array}\right] $$

There are other methods (substitution, inverses, ....), but you did not indicate what you are currently familiar with.

For example, if using substitution, here are hints:

• Use the second equation to solve for $x_1$ and substitute that in to the first.
• Solve the first equation for $x_4$ and substitute into the third equation.
• Solve the third equation for $x_3$ and substitute into the fourth equation.
• Solve the fourth equation for $x_2$ and then solve the system.
• You should get the same answers I show in my augmented matrix.

I'll let you figure out the details and answer for the solution $W$ and is $W$ a vector space.

Answer 2

I am posting my solution - mainly for the purpose of showing you how can you write down the solution; in case you have a mistake in it and want to ask others whether your solution is correct.

$\begin{pmatrix} 2 & 4 & 0 & 8 &|&10\\ 3 & 2 & 0 & 0 &|& 3\\ 3 & 0 & 2 & 6 &|&11\\ 8 & 6 & 2 &14 &|& 24 \end{pmatrix}\sim \begin{pmatrix} 1 & 2 & 0 & 4 &|& 5\\ 3 & 2 & 0 & 0 &|& 3\\ 0 &-2 & 2 & 6 &|& 8\\ 4 & 3 & 1 & 7 &|&12 \end{pmatrix}\sim \begin{pmatrix} 1 & 2 & 0 & 4 &|& 5\\ 3 & 2 & 0 & 0 &|& 3\\ 0 &-1 & 1 & 3 &|& 4\\ 4 & 3 & 1 & 7 &|&12 \end{pmatrix}\sim \begin{pmatrix} 1 & 2 & 0 & 4 &|& 5\\ 3 & 2 & 0 & 0 &|& 3\\ 0 &-1 & 1 & 3 &|& 4\\ 1 & 1 & 1 & 7 &|& 9 \end{pmatrix}\sim \begin{pmatrix} 1 & 2 & 0 & 4 &|& 5\\ 3 & 2 & 0 & 0 &|& 3\\ 0 &-1 & 1 & 3 &|& 4\\ 0 &-1 & 1 & 3 &|& 4 \end{pmatrix}\sim \begin{pmatrix} 1 & 2 & 0 & 4 &|& 5\\ 3 & 2 & 0 & 0 &|& 3\\ 0 &-1 & 1 & 3 &|& 4\\ 0 & 0 & 0 & 0 &|& 0 \end{pmatrix}\sim \begin{pmatrix} 1 & 2 & 0 & 4 &|& 5\\ 2 & 0 & 0 &-4 &|&-2\\ 0 &-1 & 1 & 3 &|& 4\\ 0 & 0 & 0 & 0 &|& 0 \end{pmatrix}\sim \begin{pmatrix} 1 & 2 & 0 & 4 &|& 5\\ 1 & 0 & 0 &-2 &|&-1\\ 0 &-1 & 1 & 3 &|& 4\\ 0 & 0 & 0 & 0 &|& 0 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0 &-2 &|&-1\\ 0 & 2 & 0 & 6 &|& 6\\ 0 &-1 & 1 & 3 &|& 4\\ 0 & 0 & 0 & 0 &|& 0 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0 &-2 &|&-1\\ 0 & 1 & 0 & 3 &|& 3\\ 0 &-1 & 1 & 3 &|& 4\\ 0 & 0 & 0 & 0 &|& 0 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0 &-2 &|&-1\\ 0 & 1 & 0 & 3 &|& 3\\ 0 & 0 & 1 & 6 &|& 7\\ 0 & 0 & 0 & 0 &|& 0 \end{pmatrix}$

If I did not made a mistake, the rows of the first matrix should be linear combinations of the rows of the result. I can use this to make a sanity check. When I look on the first three columns (because the last matrix has pivots in the first three columns) I get the coefficients 2,4,0. So I can check whether I get $(2,4,0,8,10)=2(1,0,0,-2,-1)+4(0,1,0,3,3)+0(0,0,1,6,7)$; i.e., whether the first row is linear combination with this coefficients. (If not, I must have made mistake somewhere.) I can do the same thing with other rows.

Perhaps it is worth mentioning that even if this verification works, I cannot be entirely sure that everything is correct (that I have all solutions of the system).

This can be used to find where is a mistake, as illustrated below.

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I see that you asked repeatedly in comments how to express $W$.

Notice that the last matrix corresponds to the system $$ \begin{align} x_1-2x_4&=-1\\ x_2+3x_4&=3\\ x_3+6x_4&=7 \end{align} $$ Clearly, for any choice of $x_4$ I can find $x_1$, $x_2$, $x_3$ which fulfill those equations. Namely $x_1=-1+2x_4$, $x_2=3-3x_4$, $x_3=7-6x_4$. So the set of all solutions is $$W=\{(-1+2x_4, 3-3x_4, 7-6x_4, x_4); x_4\in F\}$$ where $F$ is the field you are working with. (For example, $F=\mathbb R$ if you are solving this system over real numbers.)

The same method works in general. You can choose the variables corresponding to columns without pivot and you can calculate the remaining values.

Now you can also see that $W$ is not a subspace of $F^4$ since $(0,0,0,0) \notin W$ and every subspace contains the zero vector.

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This is an incorrect solution, I've added it here solely for the purpose of illustrating how to find the place where I made a mistake

Suppose I made computations like this: $\begin{pmatrix} 2 & 4 & 0 & 8 &|&10\\ 3 & 2 & 0 & 0 &|& 3\\ 3 & 0 & 2 & 6 &|&11\\ 8 & 6 & 2 &14 &|& 24 \end{pmatrix}\sim \begin{pmatrix} 1 & 2 & 0 & 4 &|& 5\\ 3 & 2 & 0 & 0 &|& 3\\ 0 &-2 & 2 & 6 &|& 8\\ 4 & 3 & 1 & 7 &|&12 \end{pmatrix}\sim \begin{pmatrix} 1 & 2 & 0 & 4 &|& 5\\ 3 & 2 & 0 & 0 &|& 3\\ 0 &-1 & 1 & 3 &|& 4\\ 4 & 3 & 1 & 7 &|&12 \end{pmatrix}\sim \begin{pmatrix} 1 & 2 & 0 & 4 &|& 5\\ 3 & 2 & 0 & 0 &|& 3\\ 0 &-1 & 1 & 3 &|& 4\\ 1 & 1 & 1 & 7 &|& 9 \end{pmatrix}\sim \begin{pmatrix} 1 & 2 & 0 & 4 &|& 5\\ 3 & 2 & 0 & 0 &|& 3\\ 0 &-1 & 1 & 3 &|& 4\\ 0 &-1 & 1 & 3 &|& 4 \end{pmatrix}\sim \begin{pmatrix} 1 & 2 & 0 & 4 &|& 5\\ 3 & 2 & 0 & 0 &|& 3\\ 0 &-1 & 1 & 3 &|& 4\\ 0 & 0 & 0 & 0 &|& 0 \end{pmatrix}\sim \begin{pmatrix} 1 & 2 & 0 & 4 &|& 5\\ 2 & 0 & 0 &-4 &|&-2\\ 0 &-1 & 1 & 3 &|& 4\\ 0 & 0 & 0 & 0 &|& 0 \end{pmatrix}\sim \begin{pmatrix} 1 & 2 & 0 & 4 &|& 5\\ 1 & 0 & 0 &-2 &|&-1\\ 0 &-1 & 1 & 3 &|& 4\\ 0 & 0 & 0 & 0 &|& 0 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0 &-2 &|&-1\\ 0 & 2 & 0 & 2 &|& 6\\ 0 &-1 & 1 & 3 &|& 4\\ 0 & 0 & 0 & 0 &|& 0 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0 &-2 &|&-1\\ 0 & 1 & 0 & 1 &|& 3\\ 0 &-1 & 1 & 3 &|& 4\\ 0 & 0 & 0 & 0 &|& 0 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0 &-2 &|&-1\\ 0 & 1 & 0 & 1 &|& 3\\ 0 & 0 & 1 & 4 &|& 7\\ 0 & 0 & 0 & 0 &|& 0 \end{pmatrix}$

When I try to check my solution, I see that $2(1,0,0,-2,-1)+4(0,1,0,1,3)=(2,4,0,0,10)\ne (2,4,0,8,10)$. So I know that there is a mistake somewhere.

I can do the same verification for each matrix I obtained in the process.

Now when I look at these two matrices
$\begin{pmatrix} 1 & 2 & 0 & 4 &|& 5\\ 1 & 0 & 0 &-2 &|&-1\\ 0 &-1 & 1 & 3 &|& 4\\ 0 & 0 & 0 & 0 &|& 0 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0 &-2 &|&-1\\ 0 & 2 & 0 & 2 &|& 6\\ 0 &-1 & 1 & 3 &|& 4\\ 0 & 0 & 0 & 0 &|& 0 \end{pmatrix}$
I found out that the verification does not work for the first one, but it works for the second. This means that in this step there must be a mistake. And the second row in the later matrix is indeed incorrect.

In general, if I want to find where I made a mistake, I can look for two consecutive matrices such that for one of them the verification works and for the other it does not. (This works whenever I am trying to find a matrix in reduced row echelon form, it is not specific to solving linear systems.)

I should probably add the caveat that in this way we find the mistake which is closest to the end of the computation. So if I made several mistakes in my computations, this will be rather cumbersome. (I would have to do the same thing several times.)