Dtyjlui

I am just creating random sets of vectors to try and practice solving systems of equations. I thought I had been doing well, but I came up with a set of vectors that keep stumping me. In fact, I've been trying to figure out where I went wrong for almost two hours, yet I still have no clue.

I tried the following:

$$x_1begin{bmatrix}
-1 \
-2 \
-3
end{bmatrix}+x_2begin{bmatrix}
1 \
3 \
5
end{bmatrix}+x_3begin{bmatrix}
4 \
2 \
0
end{bmatrix}
=
begin{bmatrix}
a \
b \
c
end{bmatrix}$$

Every time I try to solve this I end up stuck at a similar place.

Here is one example of what I have tried:

$$begin{eqnarray}
-x_1&+x_2&+4x_3&=a\
-2x_1&+3x_2&+2x_3&=b\
-3x_1&+5x_2&&=c
end{eqnarray}$$

Then for example, I try to eliminate the $x_3$ constant from equations 1 and 2.

$$begin{eqnarray}
-x_1&+x_2&+4x_3&=a\
4x_1&-6x_2&-4x_3&=-2b\
3x_1&-5x_2&&=a-2b
end{eqnarray}$$

Then, I try to eliminate the $x_2$ constant from the new equation and the original equation 3.

$$begin{eqnarray}
3x_1&-5x_2&=a-2b\
-3x_1&+5x_2&=c\
end{eqnarray}$$

but I end up with:

$$0+0=a-2b+c$$

I have tried eliminating different factors, but I cannot figure out what I'm doing wrong.

Guide:

Not every system has a unique solution. For your system to have a solution, we need $a-2b+c=0$.

If we have $a-2b+c=0$, now we can let $x_1=t$, from $-3x_1+5x_2=c$, we can solve for $x_2$. Now, having $x_1$ and $x_2$, we can use the equation to solve for $x_3$. It always have infinitely many solutions if $a-2b+c=0$.

The vectors
$$
begin{bmatrix}
-1 \
-2 \
-3
end{bmatrix},
begin{bmatrix}
1 \
3 \
5
end{bmatrix},
text{ and } begin{bmatrix}
4 \
2 \
0
end{bmatrix}
$$
are linearly dependent. Thus, your linear system either has no solutions or infinitely many solutions. So you won't be able to determine the numbers $x_1, x_2$, and $x_3$ uniquely.

And it won't be surprising if you discover that there is some condition such as $a - 2b + c = 0$ which must be satisfied in order for a solution to exist. For a solution to exist, the vector $(a,b,c)$ must belong to the span of your three vectors, and the span of the three given vectors is only a 2-dimensional subspace of $mathbb R^3$.

When you are trying to solve a linear system with the determinant of coefficient equal zero then you either have no solution or you have infinitely many solutions.

In your case the determinant of $$ det begin {bmatrix} -1&1&4\-2&3&2\-3&5&0end {bmatrix}=0$$

Thus depending on values of $a,b,c$ you either get infinitely many solutions or no solutions.

When trying to solve a system of equations, one of the the following possibilities can arise:

1- The system is "inconsistent". This means that it has not solution.
2- There is a unique solution. (Only 1)
3- There are infinitely many solutions.

As pointed out by Siong, you need to have a-2b+c = 0 in order to have any kind of solution.

The reason is that if a-2b+c is not equal to zero, then the last row of your matrix basically says that 0*x_1 + 0*x_2 + 0*x_3 equals something that is non-zero. This is not possible.

So you have to assume that a-2b+c is zero.

This then means that the system will have a solution. We need to dig deeper to determine if it has a unique solution, or infinitely many solutions.

It turns out in this situation a "free variable" can be set up, leading to infinitely many solutions.

To fully understand solving these equations, you would want to get familiar with the concepts of:

• Reduced Row Echelon Form


• Pivot Columns


• Basic Variables


• Free Variables

As a bit of self-promotion, I've written an app (for Android) that explains how to solve common Linear Algebra problems in easy-to-understand language. If you're interested, please look up "Linear Algebra Patterns" on Google Play Store.

Cheers,
Richard