# Thread: Two difficult linear algebra exercises (systems of equations)

1. ## Two difficult linear algebra exercises (systems of equations)

1.-Consider the system of equations

$x1 - x2 + 2x3 = 1$
$2x1 + 2x3 = 1$
$x1 - 3x2 + 4x3 = 2.$

Does this system have a solution? If so, describe explicitly all solutions.

2.-Give an example of a system of two linear equations in two unknowns which
has no solution.

Thanks so much for your help

2. To find out if it has a solution, set it up in matrix form (column form) and reduce to row echelon form. So:

$\left[ \begin{array}{cccc} 1 & 2 & 1 & 1 \\ -1 & 0 & -3 & 1 \\ 2 & 2 & 4 & 2 \end{array} \right]$

This is the matrix if column form with the augmented solutions. Now if you end up with some row where 0 = a number that isn't 0, there is no solution.

For the second problem:

$x_1 + x_2 = 1$
$2x_1 + 2x_2 = 3$

In this case you know there is no solution since if you were to cancel out $x_1$ by multiplying the top row by $-2$ and adding the two equations, you get:
$0 = 1$
Which we know is not true.

3. It is not necessary to use a matrix form (although that is more "sophisticated" and may well be easier). A more "basic" way to do the first problem is to multiply the first equation, $x_1- x2+ 2x_3= 1$, by 3 to get $3x_1- 3x_2+ 6x_3= 3$ and subtract the third equation, $x_1- 3x_2+ 4x_3= 2$, from it to eliminate $x_2$ and get $2x_1+ 2x_3= 1$. That is identical to the second equation and is equivalent to $x_1+ x_3= \frac{1}{2}$. Choose any value you want for, say, $x_1$, use that equation to find $x_3$ and then use either of the other two equations to find $x_2$.