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

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

    1.-Consider the system of equations


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

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

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

    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:

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

    In this case you know there is no solution since if you were to cancel out $\displaystyle x_1$ by multiplying the top row by $\displaystyle -2$ and adding the two equations, you get:
    $\displaystyle 0 = 1$
    Which we know is not true.
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  3. #3
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    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, $\displaystyle x_1- x2+ 2x_3= 1$, by 3 to get $\displaystyle 3x_1- 3x_2+ 6x_3= 3$ and subtract the third equation, $\displaystyle x_1- 3x_2+ 4x_3= 2$, from it to eliminate $\displaystyle x_2$ and get $\displaystyle 2x_1+ 2x_3= 1$. That is identical to the second equation and is equivalent to $\displaystyle x_1+ x_3= \frac{1}{2}$. Choose any value you want for, say, $\displaystyle x_1$, use that equation to find $\displaystyle x_3$ and then use either of the other two equations to find $\displaystyle x_2$.
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