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Math Help - Linear Algebra: Matrices and Systems Equations

  1. #1
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    Linear Algebra: Matrices and Systems Equations

    Hello. Could anyone please help me step by step in solving the following homework problems. I tried to work on it twice (especially the first problem) but I just can't seem to find my way around it.

    1)For the following system of equations, use Gaussian elimination to obtain an equivalent system whose coefficient matrix is in row echelon form. Indicate whether the system is consistent or not. If system is consistent and involves variables, transform it to reduced row echelon form and find all solutions for variables.

    1x _1 + 2x_2 - 3x_3 + 1x_4 = 1
    -1x_1 - 1x_2 + 4x_3 - 1x_4 = 6
    -2x_1 - 4x_2 + 7x_3 - 1x_4 = 1

    Solution: After adding and multiplying rows this is the matrix i Got so far,

    1 0 -1 1 -13
    0 1 -1 0 7
    0 0 1 1 3

    I do not know where to go from here
    the final answer in the back of the book is (2-6a, 4+a, 3-a, a)

    2) consider a linear system whose augmented matrix is of the form

    1 2 1 0
    2 5 3 0
    -1 1 β 0

    (a) Is it possible for the system to be inconsistent? Why or Why not?
    (b) For what values of β will the system have infinitely many solutions?

    for (a) I said the system will be inconsistent if β is not equal to 2.
    for (b) I said β will have to equal to 2 in order for the system to have infinitely many solutions

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  2. #2
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    Hello, googoogaga!

    1) For the following system of equations, use Gaussian elimination
    to obtain an equivalent system whose coefficient matrix is in row echelon form.
    Indicate whether the system is consistent or not.
    If system is consistent and involves variables, transform it to reduced row echelon form
    and find all solutions for variables.

    . . \begin{array}{ccc}w + 2x - 3y + z & = & 1 \\<br />
\text{-}w  - x + 4y - z  & = &  6 \\<br />
\text{-}2w  - 4x + 7y - z & = & 1 \end{array}

    Solution: After adding and multiplying rows,

    . . this is the matrix i got so far: . \begin{pmatrix}<br />
1 & 0 & \text{-}1 & 1 & | & \text{-}13 \\<br />
0 & 1 & \text{-}1 &  0 & | & 7 \\<br />
0 & 0 & 1 & 1 & | & 3\end{pmatrix} .
    there may be an error

    I do not know where to go from here . . . .
    keep going

    The final answer in the back of the book is:
    . . . . (2-6\alpha,\: 4 + \alpha,\:3-\alpha,\:\alpha)

    We have: . \begin{pmatrix}1 & 2 & \text{-}3 & 1 & | & 1 \\<br />
\text{-}1 & \text{-}1 & 4 & \text{-}1 & | & 6 \\<br />
\text{-}2 & \text{-}4 & 7 & \text{-}1 & | & 1\end{pmatrix}

    \begin{array}{c} \\ R_2+R_1 \\ R_3+2\!\cdot\!R_1\end{array}\;<br />
\begin{pmatrix}1 & 2 & \text{-}3 & 1 & | & 1 \\<br />
0 & 1 & 1 & 0 & | & 7 \\<br />
0 & 0 & 1 & 1 & | & 3\end{pmatrix}

    \begin{array}{c}R_1-2\!\cdot\!R_2 \\ R_2-R_3 \\ \\ \end{array}\;<br />
\begin{pmatrix}1 & 0 & \text{-}5 & 1 & | & \text{-}13 \\<br />
0 & 1 & 0 & \text{-}1 & | & 4 \\<br />
0 & 0 & 1 & 1 & | & 3\end{pmatrix}

    \begin{array}{c}R_1+5\!\cdot\!R_3 \\ \\ \end{array}\;<br />
\begin{pmatrix}1 & 0 & 0 & 6 & | & 2 \\<br />
0 & 1 & 0 & \text{-}1 & | & 4 \\<br />
0 & 0 & 1 & 1 & | & 3 \end{pmatrix}


    We have: . \begin{array}{ccccccc}w + 6z & = & 2 & \;\Rightarrow\; & w & = & 2 - 6z \\<br />
x - z & = & 4 & \Rightarrow & x & = & 4-z \\<br />
y + z & = & 3 & \Rightarrow & y & = & 3-z \\<br />
 &  &  &  & z & = & z \end{array}

    On the right side, replace z with a parameter \alpha.

    Therefore: . \begin{Bmatrix}w & = & 2 - 6\alpha \\ x & = & 4 - \alpha \\ y & = & 3 - \alpha \\ z & = & \alpha\end{Bmatrix}



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