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Math Help - Matrix as product of elementary matrices

  1. #1
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    Matrix as product of elementary matrices

    Hey guys I need to express this matrix as the product as elementary matrices

    <br />
\begin{pmatrix}<br />
1 & 0  & -2 \\ <br />
0 & 4 & 3 \\<br />
0 & 0 & 1<br />
\end{pmatrix}<br />

    I get this answer but the book has a different answer can someone tell me where I went wrong.

    <br />
\begin{pmatrix}<br />
1 & 0  & 0 \\ <br />
0 & \frac1{4} & 0 \\<br />
0 & 0 & 1<br />
\end{pmatrix}<br />
\begin{pmatrix}<br />
1 & 0  & 0 \\ <br />
0 & 1 & \frac3{4} \\<br />
0 & 0 & 1<br />
\end{pmatrix}<br />
\begin{pmatrix}<br />
1 & 0  & 2 \\ <br />
0 & 1 & 0 \\<br />
0 & 0 & 1<br />
\end{pmatrix}<br />

    Thanks
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  2. #2
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    Quote Originally Posted by evant8950 View Post
    Hey guys I need to express this matrix as the product as elementary matrices

    <br />
\begin{pmatrix}<br />
1 & 0  & -2 \\ <br />
0 & 4 & 3 \\<br />
0 & 0 & 1<br />
\end{pmatrix}<br />

    I get this answer but the book has a different answer can someone tell me where I went wrong.

    <br />
\begin{pmatrix}<br />
1 & 0  & 0 \\ <br />
0 & \frac1{4} & 0 \\<br />
0 & 0 & 1<br />
\end{pmatrix}<br />
\begin{pmatrix}<br />
1 & 0  & 0 \\ <br />
0 & 1 & \frac3{4} \\<br />
0 & 0 & 1<br />
\end{pmatrix}<br />
\begin{pmatrix}<br />
1 & 0  & 2 \\ <br />
0 & 1 & 0 \\<br />
0 & 0 & 1<br />
\end{pmatrix}<br />

    Thanks
    If you multiple your answer out, you will see it doesn't yield the original matrix.
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  3. #3
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    The book has

    \displaystyle\begin{bmatrix}1&0&0\\0&4&0\\0&0&1\en  d{bmatrix}\cdot\begin{bmatrix}1&0&0\\0&1&\frac{3}{  4}\\0&0&1\end{bmatrix}\cdot\begin{bmatrix}1&0&-2\\0&1&0\\0&0&1\end{bmatrix}

    Correct?
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  4. #4
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    No the book has:
    <br />
\displaystyle\begin{bmatrix}1&0&0\\0&4&0\\0&0&1\en  d{bmatrix}\cdot\begin{bmatrix}1&0&0\\0&1&3\\0&0&1\  end{bmatrix}\cdot\begin{bmatrix}1&0&-2\\0&1&0\\0&0&1\end{bmatrix}<br />
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  5. #5
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    Quote Originally Posted by evant8950 View Post
    No the book has:
    <br />
\displaystyle\begin{bmatrix}1&0&0\\0&4&0\\0&0&1\en  d{bmatrix}\cdot\begin{bmatrix}1&0&0\\0&1&3\\0&0&1\  end{bmatrix}\cdot\begin{bmatrix}1&0&-2\\0&1&0\\0&0&1\end{bmatrix}<br />
    Multiplying that out doesn't yield the original matrix though.
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  6. #6
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    I'm sorry the book has the solutions in this order:

    <br />
\displaystyle\begin{bmatrix}1&0&-2\\0&1&0\\0&0&1\end{bmatrix}\cdot\begin{bmatrix}1&  0&0\\0&1&3\\0&0&1\end{bmatrix}\cdot\begin{bmatrix}  1&0&0\\0&4&0\\0&0&1\end{bmatrix}

    I multiplied this out and it gave the right solution.
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  7. #7
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    Quote Originally Posted by evant8950 View Post
    I'm sorry the book has the solutions in this order:

    <br />
\displaystyle\begin{bmatrix}1&0&-2\\0&1&0\\0&0&1\end{bmatrix}\cdot\begin{bmatrix}1&  0&0\\0&1&3\\0&0&1\end{bmatrix}\cdot\begin{bmatrix}  1&0&0\\0&4&0\\0&0&1\end{bmatrix}

    I multiplied this out and it gave the right solution.
    When you doing the elementary row operations, you would think 2 times row 3 plus row 1 would work, but think of it as the value minus your operation. Therefore, you have -2-(-2) to obtain 0.

    From the above explantion, you will see we have 3-3 which is 0 so good.

    And 4*1 gives us the 4.

    Do you understand now?
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  8. #8
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    Quote Originally Posted by evant8950 View Post
    Hey guys I need to express this matrix as the product as elementary matrices

    <br />
\begin{pmatrix}<br />
1 & 0  & -2 \\ <br />
0 & 4 & 3 \\<br />
0 & 0 & 1<br />
\end{pmatrix}<br />

    I get this answer but the book has a different answer can someone tell me where I went wrong.

    <br />
\begin{pmatrix}<br />
1 & 0  & 0 \\ <br />
0 & \frac1{4} & 0 \\<br />
0 & 0 & 1<br />
\end{pmatrix}<br />
\begin{pmatrix}<br />
1 & 0  & 0 \\ <br />
0 & 1 & \frac3{4} \\<br />
0 & 0 & 1<br />
\end{pmatrix}<br />
\begin{pmatrix}<br />
1 & 0  & 2 \\ <br />
0 & 1 & 0 \\<br />
0 & 0 & 1<br />
\end{pmatrix}<br />

    Thanks
    The "KEY" idea is that multipying by an elemtry matrix is just like doing a row operation.

    So you start with the identity

    \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0& 1\end{bmatrix}

    The first row op would be to multiply row 2 by 4 this gives

    E_1=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0& 1\end{bmatrix}

    Now you want to take 3R_3+R_2=R_2 This gives

    E_2=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 3 \\ 0 & 0& 1\end{bmatrix}

    and finally for the last one

    E_3=\begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 0 \\ 0 & 0& 1\end{bmatrix}
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  9. #9
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    How do i get 1 where 4 is if I don't multiple by 1/4?

    Thanks
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  10. #10
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    Quote Originally Posted by evant8950 View Post
    How do i get 1 where 4 is if I don't multiple by 1/4?

    Thanks
    You need to get a 4 there. What times 1 equals 4?
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  11. #11
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    So I am doing the inverse row operation to the identity matrix of what I would do as a row operation on the matrix itself?
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  12. #12
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    Quote Originally Posted by evant8950 View Post
    So I am doing the inverse row operation to the identity matrix of what I would do as a row operation on the matrix itself?
    You are doing a LU Factorization.
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  13. #13
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    I think I am confused. Am I using elementary matrices to get A in reduced row echelon form? Then multiplying the elementary matrices to get A?

    Thanks for the help
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  14. #14
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    Quote Originally Posted by evant8950 View Post
    I think I am confused. Am I using elementary matrices to get A in reduced row echelon form? Then multiplying the elementary matrices to get A?

    Thanks for the help
    If A is nonsingular, then A is row equivalent to I.

    A=E_kE_{k-1}\cdots E_1I
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