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Math Help - Invertible matrix

  1. #1
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    Invertible matrix

    I have to determine all values of h for which A is invertible and I really don't know what should be my first step( If anyone could guide me through this that would be awesome. Here's the matrix:

    1 1 0 1
    1 0 0 1
    0 1 2
    h + 1
    0 1 1
    h

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  2. #2
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    1 1 0 1
    1 0 0 1
    0 1 2 h + 1
    0 1 1 h

    Sorry if this is confusing the fourth column is 1,1,h+1, h
    the first column is 1,1,0,0
    second- 1,0,1,1
    third - 0,0,2,1
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  3. #3
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    You mean then that the matrix is
    \begin{bmatrix}1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 2 & h+ 1 \\ 0 & 1 & 1 & h\end{bmatrix}

    I see two ways to do that. One is to use the fact that a matrix is invertible if and only if its determinant is non-zero. The other is to row-reduce this to triangular form and use the fact that a matrix is invertible if and only if, reduced to triangular form, it has no zeros on its main diagonal. Since a simple way of determining the determinant of a matrix is to reduce to triangular form, those are essentially the same.
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  4. #4
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    thank you ! now i have to figure out how to reduce it.
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  5. #5
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    Quote Originally Posted by anna123456 View Post
    thank you ! now i have to figure out how to reduce it.
    Quiet easy,
    R_2->(-1)R_1+R_2
    R_3->(-1)R_1+R_3
    then
    R_3->(-1)R_2+R_3
    R_4->(-1)R_2+R_4 , R_2,_3,_4... are row operations
    And so on...until you get to reduced row echelon form.
    Hope it helps
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  6. #6
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    Thank you ! Ok here's what I got:
    <br />
\begin{bmatrix}2/h & -3/h+1 & 1/h & -2/h \\ 1 & -1 & 0 & 0 \\ 2/h & -3/h & 1/h & -2/h \\ 1 & -2 & 1 & -1\end{bmatrix}<br />
    That's the inverse(A^-1)
    Now I'm not really sure what to do with this. How do I figure out what values of h make this matrix invertible?
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  7. #7
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    Quote Originally Posted by anna123456 View Post
    Thank you ! Ok here's what I got:
    <br />
\begin{bmatrix}2/h & -3/h+1 & 1/h & -2/h \\ 1 & -1 & 0 & 0 \\ 2/h & -3/h & 1/h & -2/h \\ 1 & -2 & 1 & -1\end{bmatrix}<br />
    That's the inverse(A^-1)
    Now I'm not really sure what to do with this. How do I figure out what values of h make this matrix invertible?
    I think all you had to do is get the
    \begin{bmatrix}1 & 1 & 0 & 1 \\ 1 & 0 & 0  & 1 \\ 1 & 0 & 2 & h+ 1 \\ 0 & 1 & 1 &  h\end{bmatrix}
    Matrix to rref(reduced row echelon form) and find it's det.
    detA(in rref)= a_1,_1*a_2,_2*a_3,_3*a_4,_4
    and find for which h detA doesn't equal to 0.
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  8. #8
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    We haven't learnt this formula yet so I can't use it(
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  9. #9
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    Quote Originally Posted by HallsofIvy View Post
    You mean then that the matrix is
    \begin{bmatrix}1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 2 & h+ 1 \\ 0 & 1 & 1 & h\end{bmatrix}

    I see two ways to do that. One is to use the fact that a matrix is invertible if and only if its determinant is non-zero. The other is to row-reduce this to triangular form and use the fact that a matrix is invertible if and only if, reduced to triangular form, it has no zeros on its main diagonal. Since a simple way of determining the determinant of a matrix is to reduce to triangular form, those are essentially the same.
    Is there any way to find the determinant without using any formulas? when I reduce my matrix to echelon form it looks like
    <br />
\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & h \\ 0 & 0 & 1 & 0\end{bmatrix}<br />
    so is my answer all values of h except for 0 make this matrix invertible?
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  10. #10
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    I'm really stuck ( Because it seems to easy to just say that h=/= 0.
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  11. #11
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    Ok I think this looks a bit less complicated but the only answer i can come up with is h=/= 0...
    <br />
\begin{bmatrix}1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 2 & h+ 1 \\ 0 & 1 & 1 & h\end{bmatrix}<br />
    ---> Switch /R1 and R2

    <br />
\begin{bmatrix}1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 0 & 2 & h+ 1 \\ 0 & 1 & 1 & h\end{bmatrix}<br />
    --->R2-R3
    <br />
\begin{bmatrix}1 & 0 & 0 & 1 \\ 1 & 1 & -2 & -h \\ 1 & 0 & 2 & h+ 1 \\ 0 & 1 & 1 & h\end{bmatrix}<br />
    ---> R3-R1
    <br />
\begin{bmatrix}1 & 0 & 0 & 1 \\ 1 & 1 & -2 & -h \\ 0 & 0 & 2 & h \\ 0 & 1 & 1 & h\end{bmatrix}<br />
    ----> R2+R3
    <br />
\begin{bmatrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & h \\ 0 & 1 & 1 & h\end{bmatrix}<br />
    --->R4-R2
    <br />
\begin{bmatrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & h \\ 0 & 0 & 1 & h\end{bmatrix}<br />
    --->R3- R4
    <br />
\begin{bmatrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & h\end{bmatrix}<br />
    ---> R4-R3
    <br />
\begin{bmatrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & h\end{bmatrix}<br />
    --->R1- R4/h
    <br />
\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & h\end{bmatrix}<br />
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  12. #12
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    Quote Originally Posted by anna123456 View Post
    Is there any way to find the determinant without using any formulas? when I reduce my matrix to echelon form it looks like
    <br />
\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & h \\ 0 & 0 & 1 & 0\end{bmatrix}<br />
    so is my answer all values of h except for 0 make this matrix invertible?
    One more "row operation" would be to swap the third and fourth rows.

    That will give you <br />
\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & h\end{bmatrix}<br />
    Now you also need to note that
    1) If you "add a multiple of one row to another" the determinant of a matrix
    remains unchanged.

    2) if you "multiply one row by a number", the determinant of a matrix is multiplied by that number.

    3) if you "swap two rows", the determinant of a matrix is multiplied by -1.

    Since you have not "multiplied one row by a number", the determinant of your original matrix must be \pm the determinant of this matrix: that is, \pm h.

    The determinant of your original matrix is non-zero if and only if h is non-zero.
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