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Simple showing inverse of matrix also upper triangular

  1. Nov 11, 2015 #1
    I'm trying to show that A be a 3 x 3 upper triangular matrix with non-zero determinant . Show by explicit computation that A^{-1}(inverse of A) is also upper triangular. Simple showing is enough for me.

    \begin{bmatrix}\color{blue}a & \color{blue}b & \color{blue}c \\0 & \color{blue}d & \color{blue}e \\ 0 & 0 &\color{blue}f\end{bmatrix}

    Can someone explain and show it?
     
  2. jcsd
  3. Nov 11, 2015 #2

    fresh_42

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    You can explicitly calculate the inverse as

    ##A^{-1} = \frac{1}{det A} \cdot (((-1)^{i+j} det A_{ij})_{i,j})^t = \frac{1}{det A} \cdot ((-1)^{i+j} det A_{ij})_{j,i}##

    where ##A_{ij}## denotes the matrix you get from ##A## when you cross off the i-th row and j-th column. ##B^t## denotes the transposed matrix of a matrix B.

    (compare: https://de.wikipedia.org/wiki/Inverse_Matrix Sorry, I haven't found an the English version of it.)
     
  4. Nov 23, 2015 #3

    HallsofIvy

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    The inverse of a matrix is its adjugate divided by its determinant. The "adjugate" of this matrix is
    [tex]\left|\begin{array}{ccc}df & -bf & be- cf \\ 0 & af & ae \\ 0 & 0 & ad \end{array}\right|[/tex]
     
  5. Nov 23, 2015 #4

    fresh_42

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    almost
     
  6. Nov 24, 2015 #5

    HallsofIvy

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    Right- I missed a negative sign:
    [tex]\left|\begin{array}{ccc}df & -bf & be- cf \\ 0 & af & -ae \\ 0 & 0 & ad \end{array}\right|[/tex].
     
  7. Nov 24, 2015 #6

    fresh_42

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    I meant (1,3). It has to be ##be - cd## instead as long as I'm not totally confused. But to be honest: beside laziness that's why I shirked from it. I'd have messed it up even more and would have had to perform the control calculation.
     
  8. Nov 24, 2015 #7

    HallsofIvy

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    The "upper right corner" of the adjugate is [tex]\left|\begin{array}{cc}b & c \\ d & e\end{array}\right |= be- cd [/tex], Yes, you are right.
     
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