Simple showing inverse of matrix also upper triangular

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    Inverse Matrix
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Discussion Overview

The discussion revolves around demonstrating that the inverse of a 3x3 upper triangular matrix with a non-zero determinant is also upper triangular. Participants are engaged in explicit computation and exploration of the properties of matrix inverses, particularly focusing on the adjugate and determinant.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant requests a simple computation to show that the inverse of an upper triangular matrix is also upper triangular.
  • Another participant provides a formula for the inverse of a matrix involving the determinant and the adjugate, referencing the transposed matrix of cofactors.
  • Several participants discuss the adjugate of the matrix, with one presenting a specific form of the adjugate matrix and others correcting and refining the expression for its elements.
  • There is a correction regarding a negative sign in the adjugate matrix, with participants acknowledging and discussing the need for accuracy in the calculations.
  • One participant expresses uncertainty about the calculations and admits to avoiding the problem due to potential mistakes.
  • Another participant confirms the calculation of the upper right corner of the adjugate matrix, agreeing with the previous statements about its elements.

Areas of Agreement / Disagreement

Participants are engaged in a collaborative exploration of the problem, with some corrections and refinements made along the way. However, there is no consensus on the final form of the adjugate or the complete computation of the inverse.

Contextual Notes

There are unresolved details regarding the calculation of the adjugate and its elements, as well as the overall computation of the inverse. Some assumptions about the properties of determinants and adjugates are implicit in the discussion.

snrwww
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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?
 
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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.)
 
The inverse of a matrix is its adjugate divided by its determinant. The "adjugate" of this matrix is
\left|\begin{array}{ccc}df & -bf & be- cf \\ 0 & af & ae \\ 0 & 0 & ad \end{array}\right|
 
HallsofIvy said:
The inverse of a matrix is its adjugate divided by its determinant. The "adjugate" of this matrix is
\left|\begin{array}{ccc}df & -bf & be- cf \\ 0 & af & ae \\ 0 & 0 & ad \end{array}\right|
almost
 
Right- I missed a negative sign:
\left|\begin{array}{ccc}df & -bf & be- cf \\ 0 & af & -ae \\ 0 & 0 & ad \end{array}\right|.
 
HallsofIvy said:
Right- I missed a negative sign:
\left|\begin{array}{ccc}df & -bf & be- cf \\ 0 & af & -ae \\ 0 & 0 & ad \end{array}\right|.
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.
 
The "upper right corner" of the adjugate is \left|\begin{array}{cc}b & c \\ d & e\end{array}\right |= be- cd, Yes, you are right.
 

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