Simple showing inverse of matrix also upper triangular

[snrwww]

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?

 

[fresh_42]

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.)

 

[HallsofIvy]

The inverse of a matrix is its adjugate divided by its determinant. The "adjugate" of this matrix is
$$\left|\begin{array}{ccc}df & -bf & be- cf \\ 0 & af & ae \\ 0 & 0 & ad \end{array}\right|$$

 

[fresh_42]

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

 

[HallsofIvy]

Right- I missed a negative sign:
$$\left|\begin{array}{ccc}df & -bf & be- cf \\ 0 & af & -ae \\ 0 & 0 & ad \end{array}\right|$$.

 

[fresh_42]

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.

 

[HallsofIvy]

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.