Inverse of Matrix

1. Mar 4, 2016

Telemachus

• Member warned about posting homework with no effort shown
1. The problem statement, all variables and given/known data

I have to find the inverse for this generic matrix (the dimensions are not specified, but I assume its a square matrix, I don't know if that is necessary).

$A=\left [ \begin{matrix} 1 & -1 & -1 & -1 & \dots & -1 & -1 \\ 0 & 1 & -1 & -1 & \dots & -1 & -1 \\ 0 & 0 & 1 & -1 & \dots & -1 & -1 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \dots & 0 & 1 \\ \end{matrix} \right]$

I think there must be a clever and fast way for calculating $A^{-1}$, but I don't know how to do it.

2. Mar 4, 2016

SteamKing

Staff Emeritus
It is necessary that matrix A be square for an inverse to exist.

It's not clear why you need to calculate the inverse. Are you trying to solve a system of linear equations?

In any event, the matrix A is a special kind of triangular matrix called an atomic triangular matrix, and there is a simple and easy method of calculating its inverse.

https://en.wikipedia.org/wiki/Triangular_matrix

3. Mar 4, 2016

Telemachus

No, it's just an exercise. It asks explicitly to give the inverse Matrix.

Thank you verymuch :)

One detail, it's not an atomic matrix, is just upper triangular.

Last edited: Mar 4, 2016
4. Mar 4, 2016

SteamKing

Staff Emeritus
I see that now. Sorry.

5. Mar 4, 2016

Telemachus

No problem. I did it my way, I think its okey. I've assumed an $n \times n$ matrix, and found:

$A^{-1}=\left [ \begin{matrix} 1 & 1 & 2 & 3 & \dots & n-2 & n-1 \\ 0 & 1 & 1 & 2 & \dots & n-3 & n-2 \\ 0 & 0 & 1 & 1 & \dots & n-4 & n-3 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \dots & 0 & 1 \\ \end{matrix} \right]$

This is not right, I get trouble after the third column.

Last edited: Mar 4, 2016
6. Mar 4, 2016

Staff: Mentor

Instead of jumping right in to n x n matrices, start with smaller matrices. I looked at the case with a 3 x 3 matrix, and then with a 4 x 4 matrix. There's a definite pattern that develops.

7. Mar 4, 2016

Telemachus

Yes, I've tried that way, but wasn't that clear to me. The case for the $3\times3$ is contained in the first 3 rows and columns.

$A_{3\times3}^{-1} =\left [ \begin{matrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{matrix} \right]$

Then for the $4\times 4$:
$A_{4\times4}^{-1}= =\left [ \begin{matrix} 1 & 1 & 2 & 4\\ 0 & 1 & 1 & 2 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right]$

And for $5\times5$

$A_{5\times5}^{-1}=\left [ \begin{matrix} 1 & 1 & 2 & 4 & 8\\ 0 & 1 & 1 & 2 & 4 \\ 0 & 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 \\ \end{matrix} \right]$

Then:

$A_{n\times n}^{-1}=\left [ \begin{matrix} 1 & 1 & 2 & 3 & \dots & 2^{n-3} & 2^{n-2} \\ 0 & 1 & 1 & 2 & \dots & 2^{n-4} & 2^{n-3} \\ 0 & 0 & 1 & 1 & \dots & 2^{n-5} & 2^{n-4} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \dots & 0 & 1 \\ \end{matrix} \right]$

Last edited: Mar 4, 2016
8. Mar 4, 2016

Ray Vickson

$$\left[ \begin{matrix} 1 & -1 & -1 & -1 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 \end{matrix} \right]$$
not what was intended? When did the switch occur?

9. Mar 4, 2016

Staff: Mentor

The original matrix was n X n, not 4 X 4.

10. Mar 4, 2016

Staff: Mentor

Your last matrix isn't following the pattern of the previous ones. Check the first row in the matrix just above.

11. Mar 4, 2016

Ray Vickson

Yes, I know. But the OP did some 3x3 and 4x4 examples, and I was just looking at the 4x4 case.

12. Mar 4, 2016

Telemachus

Its done. I did as Mark said, by trying some tractable dimensions, and then generalized it. Thanks.

13. Mar 4, 2016

Telemachus

That 3 was just a typo. Sorry.