Find Inverse of Matrix Homework Statement

[Telemachus]

Homework Statement

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.

Thanks in advance.

 

[SteamKing]

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.
See this article for the details:

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

 

[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.

 

[SteamKing]

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.

I see that now. Sorry.

 

[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.

 

[Mark44]

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.

Instead of jumping right into 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.

 

[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:

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

 

[Ray Vickson]

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:

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

So, was your original matrix
$$\left[ \begin{matrix} 1 & -1 & -1 & -1 \\<br /> 0 & 1 & -1 & -1 \\<br /> 0 & 0 & 1 & -1 \\<br /> 0 & 0 & 0 & 1<br /> \end{matrix} \right]$$
not what was intended? When did the switch occur?

 

[Mark44]

So, was your original matrix
$$\left[ \begin{matrix} 1 & -1 & -1 & -1 \\<br /> 0 & 1 & -1 & -1 \\<br /> 0 & 0 & 1 & -1 \\<br /> 0 & 0 & 0 & 1<br /> \end{matrix} \right]$$
not what was intended? When did the switch occur?

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

 

[Mark44]

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:

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

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

 

[Ray Vickson]

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

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

 

[Telemachus]

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

 

[Telemachus]

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

That 3 was just a typo. Sorry.