MHB Inverse of Giant $7 \times 7$ Matrix: Tips & Tricks

[A.Magnus]

How do I go about find out the inverse of this giant $7 \times 7$ matrix? Do I need to evaluate it using the tedious Gauss-Jordan method? Any property, short-cut, trick that I can take advantage of?

$$\begin{pmatrix}
0 &1 &0 &0 &0 &0 &0\\
0 &0 &0 &1 &0 &0 &0\\
0 &0 &0 &0 &0 &0 &1\\
0 &0 &0 &0 &0 &1 &0\\
0 &0 &1 &0 &0 &0 &0\\
0 &0 &0 &0 &1 &0 &0\\
1 &0 &0 &0 &0 &0 &0\\
\end{pmatrix}$$

Your time and gracious help is very much appreciated? Thank you. ~MA

 

[Greg]

Lets call the matrix you gave $A$. What is the transpose of $A$? That's your answer but I'm not immediately sure how to prove that.

 

[Joppy]

Are you supposed to be finding a shortcut or fast method? It would seem unlikely that finding the inverse of a matrix of this size would be a homework task to be done by hand?

 

[I like Serena]

Hey MaryAnn! ;)

Are you familiar, or do your notes say something about an Orthogonal matrix?

 

[A.Magnus]

Hey MaryAnn! ;)

Are you familiar, or do your notes say something about an Orthogonal matrix?

No, we don't have it. I checked textbook's index but could not find any orthogonal matrix entry. But anyway, I found out the least tedious solution: Assuming the matrix is $X$ and its inverse is $Y$, then $XY = I$. Then using the definition of multiplication of matrix, I expanded the seven entries of the diagonal of $I$ (since they are all non-zero) to come up with the $y_{ij}$ that are non-zero. The rest of $y_{ij}$ are zero. Thank you though for pitching in, thanks for your gracious help and time. ~MA

- - - Updated - - -

Lets call the matrix you gave $A$. What is the transpose of $A$? That's your answer but I'm not immediately sure how to prove that.

I did that, but its transpose did not take me anywhere. Thank you though for your gracious help, and time. ~MA

- - - Updated - - -

Are you supposed to be finding a shortcut or fast method? It would seem unlikely that finding the inverse of a matrix of this size would be a homework task to be done by hand?

I think I found the least tedious solution, see my response to I Like Serena. Thank you though for your gracious help, and for your time. ~MA

 

[I like Serena]

No, we don't have it. I checked textbook's index but could not find any orthogonal matrix entry. But anyway, I found out the least tedious solution: Assuming the matrix is $X$ and its inverse is $Y$, then $XY = I$. Then using the definition of multiplication of matrix, I expanded the seven entries of the diagonal of $I$ (since they are all non-zero) to come up with the $y_{ij}$ that are non-zero. The rest of $y_{ij}$ are zero. Thank you though for pitching in, thanks for your gracious help and time. ~MA

That works as well.
For the record, your matrix is orthogonal because each column vector has unit length, and because each pair of column vectors is orthogonal.
As a consequence the inverse is simply the transpose as greg1313 mentioned. (Mmm)

 

[HallsofIvy]

It doesn't look to me like "Gauss-Jordan" will be all that tedious. You should be able to do some obvious "swaps" of two rows to get the matrix to the identity matrix,

 

[A.Magnus]

That works as well.
For the record, your matrix is orthogonal because each column vector has unit length, and because each pair of column vectors is orthogonal.
As a consequence the inverse is simply the transpose as greg1313 mentioned. (Mmm)

Thank you again for your gracious help. ~MA

- - - Updated - - -

It doesn't look to me like "Gauss-Jordan" will be all that tedious. You should be able to do some obvious "swaps" of two rows to get the matrix to the identity matrix,

Thank you for your gracious comment. ~MA