Matrix Theory (Matrix Inverse test question)

mateomy
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I was asked this question on a test I just took and I was kinda stumped. I worked it out but I think I did it incorrectly so I was hoping for some input. Thanks.


'A' is a 3x3 matrix and 'I' is the identity matrix for a 3x3. Find the Inverse of 'A' given:

<br /> (\mathbf{A}\,+\,\mathbf{I})^{2}\,=\,\mathbf{0}<br />

Where '0' is the zero matrix.

So what I did (probably ignorantly) was treat it like any standard algebraic equation; took the radical of both sides, and subtracted the Identity from each side. So I had the zero matrix minus the standard identity 3x3. So I ended up getting a negative identity matrix on the right hand side. So I can see that 'A' equals the negative of an identity matrix. So to find the identity of that I put my 'A' next to and identity to solve in the standard fashion, for the identity of A. Which just happened to be, again, a negative of a 3x3 identity matrix.

Sorry I didn't feel like Latex'ing out all the matrix work but I think it can be followed by what I wrote.

Thanks again.
 
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mateomy said:
I was asked this question on a test I just took and I was kinda stumped. I worked it out but I think I did it incorrectly so I was hoping for some input. Thanks.


'A' is a 3x3 matrix and 'I' is the identity matrix for a 3x3. Find the Inverse of 'A' given:

<br /> (\mathbf{A}\,+\,\mathbf{I})^{2}\,=\,\mathbf{0}<br />

Where '0' is the zero matrix.

So what I did (probably ignorantly) was treat it like any standard algebraic equation; took the radical of both sides, and subtracted the Identity from each side. So I had the zero matrix minus the standard identity 3x3. So I ended up getting a negative identity matrix on the right hand side. So I can see that 'A' equals the negative of an identity matrix. So to find the identity of that I put my 'A' next to and identity to solve in the standard fashion, for the identity of A. Which just happened to be, again, a negative of a 3x3 identity matrix.

Sorry I didn't feel like Latex'ing out all the matrix work but I think it can be followed by what I wrote.

Thanks again.

If B is a matrix and B^2 = 0, that does NOT imply that B must be zero; matrices do not behave like real numbers in all respects. However, you _can_ use standard algebra to expand (A+I)^2, because the matrices A and I commute under multiplication (that is, AI = IA), and you _can_ use AB+AC = A(B+C), etc.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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