How to Multiply Matrices I and Calculate the Inverse of Matrix B?

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Peter G.
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Hi,

1.Show that B satisfies the equation (B-pI)(B-qI) = 0

2.Hence, or otherwise, show that B-1 = 0.5(3I - B)

In these kind of questions I don't know what they are testing me for! Let's take the first one as an example: The only skill they can possibly try to asses is whether I know how to multiply the matrix I by values I found previously (p and q). Other than that, all I can do to "answer" the question is performing the multiplication and showing it equals zero?

For the second one it is even worst... I know how to calculate the inverse of B and when I do it does in fact equal 0.5(3I - B), but, what should I put on paper? Calculate the inverse the regular way and then perform 0.5(3I - B) and show the results are equal?

I'm not sure if I was able to convey my doubt clearly... If the problem is due to lack of information in questions 1 and 2 I can add more information or rephrase my query.

Thanks,
Peter G.
 
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Is that all the question gives? Is B just a general nxn matrix or is it something given? (since you said you found values of p and q)

You would just need to show that B, whatever that is will make that equation zero.

For the second one, if you pre-multiply both sides B you might be able to factorize it in the form given in 1.
 
Yeah, B is a 2x2 matrix. So I basically just multiply everything out and show it equals zero?

For the second one you mean if I expand the first equation I can get the second one?

Thanks once again,
Peter G.
 
Peter G. said:
Yeah, B is a 2x2 matrix. So I basically just multiply everything out and show it equals zero?
Are you given a specific matrix B and values for p and q?
Peter G. said:
For the second one you mean if I expand the first equation I can get the second one?
Multiply 0.5(3I - B) by B for the matrix you are given (assuming you know B). If the expression simplifies to I, then 0.5(3I - B) is the inverse of B.
 
Hi,

Yeah, I have a specific value for both the matrix B, p and q.

I got the second one now, thanks!