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themadhatter1
Jul6-10, 12:04 PM
1. The problem statement, all variables and given/known data
Use the matrix capabilities of a graphing utility to find:
f(A)=a_{0}I_{n}+a_{1}A+a_{2}A^2+\cdots+a_{n}A^n

1.
f(x)=x^2-5x+2
A=\left[\begin{array}{cc}2&0\\4&5\end{array}\right]

2. Relevant equations



3. The attempt at a solution

Well, I know the answer is
\left[\begin{array}{cc}-4&0\\8&2\end{array}\right]
However, I don't know how to get it.

I would think you would do A^2-5A+2 however you cant add a constant to a matrix. I'm not sure exactly what I'm supposed to do.

Gear300
Jul6-10, 12:11 PM
The constant 2 might be

\left[\begin{array}{cc}2&2\\2&2\end{array}\right]

themadhatter1
Jul6-10, 12:15 PM
Nope, thats not it. Just tried it and it's wrong, not sure how it would be it though.

Gear300
Jul6-10, 12:27 PM
A constant by itself might signify that it is in operation with an identity. So 2 might be


\left[\begin{array}{cc}2&0\\0&2\end{array}\right]

themadhatter1
Jul6-10, 12:31 PM
oh, ok. That's turns out to be right. Thanks!

Mark44
Jul6-10, 01:04 PM
A constant by itself might signify that it is in operation with an identity. So 2 might be


\left[\begin{array}{cc}2&0\\0&2\end{array}\right]


Right. The polynomial is f(A) = A2 - 5A + 2I.

themadhatter1
Jul6-10, 04:42 PM
How exactly did you derive 'I's value?

Mark44
Jul6-10, 05:04 PM
I is the 2 x 2 identity matrix, defined as
\left[\begin{array}{cc}1&0\\0&1\end{array}\right]

Since A is given as a 2 x 2 matrix, the appropriate identity matrix must also be 2 x 2. If A were given as a 3 x 3 matrix, you would need to use the 3 x 3 identity matrix, which is defined as
\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]

The form of the identity matrix to use depends on the size of the square matrices being used in the problem.

themadhatter1
Jul6-10, 05:11 PM
Ahh.. ok. I didn't know what an identity matrix was before. But now I know. Interesting.