How Can I Prove That B^3 Equals 7B Minus 6I Using Matrix B?

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Homework Help Overview

The discussion revolves around proving that the matrix B satisfies the equation B^3 = 7B - 6I, given that it satisfies the quadratic equation B^2 - 3B + 2I = 0. The matrix B is defined with specific numerical entries.

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  • Mixed

Approaches and Questions Raised

  • Participants explore various methods to demonstrate the relationship between B^3 and the quadratic equation. Some suggest using factorization techniques similar to those used in previous math classes, while others propose direct multiplication of the matrix B.

Discussion Status

Participants are actively engaging with the problem, offering different perspectives and methods. Some have provided partial calculations and insights, while others express confusion about the next steps. There is no explicit consensus on the approach to take, but several productive lines of reasoning are being explored.

Contextual Notes

There is a mention of a deadline for the homework, indicating time constraints. Participants are also discussing the requirement to use the quadratic equation to prove the relationship, which adds a layer of complexity to the problem.

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Homework Statement


Show that the matrix B= 4 -2 -2
1 1 -1
2 -2 0
satisfies the quadratic equation
B^2-3B+2I=0

Hence show that B^3=7B-6I


The Attempt at a Solution


Now i have already worked out the quadratic equation and the matirx B satisfies the equaition B^2-3B+2I=0.

My problem is i can't seem to prove the next part i am stuck here is how for i got


B^3=7B-6I
multiplying by B^2 gives

B^4 - 3B^3 + 2B^2I = O

multiply b B^-1 (B inverse) gives

IB^3 - 3IB^2 = 2IB =0

this is where i am stuck.. thanks for any help given...question dued tomorrow.
 
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You said you showed that B satisfies B^2 - 3B + 2I = 0. Now you want to show that the same matrix satisfies B^3 - 7B + 6I = 0.

These matrix equations work much like the equations you worked with earlier in your math classes with regard to factorization. For example, your first equation factors into
(B - 2I)(B - I) = 0. Instead of doing what you did before (multiplying B by itself, subtracting 3*B, and then adding 2*I), you could have calculation B - 2I and B - I and multiplied these two matrices together.

As it turns out, your second equation can also be factored, and one of the factors is B^2 - 3B + 2I.
 
You are given B so just multiply to get B3. Then calculuate 7B- 6I and see if they are the same!
 
yes i agree but we can't do it that we we have to use the quadratic equation to prove that B^3 = 7B-6I
 
Then do the obvious thing: divide x3- 7x+ 6 by x2- 3x+ 2.
 
B^2=3b-2i
b^3=3b^2-2b=3(3b-2i)-2b=7b-6i
 
and how will that help me i get x+3 :|
 
Thanks a lot cloudyhill that helps greatly!
 
That tells you that B^3-7B+6I=(B+3I)(B^2-3B+2I) but you already showed that B^2-3B+2I=0, sooo B^3-7B+6I=\ldots?:wink:
 

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