# Matrix Algebra - University Maths?

I have no idea how to approach this question and my textbook is less than useful, so I'd appreciate any help especially if you can show step by step what you're doing.

Suppose that A is a square matrix of size n x n that satisfies the matrix equation:

A^14 A^T A^14 + 72A^5 (A^T)^2 A^5 + 16A A^T A^3 - 7I = 0

Show that the matrix is invertible and find an expression for A^-1 in terms of A.

Thanks.

Relevance
• 9 years ago

If you add 7I to both sides and divide by 7 you learn

(1/7) A^14 A^T A^14 + (72/7) A^5 (A^T)^2 A^5 + (16/7) A A^T A^3 = I.

Label this equation by (*).

Note that by the distributive law, you can factor an A out of the left hand side of (*). Since order of matrix multiplication matters in general, one should distinguish between factoring it out on the left (writing the left hand side as A * [another matrix]) or factoring it out on the right (writing the left hand side as [another matrix] * A). Both operations are possible here and they lead to different but equivalent results.

If you factor an A out on the left from the left hand side of (*), you learn that

A [(1/7) A^13 A^T A^14 + (72/7) A^4 (A^T)^2 A^5 + (16/7) A^T A^3] = I

which shows that the matrix M = (1/7) A^13 A^T A^14 + (72/7) A^4 (A^T)^2 A^5 + (16/7) A^T A^3 satisfies AM = I. Since A is a square matrix, some theory that tells you that this means A is invertible (ie, that MA = I also), and that M is A^(-1). So

(1/7) A^13 A^T A^14 + (72/7) A^4 (A^T)^2 A^5 + (16/7) A^T A^3

is one formula for A^(-1).

If on the other hand you factor an A out on the right from the left hand side of (*), you learn

[(1/7) A^14 A^T A^13 + (72/7) A^5 (A^T)^2 A^4 + (16/7) A A^T A^2] A = I.

This shows that the matrix N = (1/7) A^14 A^T A^13 + (72/7) A^5 (A^T)^2 A^4 + (16/7) A A^T A^2 satisfies NA = I, and again from the theory of square matrices you can deduce that A is invertible (so that AN = I also) and so that N = A^(-1). So

(1/7) A^14 A^T A^13 + (72/7) A^5 (A^T)^2 A^4 + (16/7) A A^T A^2

is another formula for A^(-1).

Note that these two formulas are not obviously the same! There is no reason for A and A^T to commute, in general, so the product A^14 A^T A^13 could very well be different from the product A^13 A^T A^14, and similarly with the other terms--- the individual terms in the formulas for M and N could very well be different. But thanks to the general theory of invertible square matrices--- whenever A is a square matrix, and M and N are square matrices with AM = I and NA = I, it must be that A is invertible and M = N = A^(-1)--- it is guaranteed that the two formulas for M and N must give the same result, and give you A^(-1).

[If A were just an arbitrary matrix--- ie, not one assumed to satisfy the original equation--- then the matrices specified by the formulas for M and N will in general be different.]

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4 years ago

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