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# Matrices with repeated exponents

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If A is a matrix and x, y are vectors such that neither is a multiple of the other and Ax = y and Ay = x + 2y, how do we find a and b when we have that (A^5) x = ax + by?

Apr 15, 2024

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We can solve for a and b by utilizing the given information about A and its effect on x and y, along with the property of matrix multiplication.

Here's how we proceed:

Analyze the equation (A^5)x = ax + by:

This equation states that applying matrix A to vector x five times consecutively (A raised to the power of 5) results in a linear combination of x and y, with coefficients a and b.

We are given that Ax = y and Ay = x + 2y. These equations define how A transforms x and y.

Express (A^5)x in terms of x and y:

We can't directly expand (A^5) as it's a high power. However, we can use the given information about A iteratively.

Substitute from the first given equation: (A^2)x = Ay.

Use the second given equation: (A^2)x = x + 2y.

Similarly, we can continue:

(A^3)x = A((A^2)x) = A(x + 2y) = Ax + 2Ay (using the definition of matrix multiplication).

Substitute from the first given equation: (A^3)x = y + 2(x + 2y) = 3x + 4y.

We can repeat this process further, but the pattern should be clear.

Find an expression for (A^4)x and (A^5)x:

Following the established pattern, we can see that:

(A^4)x = 3(A^3)x = 3(3x + 4y) = 9x + 12y.

(A^5)x = 3(A^4)x = 3(9x + 12y) = 27x + 36y.

Substitute (A^5)x in the original equation:

The original equation is: (A^5)x = ax + by.

Substitute the expression we found for (A^5)x: 27x + 36y = ax + by.

Solve for a and b:

We want to isolate a and b. Since x and y are not multiples of each other, we can treat them as independent variables.

If we set y = 0, the equation becomes 27x = ax, which implies a = 27.

If we set x = 0, the equation becomes 36y = by, which implies b = 36.

Therefore, in the equation (A^5)x = ax + by, the values of a and b are:

a = 27

b = 36

Apr 15, 2024