+50 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?
+174 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.
Utilize the information about A:
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.
Start with the first equation: (A^2)x = A(Ax) = A(y).
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
