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Let \(\mathbf{A}\) be a matrix,  and let x and y be linearly independent vectors such that

\(\mathbf{A} \mathbf{x} = \mathbf{y}, \mathbf{A} \mathbf{y} = \mathbf{x} + 2\mathbf{y}\)
Then we have that 

\(\mathbf{A}^5 \mathbf{x} = a \mathbf{x} + b\mathbf{y}\)
for some scalars a and b. Find the ordered pair (a,b).

 Feb 19, 2019

Best Answer 

 #1
avatar+4434 
+3

\(A x = y\\ Ay = x + 2y\)

 

\(A^2x =AAx = Ay = x+2y\)

 

\(A^3x=AA^2x = A(x+2y) = \\ Ax + 2Ay = y + 2(x+2y) = \\ 2x+5y\)

 

\(A^4x = A(2x+5y) = 2y + 5(x+2y) = 5x+12y\)

 

\(A^5 x = A(5x+12y) = 5y + 12(x+2y) = 12x + 29y\)

 

\((a,b) = (12, 29)\)

.
 Feb 19, 2019
edited by Rom  Feb 19, 2019
 #1
avatar+4434 
+3
Best Answer

\(A x = y\\ Ay = x + 2y\)

 

\(A^2x =AAx = Ay = x+2y\)

 

\(A^3x=AA^2x = A(x+2y) = \\ Ax + 2Ay = y + 2(x+2y) = \\ 2x+5y\)

 

\(A^4x = A(2x+5y) = 2y + 5(x+2y) = 5x+12y\)

 

\(A^5 x = A(5x+12y) = 5y + 12(x+2y) = 12x + 29y\)

 

\((a,b) = (12, 29)\)

Rom Feb 19, 2019
edited by Rom  Feb 19, 2019

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