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Let 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).

Sep 25, 2020

#1
+25569
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Let 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).

$$\begin{array}{|rcll|} \hline Ax &=& y \quad | \quad \cdot A \\ A^2x &=& Ay \quad | \quad Ay = x+2y \\ \mathbf{A^2x} &=& \mathbf{x+2y} \quad | \quad \cdot A \\ A^3x &=& A(x+2y) \\ A^3x &=& Ax+2Ay \\ A^3x &=& y+2(x+2y) \\ A^3x &=& y+2x+4y \\ \mathbf{A^3x} &=& \mathbf{2x+5y} \quad | \quad \cdot A \\ A^4x &=& A(2x+5y) \\ A^4x &=& 2Ax+5Ay \\ A^4x &=& 2y+5(x+2y) \\ A^4x &=& 2y+5x+10y \\ \mathbf{A^4x} &=& \mathbf{5x+12y} \quad | \quad \cdot A \\ A^5x &=& A(5x+12y) \\ A^5x &=& 5Ax+12Ay \\ A^5x &=& 5y+12(x+2y) \\ A^5x &=& 5y+12x+24y \\ \mathbf{A^5x} &=& \mathbf{12x+29y} \\ \hline \end{array}$$

Sep 28, 2020