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Possible Duplicate:
Solve $A^nx=b$ for an idempotent matrix

It is given that:

$$ A = \begin{bmatrix} 2 & 3 & -4 \\ 0 & 1 & 0 \\ 1/2 & 3/2 & -1 \end{bmatrix} $$

and

$$ b = \begin{bmatrix} 1 & 0 & 0 \\ \end{bmatrix} $$

Solve $A^n x = b$ for each positive integer n.

$A$ is idempotent so $A^n$ for positive integers of $n$ will still be equal to $A$. So the equation simply becomes $Ax = b$.

However, unless I am mistaken, $A$ is not invertible so I cannot solve for $x$ by multiplying both sides of the equation by $A^{-1}$.

What should I do instead?

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marked as duplicate by N. S., Bombyx mori, user53153, Austin Mohr, Paul Jan 28 '13 at 3:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Is $A$ idempotent? It does not seem so! $\endgroup$ – Maesumi Jan 28 '13 at 3:28
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Hint: Look at the second equation, it gives the value of $x_2$. Now use substituition.

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If you find $Ax=b$ with $b=[1,0,0]$, then the system has no solution. The reason is the first row is a linear combination of second and third row, so if both of them are $0$ then the first one must be $0$ as well.

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