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Let $$A=\begin{bmatrix}2& 3& -4\\ 0& 1 & 0\\ 0.5& 1.5 &-1\end{bmatrix},~ b=\begin{bmatrix}1\\ 0\\ 0\end{bmatrix}.$$ Show that $A$ is idempotent and solve the matrix equation $$A^nx=b$$ for each positive integer $n$.

Alex Becker
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Nighthawk
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  • What is idempotent for you? – Sigur Jan 26 '13 at 23:04
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    The question defines a matrix as idempotent if A^2=A – Nighthawk Jan 26 '13 at 23:05
  • @Nighthawk Ever considered the possibility that there is no solution? Perhaps if you look at the range space of A? In particular the third column is a multiple of the first ... – Calvin Lin Jan 26 '13 at 23:13
  • I stated the equation was inconsistent, however, my prof said it was possible to solve it. – Nighthawk Jan 26 '13 at 23:20
  • @Nighthawk Maybe he said that there exists a set of solutions? The set of solutions of an inconsistent system of equations is the empty set and that's your answer. – Git Gud Jan 26 '13 at 23:22

1 Answers1

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Is there a specific part where you're having trouble?

For the first part, do exactly as the question asks -- multiply $A$ by itself and check that you get $A$ back.

For the second part, if $A^2=A$, then $A^n = \ldots$?

EDIT: You can solve the system using Gaussian elimination. Starting from $$\left[\begin{array}{ccc}2 & 3 & -4\\0 & 1 & 0\\1/2 & 3/2 & -1\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\0\\0\end{array}\right]$$ we can add $-4$ times the last row to the first to get $$\left[\begin{array}{ccc}0 & -3 & 0\\0 & 1 & 0\\1/2 & 3/2 & -1\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\0\\0\end{array}\right],$$ and now adding 3 times the second row to the first gives $$\left[\begin{array}{ccc}0 & 0 & 0\\0 & 1 & 0\\1/2 & 3/2 & -1\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\0\\0\end{array}\right]$$ and the system has no solution.

user7530
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    The Second part, I have shown A is idempotent and I know A^n will be A. But I can't solve the matrix equation – Nighthawk Jan 26 '13 at 23:09
  • @Nighthawk You can't solve $Ax=b \space$ after doing everything else? I find that hard to believe. – Git Gud Jan 26 '13 at 23:10
  • @Nighthawk - So your problem is how to solve $Ax=b$. You can do this by ranking $A$ with $b$ written along side, do you know this method ? how were you thought to solve $Ax=b$, where is it a problem for you ? – Belgi Jan 26 '13 at 23:13
  • @GitGud The OP cannot solve $Ax=b$ because the equation has no solution. – user1551 Jan 26 '13 at 23:14
  • system is inconsistent, singular and the determinate is 0 so I can't use cramers rule. My professor says that it is possible to solve it but these are the only methods I have learned so far. – Nighthawk Jan 26 '13 at 23:15
  • @Nighthawk Unless you made a mistake in your computations "system is inconsistent" is the solution to the problem... – N. S. Jan 26 '13 at 23:18
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    @Nighthawk Your professor is wrong (and it won't be the only time that he states something false). You can use the augmented matrix method to show that no solution is possible. – Calvin Lin Jan 26 '13 at 23:19