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I f $$A= \begin{pmatrix}1& 0 & 0 \\ 1 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix}$$ then $ A^{50} $ is

  1. $$ \begin{pmatrix}1& 0 & 0 \\ 50 & 1 & 0\\ 50 & 0 & 1 \end{pmatrix}$$

  2. $$\begin{pmatrix}1& 0 & 0 \\ 48 & 1 & 0\\ 48 & 0 & 1 \end{pmatrix}$$

  3. $$\begin{pmatrix}1& 0 & 0 \\ 25 & 1 & 0\\ 25 & 0 & 1 \end{pmatrix}$$

  4. $$\begin{pmatrix}1& 0 & 0 \\ 24 & 1 & 0\\ 24 & 0 & 1\end{pmatrix}$$

I am stuck on this problem. Can anyone help me please...............

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  • 3
    $\begingroup$ Look for a pattern with $A^2,A^3,$ etc. $\endgroup$ – Daryl Dec 30 '12 at 6:56
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Simplify your problem as follows. The Jordan Normal form of your matrix $A$ is

$$J = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1& 1 \\ 0& 0& 1 \end{pmatrix},$$

where the matrix $P$ is such that $P^{-1}AP = J$ is given by

$$P = \begin{pmatrix} 0 & 0& 2 \\ -1 & 1 & 1 \\ 1 & 1 &0 \end{pmatrix}.$$

Then $A^{50} = PJ^{50}P^{-1}$. Now what is $J^{50}$?

Edit: Here's how you compute the Jordan Normal form of $A$ and the matrix $P$. First note that it has eigenvalues -1 and 1(with multiplicity 2). The Jordan normal form of $A$ is not diagonal because $A$ is not diagonalisable. It has eigenvectors $(0,1,-1)$ and $(0,1,1)$.

Now by my answer here we already know what the Jordan Normal Form of $A$ looks like, i.e. we know the matrix $J$. How do we obtain the matrix $P$? We now write $J$ like this:

$$J = \left(\begin{array}{c|cc} -1 & 0 & 0 \\ \hline 0 & 1& 1 \\ 0& 0& 1 \end{array}\right).$$

I have put in the grid to emphasize the fact that the first column of the matrix $P$ has to be the eigenvector $(0,-1,1)$ associated to the eigenvalue $-1$. The second column of $P$ looking at the second column of $J$ must now be the eigenvector associated to the eigenvalue $1$. What about the third column of $P$? This is of course given by a basis element for the kernel of $(A - I)^2$. We find that

$$(A - I)^2 = \begin{pmatrix} 0 & 0 & 0 \\ -1 & 2 & -2 \\ 1 & -2 & 2 \end{pmatrix} $$

which has a basis for its kernel the vector $(2,1,0)$ and so this completes the computation of $P$.

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  • $\begingroup$ @ BenjaLim sir, I can't find third column of P????? $\endgroup$ – user45799 Dec 30 '12 at 7:24
  • $\begingroup$ @Prasanta See my edit above. $\endgroup$ – user38268 Dec 30 '12 at 10:11
  • $\begingroup$ Thank you very much, @ BenjaLim sir! $\endgroup$ – user45799 Dec 30 '12 at 14:53
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You should learn BenjaLim's answer, which provides a general method for dealing with this kind of problems. However, here is a simple answer just for fun. Note that $$ A^2= \begin{pmatrix} 1&0&0\\ 1&1&0\\ 1&0&1\end{pmatrix} =I+\underbrace{\begin{pmatrix} 0&0&0\\ 1&0&0\\ 1&0&0\end{pmatrix}}_{L} $$ and $L^2=0$. Therefore $$ A^{50} = (I+L)^{25} = I+{25\choose 1}L+\sum_{k=2}^{25}{25\choose k}L^k=I+25L $$ and hence the answer is 3.

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The answer is 3. $$\begin{pmatrix}1& 0 & 0 \\ 25 & 1 & 0\\ 25 & 0 & 1 \end{pmatrix}$$

Just compute $A^2$ , $A^3$ , $A^4$ and $A^5$ and you will understand the repeated pattern.

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