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It says: If [tex]x_{n+1} = Ax_{n}[/tex], write an expression for [tex]x_{n}[/tex]. The matrix A = [tex]\left(\begin{array}{ccc}2&0&0\\1&3&0\\-3&5&4\end{array}\right)[/tex]

From what I understand, this question wants me to do an eigenvector decomposition so this is what I came up with (after finding the eigenvectors and eigenvalues):

[tex]x_{n} = c_{1}(4 * \left(\begin{array}{c}0\\0\\1\end{array}\right)) + c_{2}(3 * \left(\begin{array}{c}0\\\frac{-1}{5}\\1\end{array}\right)) + c_{3}(2 * \left(\begin{array}{c}\frac{1}{4}\\\frac{-1}{4}\\1\end{array}\right))[/tex]

Is that what the question was asking for??

The other question (or help I need) is this one: Verify that there are infinitely many least squares solutions which are given by x = [tex]\left(\begin{array}{c}\frac{2}{7}\\\frac{13}{84}\\0\end{array}\right) + t \left(\begin{array}{c}\frac{-1}{7}\\\frac{5}{7}\\1\end{array}\right)[/tex]

It's talking about this system...

[tex]\left(\begin{array}{ccc}3&1&1\\2&-4&10\\-1&3&-7\end{array}\right) \left(\begin{array}{c}x_{1}\\x_{2}\\x_{3}\end{array}\right) = \left(\begin{array}{c}2\\-2\\1\end{array}\right)[/tex]

I multiplied both sides by the inverse of the matrix and got this system:

[tex]\left(\begin{array}{ccc}14&-8&30\\-8&26&-60\\30&-60&150\end{array}\right) \left(\begin{array}{c}x_{1}\\x_{2}\\x_{3}\end{array}\right) = \left(\begin{array}{c}-1\\13\\-25\end{array}\right)[/tex]

and I then row reduced it to..

[tex]\left(\begin{array}{cccc}14&-8&30&1\\0&\frac{150}{7}&\frac{-300}{7}&\frac{95}{7}\\0&0&0&0\end{array}\right)[/tex]

I thought I was going on the right track since I can see that this has infinitely many solutions but when I tried to find x...

[tex]x_{3} = t[/tex]

[tex]x_{2} = \frac{19}{30} + 2t[/tex]

[tex]x_{1} = \frac{13}{30} - t[/tex]

which definitely does not verify the above question. Can someone tell me where I went wrong? Thanks :)