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Find an upper triangular matrix $A$ such that $A^3=\begin{pmatrix}8&-57\\0&27\end{pmatrix}$. I tried to solve this problem using Cayley–Hamilton Theorem, but I am unable to solve that.

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    $\begingroup$ Well, $A$ must be of the form $A=\begin{pmatrix}2&a\\0&3\end{pmatrix}$; now try to find the value of $a$ that works. $\endgroup$ – gniourf_gniourf Feb 9 '16 at 14:18
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We have $$ A^3 = \pmatrix{8&-57\\0&27} $$ We calculate $$ A = \pmatrix{a&b\\0&c} \implies\\ A^3 = \pmatrix{a^3 & * \\0 & c^3} $$ where $*$ is an entry I don't care about yet. Now, find a valid $a,c$, and use these to find $b$ (by actually computing $*$).

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We have $$ \begin{pmatrix} 2 & b \\ 0 & 3 \end{pmatrix}^3-\begin{pmatrix} 8 & -57 \\ 0 & 27 \end{pmatrix} =\begin{pmatrix} 0 & 19(b+3) \\ 0 & 0 \end{pmatrix}. $$ In general we have $$ \begin{pmatrix} a & b \\ 0 & d \end{pmatrix}^3=\begin{pmatrix} a^3 & b(a^2+ad+d^2) \\ 0 & d^3 \end{pmatrix}. $$

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