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Thread: How do I solve this? Linear Algebra.

  1. #1
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    Unhappy How do I solve this? Linear Algebra.

    Find the upper triangular matriz A such that A^3 = (8, -57 and second row 0, 27).

    Can anyone help?
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  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
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    Hi,

    Let $\displaystyle A=\left(\begin{smallmatrix}x & y\\ 0 & z\end{smallmatrix}\right)$. If you compute $\displaystyle A^3$ you'll get
    $\displaystyle A^3=\begin{pmatrix}{x}^{3} & y\left({z}^{2}+xz+x^2 \right) \cr 0 & {z}^{3}\end{pmatrix}$

    so you have to solve
    $\displaystyle \begin{pmatrix}{x}^{3} & y\left({z}^{2}+xz+x^2 \right) \cr 0 & {z}^{3}\end{pmatrix}=\begin{pmatrix}8&-57\\0 & 27 \end{pmatrix}$
    for $\displaystyle x$, $\displaystyle y$ and $\displaystyle z$.
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