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Let $M=\begin{pmatrix} 6 & 2\\ -2& -3 \end{pmatrix}$

then trace of $M^{5}$
asked in Mathematical Logic by Veteran (103k points) | 68 views
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$3^5=243?$
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eigen values of M:  $\lambda 1=$ $\frac{3+\sqrt{65}}{2}$, $\lambda 2=$ $\frac{3-\sqrt{65}}{2}$

eigen values of M5: $\lambda '1=(\frac{3+\sqrt{65}}{2})^5$ , $\lambda '2=(\frac{3-\sqrt{65}}{2})^5$

trace of M5= sum of eigen values of M5 = $(\frac{3+\sqrt{65}}{2})^5 +(\frac{3-\sqrt{65}}{2})^5$

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M = $\begin{pmatrix} 6&2 \\ -2&-3 \end{pmatrix}$

Characteristic equation is :

$\begin{vmatrix} M - \lambda I \end{vmatrix} = 0$

So after solving characteristic equation will be - 

$\lambda ^{2} = 3\lambda + 14I$

Using Caley Hamilton's theorem 

$M^{2} = 3M + 14I$

Again $M^{5} = M^{2}\times M^{3}$

$M^{5} = (3M + 14I)\times M^{3}$

           $= 3M^{4} + 14M^{3}$

            = $3(M^{2})^{2} + 14M\times M^{2}$

            $= 3(3M+14I)^{2} + 14M(3M+14I)$

            $= 3(9M^{2} + 84M + 196I) + 42M^{2} + 196M$

             $= 27M^{2} + 252M + 588I + 42M^{2} + 196M$

            $= 69M^{2} + 448M + 588I$

            $= 69(3M + 14I) + 448M+ 588I$

            $= 207M + 966I + 448M + 588I$

            $= 655M + 1554I$

            $= 655\times \begin{pmatrix} 6 &2 \\ -2&-3 \end{pmatrix} + 1554\times \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

           $= \begin{pmatrix} 3930 & 1310 \\ -1310 & -1965 \end{pmatrix} + \begin{pmatrix} 1554 & 0\\ 0 & 1554 \end{pmatrix}$

            $= \begin{pmatrix} 5484 & 1310\\ -1310 & -411 \end{pmatrix}$

Trace of $M^{5}$ is $5484-411$ $= 5073$

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