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tikki14 · Feb 19, 2012, 11:10 AM

$M = \begin{bmatrix} 2 & 2 1 & 1 \end{bmatrix} $
Solve the equation: $X^5=A, X \epsilon M_2 (R)$
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If $X = \begin{bmatrix} a & b c & d \end{bmatrix}$

$det(A)=0$
$det(X^5)=[det(X)]^5=0$ , so $det(X)=0$

Using Hamilton Cayley theorem:
$X^2 = (a d)X det(X) = (a d)X$
$X^5 = X^4 \times X = (X^2)^2 \times X = (a d)^2 X^2 \times X = (a d)^2 (a d) X^2 = (a d)^3 X^2 = (a d)^4 X$

How should I proceed next?

tikki14 · Feb 19, 2012, 11:12 AM

X^5 = (a+d)^4 X

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