Let A=pmatrix & -16 & pmatrix, 0, such that (A)=0 and +=1.

Question

Let $A=\begin{pmatrix}\alpha & -1\6 & \beta\end{pmatrix}$, $\alpha>0$, such that $\det(A)=0$ and $\alpha+\beta=1$. If $I$ denotes the $2	imes2$ identity matrix, then the matrix $(I+A)^8$ is:

Accepted Answer

$|A|=\alpha\beta+6=0\Rightarrow\alpha\beta=-6$. With $\alpha+\beta=1$, $\alpha,\beta$ are roots of $x^2-x-6=0\Rightarrow x=3,-2$. Since $\alpha>0$: $\alpha=3,\ \beta=-2$.

So $A=\begin{pmatrix}3&-1\6&-2\end{pmatrix}$, $I+A=\begin{pmatrix}4&-1\6&-1\end{pmatrix}$.

Squaring repeatedly: $(I+A)^2=\begin{pmatrix}10&-3\18&-5\end{pmatrix}$, $(I+A)^4=\begin{pmatrix}46&-15\90&-29\end{pmatrix}$, $(I+A)^8=\begin{pmatrix}766&-255\1530&-509\end{pmatrix}$.

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