Let $A=\left[\begin{array}{cc}\alpha & -1 \ 6 & \beta\end{array} ight], \alpha \gt 0$, such that $\operatorname{det}(A)=0$ and $\alpha+\beta=1$. If I denotes $2 imes 2$ identity matrix, then the matrix $(1+\mathrm{A})^8$ is:

Question

Let $A=\left[\begin{array}{cc}\alpha & -1 \ 6 & \beta\end{array}ight], \alpha \gt 0$, such that $\operatorname{det}(A)=0$ and $\alpha+\beta=1$. If I denotes $2 	imes 2$ identity matrix, then the matrix $(1+\mathrm{A})^8$ is:

Aicharya · Accepted Answer

$\begin{aligned} & |\mathrm{A}|=0 \ & \alpha \beta+6=0 \ & \alpha \beta=-6 \ & \alpha+\beta=1 \ & \Rightarrow \alpha=3, \beta=-2 \ & \mathrm{~A}=\left[\begin{array}{ll}3 & -1 \ 6 & -2\end{array}ight] \ & \mathrm{A}^2=\left[\begin{array}{ll}3 & -1 \ 6 & -2\end{array}ight]\left[\begin{array}{ll}3 & -1 \ 6 & -2\end{array}ight]=\left[\begin{array}{ll}3 & -1 \ 6 & -2\end{array}ight] \ & 	herefore \mathrm{A}^2=\mathrm{A} \ & \mathrm{A}=\mathrm{A}^2=\mathrm{A}^3=\mathrm{A}^4=\mathrm{A}^5 \ & (\mathrm{I}+\mathrm{A})^8 \ & =\mathrm{I}+{ }^8 \mathrm{C}_1 \mathrm{~A}^7+{ }^8 \mathrm{C}_2 \mathrm{~A}^6+\ldots . .+{ }^8 \mathrm{C}_8 \mathrm{~A}^8 \ & =\mathrm{I}+\mathrm{A}\left({ }^8 \mathrm{C}_1+{ }^8 \mathrm{C}_2+\ldots . .+{ }^8 \mathrm{C}_8ight) \ & =\mathrm{I}+\mathrm{A}\left(2^8-1ight) \ & =\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}ight]+\left[\begin{array}{cc}765 & -255 \ 1530 & -510\end{array}ight]\end{aligned}$ $=\left[\begin{array}{cc}766 & -255 \ 1530 & -509\end{array}ight]$

Let $A=\left[\begin{array}{cc}\alpha & -1 \\ 6 & \beta\end{array}\right], \alpha \gt 0$, such that…

Solution