Question and Answers Forum

All Questions      Topic List

Matrices and Determinants Questions

Previous in All Question      Next in All Question      

Previous in Matrices and Determinants      Next in Matrices and Determinants      

Question Number 32877 by 7991 last updated on 05/Apr/18

$${\left[\begin{array}{cc}2& 1\\ 0& 2\end{array}\right]}^{2018}=.....???$$

Answered by Joel578 last updated on 05/Apr/18

$$A=\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)$$$$A^2=\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)=\left(\begin{array}{cc}4& 4\\ 0& 4\end{array}\right)$$$$A^3=\left(\begin{array}{cc}4& 4\\ 0& 4\end{array}\right)\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)=\left(\begin{array}{cc}8& 12\\ 0& 8\end{array}\right)$$$$A^4=\left(\begin{array}{cc}8& 12\\ 0& 8\end{array}\right)\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)=\left(\begin{array}{cc}16& 32\\ 0& 16\end{array}\right)$$$$⋮$$$$A^{2018}=\left(\begin{array}{cc}{2}^{2018}& 2018.2^{2017}\\ 0& 2^{2018}\end{array}\right)$$

Commented by MJS last updated on 05/Apr/18

$$A^5=\left(\begin{array}{cc}32& 80\\ 0& 32\end{array}\right)$$$$4=2\times 2$$$$12=3\times 2^2$$$$32=4\times 2^3$$$$80=5\times 2^4$$$$A^n=\left(\begin{array}{cc}{2}^n& n\times 2^{(n-1)}\\ 0& 2^n\end{array}\right)$$$$A^{2018}=\left(\begin{array}{cc}{2}^{2018}& 2018\times 2^{2017}\\ 0& 2^{2018}\end{array}\right)$$

Commented by Joel578 last updated on 05/Apr/18

$$thankyouverymuch$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com