Question

If $A=\left[\begin{array}{cc}-4& -1\\ 3& 1\end{array}\right]$ then the determinate of the matrix $A^{2016}-2A^{2015}-A^{2014}$ is

• A. $2014$
• B. $2016$
• C. $-175$
• D. $-25$

Answer 1

Answer: (D)

$-25$

Given that

$A=\left[\begin{array}{cc}-4& -1\\ 3& 1\end{array}\right]$

$|A^{2016}-2A^{2015}-A^{2014}|=?$

$\Rightarrow |A^{2014}||A^2-2A-I|$

$=|A{|}^{2014}[A^2-2A-I]$

$A^2=A.A=\left[\begin{array}{cc}-4& -1\\ 3& 1\end{array}\right]\left[\begin{array}{cc}-4& -1\\ 3& 1\end{array}\right]$

$A^2=\left[\begin{array}{cc}16-3& 4-1\\ -12+3& -3+1\end{array}\right]=\left[\begin{array}{cc}13& 3\\ -9& -2\end{array}\right]$

So,

$|A{|}^{2014}[\left[\begin{array}{cc}13& 3\\ -9& -2\end{array}\right]-2\left[\begin{array}{cc}-4& -1\\ 3& 1\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]]$

$=|A{|}^{2014}[\left[\begin{array}{cc}13& 3\\ -9& -2\end{array}\right]-\left[\begin{array}{cc}-8& -2\\ 6& 2\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]]$

$=|A{|}^{2014}\left[\begin{array}{cc}13+8-1& 3+2-0\\ -9-6-0& -2-2-1\end{array}\right]$

$=|A{|}^{2014}\left[\begin{array}{cc}20& 5\\ -15& -5\end{array}\right]$

$|A|=\left[\begin{array}{cc}-4& -1\\ 3& 1\end{array}\right]=-4\times 1-(-1)\times 3$

$=-4+3=-1$

then

$|A{|}^{2014}\left[\begin{array}{cc}20& 5\\ -15& -5\end{array}\right]$

$(-1{)}^{2014}[-100+75]$

$(-1{)}^{2014}\times (-25)=1\times (-25)$

$=-25$

Hence this is though