Question

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

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

Answer 1

Answer: (D)

$-25$

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

Now, $A^{2016}-2A^{2015}-A^{2014}=A^{2014}(A^2-2A-I)$

$\Rightarrow |A^{2016}-2A^{2015}-A^{2014}|=|A^{2014}||A^2-2A-I|$

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

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

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

$=-25$