Question

If $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right],$ then $A^4-5A^3-A^2-4A-2I=$

• A. $0$
• B. $A$
• C. $I$
• D.
  $2I$
  

Answer 1

The correct option is B $A$

The characteristics equation of A : $|A-\lambda I|=0$
$\Rightarrow \left[\begin{array}{cc}1-\lambda & 2\\ 3& 4-\lambda \end{array}\right]=0\Rightarrow (1-\lambda )(4-\lambda )-6=0\Rightarrow {\lambda }^2-5\lambda -2=0$
By Caley_Hamiltons theorem,
$\Rightarrow A^2-5A-2I=0\cdots \left(i\right)$
Now, $A^4-5A^3-A^2-4A-2I=A^2(A^2-5A-2I)+A^2-4A-2I=A^2\left(0\right)+A^2-4A-2I-A+A=A$