Question

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

• A. $\frac{1}{5}\left[\begin{array}{ccc}4& 2& -1\\ -1& 3& 1\\ -1& 2& 4\end{array}\right]$
• B. $\frac{1}{5}\left[\begin{array}{ccc}4& -2& -1\\ -1& 3& -1\\ -1& -2& 4\end{array}\right]$
  
• C. $\frac{1}{3}\left[\begin{array}{ccc}4& 2& -1\\ -1& 3& 1\\ -1& 2& 4\end{array}\right]$
• D. $\frac{1}{3}\left[\begin{array}{ccc}4& -2& -1\\ -1& 3& -1\\ -1& -2& 4\end{array}\right]$
  

Answer 1

The correct option is B $\frac{1}{5}\left[\begin{array}{ccc}4& -2& -1\\ -1& 3& -1\\ -1& -2& 4\end{array}\right]$


Given $A=\left[\begin{array}{ccc}2& 2& 1\\ 1& 3& 1\\ 1& 2& 2\end{array}\right]$

$\lambda I=\left[\begin{array}{ccc}\lambda & 0& 0\\ 0& \lambda & 0\\ 0& 0& \lambda \end{array}\right]$

Characteristic equation $|A-\lambda I|=0$
$\Rightarrow \left|\begin{array}{ccc}2-\lambda & 2& 1\\ 1& 3-\lambda & 1\\ 1& 2& 2-\lambda \end{array}\right|=0$
$\Rightarrow (2-\lambda )\left[\right(3-\lambda \left)\right(2-\lambda )-2]-2(2-\lambda -1)+1(2-3+\lambda )=0$
$\Rightarrow {\lambda }^3-7{\lambda }^2+11\lambda -5=0$
Ever matrix satisfies it's characteristic equation
$\Rightarrow (\lambda -1)({\lambda }^2-6\lambda +5)=0$
$\Rightarrow (\lambda -1{)}^2(\lambda -5)=0$
$\Rightarrow (A-I{)}^2(A-5I)=0$
$A^3-7A^2+11A-5=0$
$\Rightarrow A^{-1}(A^3-7A^2+11A-5)=0$
$\Rightarrow A^2-7A+11I-5A^{-1}=0$
$\Rightarrow A^{-1}=\frac{1}{5}[A^2-7A+11I]$
$A^2=\left[\begin{array}{ccc}2& 2& 1\\ 1& 3& 1\\ 1& 2& 2\end{array}\right]\left[\begin{array}{ccc}2& 2& 1\\ 1& 3& 1\\ 1& 2& 2\end{array}\right]=\left[\begin{array}{ccc}7& 12& 6\\ 6& 13& 6\\ 6& 12& 7\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{5}\left[\begin{array}{ccc}4& -2& -1\\ -1& 3& -1\\ -1& -2& 4\end{array}\right]$