If $A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& -2& 4\end{array}\right] . I=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$A^{-1}=\frac{1}{6}[A^2+cA+dI]$

where c, d ∈ R, the pair of values (c, d ) are

 

Given

$A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& -2& 4\end{array}\right] ,A^{-1}=\frac{1}{6}\left[\begin{array}{ccc}6& 0& 0\\ 0& 4& -1\\ 0& 2& 1\end{array}\right]$

$A^2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& -2& 4\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& -2& 4\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 5\\ 0& -10& -14\end{array}\right]$

$cA=\left[\begin{array}{ccc}c& 0& 0\\ 0& c& c\\ 0& -2c& 4c\end{array}\right] , dI=\left[\begin{array}{ccc}d& 0& 0\\ 0& d& 0\\ 0& 0& d\end{array}\right]$

$\therefore A^{-1}=\frac{1}{6}[A^2+cA+dI]$

$\Rightarrow 6=1+c+d,$          (By equality of matrices)

$\therefore$     (–6, 11) satisfy the relation