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  4. if A = [1&0&0 0&1&1 0&-2&4],6A-1 = A2 +cA +dI, then (c,d ) is

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Q. if $A = \begin{bmatrix}1&0&0\\ 0&1&1\\ 0&-2&4\end{bmatrix},6A^{-1} = A^{2} +cA +dI$, then $(c,d ) $ is

Matrices

Solution:

Given, $6A^1 =A^2 + cA + dI$

$ \Rightarrow 6A^1 -A - A^2 A + cA + dI \cdot A$

$\Rightarrow 6I = A^3 + cA^2 + dA $

$\Rightarrow A^3 + cA^2 + dA-6I=0$

Now, $A^{2} = \begin{bmatrix}1&0&0\\ 0&1&1\\ 0&-2&4\end{bmatrix}\begin{bmatrix}1&0&0\\ 0&1&1\\ 0&-2&4\end{bmatrix}=\begin{bmatrix}1&0&0\\ 0&-1&5\\ 0&-10&14\end{bmatrix}$

$A^{3}\begin{bmatrix}1&0&0\\ 0&-1&5\\ 0&-10&14\end{bmatrix}\begin{bmatrix}1&0&0\\ 0&1&1\\ 0&-2&4\end{bmatrix}=\begin{bmatrix}1&0&0\\ 0&-1&5\\ 0&-10&14\end{bmatrix}$

$\therefore $ From equation$ \left(i\right)$, we have

$\begin{bmatrix}1&0&0\\ 0&-11&19\\ 0&-38&46\end{bmatrix}+c\begin{bmatrix}1&0&0\\ 0&-1&5\\ 0&-10&14\end{bmatrix}+d\begin{bmatrix}1&0&0\\ 0&1&1\\ 0&-2&4\end{bmatrix}-6\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}= 0$

$\Rightarrow \begin{bmatrix}1+c+d-6&0&0\\ 0&-11-c+d-6&19+5c+d\\ 0&-38-10c-2d&46+14c+4d-6\end{bmatrix}=\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{bmatrix}$

$\Rightarrow 1+c+d-6 =0, -11 -c +d -6 =0$

$\Rightarrow c+ d = 5, -c+d = 17$

$\Rightarrow c = -6 , d =11$