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Inverse of a Matrix

Let A =[aij] ...

Question

Let $A = [a_{i j}]$ be $3 \times 3$ matrix given by
$a_{i j} = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} (\frac{i + j}{2}) + \frac{| i - j |}{2}, & i f i \neq j \frac{i^{j} - (i . j)}{i^{2} + j^{2}}, & i f i = j \end{matrix} ⎫ ⎪ ⎬ ⎪ ⎭$
where $a_{i j}$ denotes element of $i^{t h}$ row & $j^{t h}$ column of matrix A.If $A = p A^{2} - q A^{- 1} - r I$ , then remainder when $(p + q + r)^{11}$ is divided by 7 is

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Solution

$A = ⎡ ⎢ ⎣ \begin{matrix} 0 & 2 & 3 2 & 0 & 3 3 & 3 & 1 \end{matrix} ⎤ ⎥ ⎦$
$| A | = 32 \Rightarrow A^{- 1}$ will exist.
Characteristic equation of matrix A is
$| A - λ I | = 0$
$λ^{3} - λ^{2} - 22 λ - 32 = 0$
So $A^{3} - A^{2} - 22 A = 32 I$
$A^{2} - A - 22 I = 32 A^{- 1}$
$A = A^{2} - 32 A^{- 1} - 22 I$
So p+q + r =55
$55^{11} = (56 - 1)^{11} = (7 K - 1)$
So remainder is 6

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