If $α$ and $β (α > β)$ are the roots of the equation $[1 5] {[\begin{array}{l} 1 3 \\ - 4 7 \end{array}]}^{2} {[\begin{array}{l} \frac{7}{19} - \frac{3}{19} \\ \frac{4}{19} \frac{1}{19} \end{array}]}^{4} {[\begin{array}{l} 1 3 \\ - 4 7 \end{array}]}^{2} [\begin{array}{l} x^{2} - 5 x + 5 \\ - 3 \end{array}] = [- 4]$ then the value of $(2 + α) (2 - β)$

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Detailed Solution

Let $A = [\begin{array}{l} 1 3 \\ - 4 7 \end{array}]$ then $A^{- 1} = [\begin{array}{l} \frac{7}{19} - \frac{3}{19} \\ \frac{4}{19} \frac{1}{19} \end{array}]$
Since $A^{- 1} A = I = A A^{- 1}, [\begin{array}{l} 1 3 \\ - 4 7 \end{array}] [\begin{array}{l} \frac{7}{19} - \frac{3}{19} \\ \frac{4}{19} \frac{1}{19} \end{array}] = I = [\begin{array}{l} \frac{7}{19} - \frac{3}{19} \\ \frac{4}{19} \frac{1}{19} \end{array}] [\begin{array}{l} 1 3 \\ - 4 7 \end{array}]$
$[1 5] [\begin{array}{l} x^{2} - 5 x + 5 \\ - 3 \end{array}] = [- 4] \Rightarrow x^{2} - 5 x + 5 - 15 = - 4 \Rightarrow x^{2} - 5 x - 6 = 0 \Rightarrow x = 6, - 1 (α, β)$