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Derivative of a Determinant

If fy = 1- ya...

Question

If $f (y) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 - y)^{a_{1} b_{1}} & (1 - y)^{a_{1} b_{2}} & (1 - y)^{a_{1} b_{3}} (1 - y)^{a_{2} b_{1}} & (1 - y)^{a_{2} b_{2}} & (1 - y)^{a_{2} b_{3}} (1 - y)^{a_{3} b_{1}} & (1 - y)^{a_{3} b_{2}} & (1 - y)^{a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣$ and $a_{i}, b_{i}$ are even for $i = 1, 2, 3,$ then $f^{'} (2)$ is.

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Solution

$f (y) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 - y)^{a_{1} b_{1}} & (1 - y)^{a_{1} b_{2}} & (1 - y)^{a_{1} b_{3}} (1 - y)^{a_{2} b_{1}} & (1 - y)^{a_{2} b_{2}} & (1 - y)^{a_{2} b_{3}} (1 - y)^{a_{3} b_{1}} & (1 - y)^{a_{3} b_{2}} & (1 - y)^{a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ \Rightarrow f^{'} (y) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a_{1} b_{1} (1 - y)^{a_{1} b_{1} - 1} & (1 - y)^{a_{1} b_{2}} & (1 - y)^{a_{1} b_{3}} a_{2} b_{1} (1 - y)^{a_{2} b_{1} - 1} & (1 - y)^{a_{2} b_{2}} & (1 - y)^{a_{2} b_{3}} a_{3} b_{1} (1 - y)^{a_{3} b_{1} - 1} & (1 - y)^{a_{3} b_{2}} & (1 - y)^{a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 - y)^{a_{1} b_{1}} & a_{1} b_{2} (1 - y)^{a_{1} b_{2} - 1} & (1 - y)^{a_{1} b_{3}} (1 - y)^{a_{2} b_{1}} & a_{2} b_{2} (1 - y)^{a_{2} b_{2} - 1} & (1 - y)^{a_{2} b_{3}} (1 - y)^{a_{3} b_{1}} & a_{3} b_{2} (1 - y)^{a_{3} b_{2} - 1} & (1 - y)^{a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 - y)^{a_{1} b_{1}} & (1 - y)^{a_{1} b_{2}} & a_{1} b_{3} (1 - y)^{a_{1} b_{3} - 1} (1 - y)^{a_{2} b_{1}} & (1 - y)^{a_{2} b_{2}} & a_{2} b_{3} (1 - y)^{a_{2} b_{3} - 1} (1 - y)^{a_{3} b_{1}} & (1 - y)^{a_{3} b_{2}} & a_{3} b_{3} (1 - y)^{a_{3} b_{3} - 1} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Putting $y = 2$
We know that $a_{i}, b_{i}$ are even so $a_{i} b_{i}$ will also be even.
$f^{'} (2) = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} b_{1} & 1 & 1 a_{2} b_{1} & 1 & 1 a_{3} b_{1} & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 1 & a_{1} b_{2} & 1 1 & a_{2} b_{2} & 1 1 & a_{3} b_{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & a_{1} b_{3} 1 & 1 & a_{2} b_{3} 1 & 1 & a_{3} b_{3} \end{matrix} ∣ ∣ ∣ ∣$
As two columns are same so,
$f^{'} (2) = 0$

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f (y) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (1 - y)^{a_{1} b_{1}} & (1 - y)^{a_{1} b_{2}} & (1 - y)^{a_{1} b_{3}} (1 - y)^{a_{2} b_{1}} & (1 - y)^{a_{2} b_{2}} & (1 - y)^{a_{2} b_{3}} (1 - y)^{a_{3} b_{1}} & (1 - y)^{a_{3} b_{2}} & (1 - y)^{a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣

a, b

i = 1, 2, 3

f^{'} (2)

A_{i}

is the area bounded by

| x - a_{i} | + | y | = b_{i}, i \in N,

a_{i + 1} = a_{i} + \frac{3}{2} b_{i}

b_{i + 1} = \frac{b_{i}}{2}, a_{1} = 0, b_{1} = 32,

A_{i}

is the area bounded by

| x - a_{i} | + | y | = b_{i},

a_{i + 1} = a_{i} + \frac{3}{2} b_{i}

b_{i + 1} = \frac{b_{i}}{2};

a_{1} = 0, b_{1} = 32,

A_{i}

is the area bounded by

| x - a_{i} | + | y | = b_{i}, i \in N,

a_{i + 1} = a_{i} + \frac{3}{2} b_{i}

b_{i + 1} = \frac{b_{i}}{2}, a_{1} = 0, b_{1} = 32,

A_{i}

’ is the area bounded by bounded by

| x - a_{i} | + | y | = b_{i}, i ϵ N, w h e r e a_{i + 1} = a_{i} + \frac{3}{2} b_{i} a n d b_{i + 1} = \frac{b_{i}}{2}, a_{1} = 0, b_{1} = 32

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