If f-y- -1-y-a-1b-1 amp-1-y-a-1b-2 amp-1-y-a-1b-3 -1-y-a-2b-1 amp-1-y-a-2b-2 amp-1-y-a-2b-3 -1-y-a-3b-1 amp-1-y-a-3b-2 amp-1-y-a-3b-3 and a- b are even for i-1- 2- 3- then find f-2-

If nbsp;f'(y)= a_1b_1(1 y)^a_1b_1 1 amp; a_2b_2(1 y)^a_1b_2 1 amp; a_3b_3(1 y)^a_1b_3 1 (1 y)^a_2b_1 amp;(1 y)^a_2b_2 amp;(1 y)^a_2b_3 (1 y)^a_3b_1 ...

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Byju's Answer

Standard XII

Mathematics

Derivative of a Determinant

If f(y)= (1-y...

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 $a, b$ are even for $i = 1, 2, 3$ . then find $f^{'} (2)$

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Solution

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

$f^{'} (2) = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} b_{1} & a_{1} b_{2} & a_{1} b_{3} 1 & 1 & 1 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a_{2} b_{1} & a_{2} b_{2} & a_{2} b_{3} 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 & 1 & 1 a_{3} b_{1} & a_{3} b_{2} & a_{3} b_{3} \end{matrix} ∣ ∣ ∣ ∣$

$f^{'} (2) = 0 + 0 + 0 = 0$

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Similar questions

Let S = {a, b, c} and T = {1, 2, 3}. Find F⁻¹ of the following functions F from S to T, if it exists.

(i) F = {(a, 3), (b, 2), (c, 1)} (ii) F = {(a, 2), (b, 1), (c, 1)}

Q. Consider f : {1, 2, 3} → { a , b , c } given by f (1) = a , f (2) = b and f (3) = c . Find f −1 and show that ( f −1 ) −1 = f .

Q. 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, b

are even for

i = 1, 2, 3

. then find

f^{'} (2)

Q. Let

S = {a, b, c}

and

T = {1, 2, 3}

. Find

F^{- 1}

of the following function F from S to T, if it exists.

(i) F = {(a, 3), (b, 2), (c, 1)}

Express the given formula in term of x.

$\frac{\frac{1}{x} + \frac{1}{y}}{\frac{1}{f}} = (2 + 4 t)$

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If f(y)=∣∣ ∣ ∣∣(1−y)a1b1(1−y)a1b2(1−y)a1b3(1−y)a2b1(1−y)a2b2(1−y)a2b3(1−y)a3b1(1−y)a3b2(1−y)a3b3∣∣ ∣ ∣∣ and a,b are even for i=1,2,3. then find f′(2)