MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Derivative of a Determinant

If $ fy= begi...

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.

Open in App

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

Suggest Corrections

thumbs-up

0

Similar questions

Y = $(x y + y z)^{2} - 2 x^{2} y^{2} z$
When $x = - 1, y = 1$ and $z = 2$ , |Y| = ___

\frac{a}{3 y} + \frac{3 b}{x} = 7

a + 1 = 2 b + 1 = x = 5,

find the value of

^{'} y^{'} .

Q. If the ordered pairs (y-1,x+2) and (x-2,2y+1) are equal, find x and y.

∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 & 1 + x & 1 1 & 1 & 1 + y \end{matrix} ∣ ∣ ∣ ∣ =

The line through P (4, 3) intersects y – axis at Q.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Derivative of a Determinant

Standard XI Mathematics

Join BYJU'S Learning Program

CrossIcon