Question

A= |(x^k,2,2),(1,x^k,2),(1,1,x^k)| and B= |(x,2),(1,x)|

Value of k for which, (d|A|)/dx = 3|B|

a)" "3

b)" "2

c)" "1

a)" "0

Solution

We can solve this using more than 1 method.

Method 1

We differentiate LHS before solving for determinant. For that we need to differentiate 1 row at a time.

Given equation, (d|A|)/dx = 3|B|

In LHS

(d|A|)/dx = (d/dx)|(x^k,2,2),(1,x^k,2),(1,1,x^k)|

=> (d|A|)/dx = |(kx^(k-1),0,0),(1,x^k,2),(1,1,x^k)| + |(x^k,2,2),(0,kx^(k-1),0),(1,1,x^k)| + |(x^k,2,2),(1,x^k,2),(0,0,kx^(k-1))|

Now solving for determinant,

=> (d|A|)/dx = kx^(k-1)(x^2k - 2) + kx^(k-1)(x^2k - 2) + kx^(k-1)(x^2k - 2)

=> (d|A|)/dx = 3kx^(k-1)(x^2k - 2)

Similary for RHS

3B = 3|(x,2),(1,x)| = 3(x^2 - 2)

Comparing LHS and RHS, we can find the value of k= 1

Method 2

We can solve for determinant 1st and then differentiate.

A= |(x^k,2,2),(1,x^k,2),(1,1,x^k)| = x^(3k) -6x^k + 6

=> (d|A|)/dx = 3kx^(3k-1) - 6kx^(k-1)

Similarly, 3B = 3|(x,2),(1,x)| = 3(x^2 - 2)

Comparing LHS and RHS, we can find the value of k= 1