Determinant problem 3 Nov

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!!

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