Solved Subjective Question on Determinant Set 4

Posted on - 04-01-2017

IIT JEE

Example.1

If f, g, h are differentiable functions of x, and D(x) = , prove that D'(x) =

Solution

D(x) =

Operating (i) R₂ → R₂ – R₁ (ii) R₃ → R₃ –4 R₂ + 2R₁ and shifting x of R₂ to R₃

D (x) =

&⇒ D¢ (x) = 0 +0 +. Hence Proved

Example.2

Prove without expansion that .

Solution

Rewriting the given determinant as under

by operating in second determinant C₃→ C₃ + bC₂ , we get

=D₁ - =D₁ +

==.

Example.3

If A, B, C are angles of a triangles then prove that = – 4

Solution

Since A + B + C = p and eⁱ^p = cosp + i sinp = -1,

e^i(B+C) = eⁱ⁽^p^-
A) = – e^-iA

e ^–i(B+C) = –e^iA

By taking e^iA , e ^iB, e^iC common from R₁ , R₂ and R₃ respectively, we have

D = -

By taking e^iA, e ^iB, e^iC common from C₁, C₂ and C₃ respectively, we have

D = = – 4.

Example.4

Consider a determinant of order three whose all elements are 1 or –1, prove that maximum value of the determinant is 4. .

Solution

Let D = be a determinant of order 3,

Where a₁, a₂, a₃, b₁, b₂, b₃, c₁, c₂, c₃ are 1 or –1

Now D = a₁(b₂c₃–b₃c₂) – a₂(b₁c₃– b₃c₁) + a₃(b₁c₂– b₂c₁)

= (a₁b₂c₃+ a₂b₃c₁+ a₃b₁c₂) – (a₁b₃c₂+ a₂b₁c₃ + a₃b₂c₁)

value of these six terms will be either –1or 1 \ D £ 6 …(1)

Since we have to find the largest value of D, therefore first of all we consider the case when a₁b₂c₃ = a₂b₃c₁= a₃b₁ c₂ = 1

&⇒ a₁a₂a₃b₁b₂b₃c₁c₂c₃ = 1×1×1 = 1 …(2)

Let x = a₁b₃c₂ y= a₂b₁c₃,z = a₃b₂c₁,

Then xyz = a₁a₂a₃b₁b₂b₃c₁c₂c₃ =1 [form (2)]

But x = 1 or –1, y = 1 or –1, z = 1 or –1

\ all of x, y,z cannot be –1

Two of x,y,z are 1 and third is –1 is also not possible

\ either x = y = z = 1 or two of x, y, z are – 1 and other is 1

when x = y = z = 1, D = 0

When two of x, y, z are –1 and third is 1, D = 3 – (–1 –1+1) = 4

This is possible Let a₁ = 1, b₃ = 1, c₂ = –1, a₂= 1, b₁ = –1, a₃ = b₂ = c₁ =1

In this case D = = 4 \ max value of D = 4.

Example.5

Prove that = x₁ x₂ x₃.

Solution

The given determinant can be written in two determinants as under

Now first can be expanded along C₁ and by taking common b₁ from C₁ in the second determinant and applying C₂→ C₂ – b₂C₁ and
C₃ → (C₃ – b₃C₁), we can get the result. .

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