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

D(x) =

Operating (i) R_{2}
→ R_{2}
– R_{1} (ii) R_{3} → R_{3} –4 R_{2}
+ 2R_{1} and shifting x of R_{2} to R_{3}

D (x) =

&⇒ D¢ (x) = 0 +0 +. Hence Proved

Prove without expansion that .

Rewriting the given determinant as under

by operating in second determinant C_{3}→ C_{3}
+ bC_{2} , we get

=D_{1} - =D_{1} +

==.

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

Since A + B + C = p and e^{i}^{p} =
cosp +
i sinp =
-1,

e^{i(B+C)} = e^{i(}^{p}^{ -
A)}
= – e^{-iA}

e ^{–i(B+C)} = –e^{iA}

By taking e^{iA} ,
e ^{iB}, e^{iC} common from R_{1} , R_{2}
and R_{3} respectively, we have

D = -

By taking e^{iA}, e
^{iB}, e^{iC} common from C_{1}, C_{2} and C_{3}
respectively, we have

D = = – 4.

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

Let D = be a determinant of order 3,

Where a_{1}, a_{2}, a_{3},
b_{1}, b_{2}, b_{3}, c_{1}, c_{2}, c_{3}
are 1 or –1

Now D
= a_{1}(b_{2}c_{3}–b_{3}c_{2}) – a_{2}(b_{1}c_{3
}– b_{3}c_{1}) + a_{3}(b_{1}c_{2 }–
b_{2}c_{1})

= (a_{1}b_{2}c_{3 }+
a_{2}b_{3}c_{1 }+ a_{3}b_{1}c_{2})
– (a_{1}b_{3}c_{2 }+ a_{2}b_{1}c_{3}
+ a_{3}b_{2}c_{1})

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 a_{1}b_{2}c_{3} = a_{2}b_{3}c_{1}=
a_{3}b_{1} c_{2} = 1

&⇒
a_{1}a_{2}a_{3}b_{1}b_{2}b_{3}c_{1}c_{2}c_{3}
= 1×1×1 = 1 …(2)

Let x = a_{1}b_{3}c_{2}
y= a_{2}b_{1}c_{3},z = a_{3}b_{2}c_{1},

Then xyz = a_{1}a_{2}a_{3}b_{1}b_{2}b_{3}c_{1}c_{2}c_{3}
=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 a_{1} = 1, b_{3}
= 1, c_{2} = –1, a_{2}= 1, b_{1} = –1, a_{3} =
b_{2} = c_{1} =1

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

Prove
that = x_{1} x_{2} x_{3}.

The given determinant can be written in two determinants as under

Now first can be expanded along C_{1} and by
taking common b_{1} from C_{1} in the second determinant and
applying C_{2 }→ C_{2} – b_{2}C_{1}
and

C_{3} → (C_{3} – b_{3 }C_{1}), we can
get the result. .