UPSC CDS 2 2021 Math Paper

Let
‌

ay−bx

c

=‌

cx−az

b

=‌

bz−cy

a

=k ... (i)
In the given condition we know that a,b and c can not be zero.
From Eq. (i), we get
ay−bx=ck
cx−az=bk
bz−cy=ak
Multiplying Eq. (ii) by c, we get
acy −bcx=c2k
Multiplying Eq. (iii) by b, we get
bcx−abz=b2k
Multiplying Eq. (iv) by a, we get
abz−acy=a2k
Adding Eq. (v), Eq. (vi) and Eq. (vii),
we get
acy −bcx+bcx−abz+abz
−acy=k(a2+b2+c2)
⇒0=k(a2+b2+c2)
Since, a,b and c are not zero.
∴k=0
Put k=0 into Eq. (ii), Eq (iii) and Eq.
(iv), we get
*1) ay=bx⇒‌

x

a

=‌

y

b

*2) cx=az⇒‌

x

a

=‌

z

c

... (ix)
⋆3) bz=cy⇒‌

z

c

=‌

y

b

...(x)
Eq. (viii) says ‌

x

a

=‌

y

b

is correct.
Now,
From Eq. (ix), x=‌

z⋅a

c

and from Eq.
(x), y=‌

z⋅b

c

Put into ‌

x+y+z

a+b+c

, we get
⇒‌‌‌

‌ z⋅a c + z⋅b c +z

a+b+c

=‌

‌ z⋅a+z⋅b+z⋅c c

a+b+c

=‌

z(a+b+c)

c

⋅‌

1

(a+b+c)

=‌

z

c

Hence, both the Statements 1 and 2 are correct.