IPMAT Indore 2022 · Quantitative Ability · Question 25

Step 1 — expand the determinant. For this cyclic matrix the standard result is

\det A=3abc-{a}^{3}-{b}^{3}-{c}^{3}

=-\left({a}^{3}+{b}^{3}+{c}^{3}-3abc \right)

Step 2 — factorise the bracket. The classic identity:

{a}^{3}+{b}^{3}+{c}^{3}-3abc=\left( a+b+c \right)\left({a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ca \right)

and the second factor can be rewritten as

\frac{1}{2}\left[{\left( a-b \right)}^{2}+{\left( b-c \right)}^{2}+{\left( c-a \right)}^{2}\right]

Step 3 — determine the sign. Since a>b>c>0:

a+b+c>0 ✔

the bracket of squares is strictly positive, because a, b, c are distinct, so at least one square is non-zero ✔

Therefore {a}^{3}+{b}^{3}+{c}^{3}-3abc>0, and

\det A=-\left( \text{positive}\right)<0

Step 4 — confirm numerically. Take a=3, b=2, c=1:

\det =3\left( 6 \right)-27-8-1=18-36=-18<0 ✓

Answer: Option 3 — is negative

Shortcut. Test one triple and trust the structure. With a,b,c=3,2,1 the determinant is -18. Since the conditions a>b>c>0 fix a single ordering pattern, the sign cannot flip — which rules out option (a) at once.

If you want the reason rather than the example: \det A=-\left({a}^{3}+{b}^{3}+{c}^{3}-3abc \right), and the bracket equals

1/2\left( a+b+c \right)\left[{\left( a-b \right)}^{2}+{\left( b-c \right)}^{2}+{\left( c-a \right)}^{2}\right]>0

for distinct positive a,b,c. The distinctness is what forces “negative” rather than “zero” — if a=b=c the determinant would vanish, which is why option (d) is offered.