IPMAT Indore 2023 · Quantitative Ability · Question 26

Step 1 — factor the matrix expression. Pull A out on the left:

{A}^{3}-3{A}^{2}-5A=A\left({A}^{2}-3A-5I \right)

Step 2 — use multiplicativity of the determinant.

\det \left({A}^{3}-3{A}^{2}-5A \right)=\det \left( A \right)\cdot \det \left({A}^{2}-3A-5I \right)=0

A product is zero when either factor is zero — so it is enough to make \det \left( A \right)=0.

Step 3 — the easy branch.

\det \left( A \right)=1\cdot a-2\cdot 3=a-6=0 \Rightarrow a=6

Since 6 is among the options, we are done. But it is worth confirming no other option works.

Step 4 — the other branch, for completeness. By the Cayley–Hamilton theorem, {A}^{2}=\left( 1+a \right)A-\left( a-6 \right)I, so

{A}^{2}-3A-5I=\left( a-2 \right)A-\left( a-1 \right)I

Taking the determinant of this 2\times 2 matrix gives

-7{a}^{2}+27a-25=0 \Rightarrow 7{a}^{2}-27a+25=0

whose discriminant is 729-700=29 — not a perfect square, so its roots are irrational and none of 4, 5 or 1 qualifies.

The complete factorisation is

\det \left({A}^{3}-3{A}^{2}-5A \right)=-\left( a-6 \right)\left( 7{a}^{2}-27a+25 \right)

confirming a=6 is the only option that works.

Answer: Option 3 — 6

Shortcut. Factor out A and stop. \det \left( A \right)\cdot \det \left( \cdots \right)=0 is satisfied the moment \det \left( A \right)=0, and that is a one-line condition:

a-6=0 \Rightarrow a=6

Do not expand {A}^{3}. Cubing a 2\times 2 matrix symbolically is several minutes of error-prone work for an answer you can read off in ten seconds.

The structural lesson: whenever a determinant condition involves a polynomial in A with no constant term, factor A out first. The question is then almost always answered by the simplest branch, \det \left( A \right)=0 — and the options confirm it, since only one of them is a root of anything rational.