NVAHardJEE 2026Inverse & Adjoint of a Matrix

JEE Mathematics 2026 Question with Solution

For some α,βR\alpha, \beta \in R, let A=(α212)A = \begin{pmatrix} \alpha & 2 \\ 1 & 2 \end{pmatrix} and B=(11β1)B = \begin{pmatrix} 1 & 1 \\ \beta & 1 \end{pmatrix} be such that A24A+2IB23B+I=OA^2-4A+2I-B^2-3B+I=O. Then (det(adj(A3B3)))2(\det(adj(A^3-B^3)))^2 is equal to:

Answer

Correct answer:50625

Step-by-step solution

Standard Method

Given:

  • A=(α212)A = \begin{pmatrix} \alpha & 2 \\ 1 & 2 \end{pmatrix}
  • B=(11β1)B = \begin{pmatrix} 1 & 1 \\ \beta & 1 \end{pmatrix}
  • The stated matrix relation is interpreted in the solution as
A24A+3I=B2+3BA^2 - 4A + 3I = B^2 + 3B

Find: (det(adj(A3B3)))2(\det(\operatorname{adj}(A^3-B^3)))^2.

The extracted solution itself states that the question is flawed and leads to a contradiction, so the published numerical answer is obtained only after assuming an intended corrected problem.

Compute the left-hand side:

A2=(α2+22α+4α+26)A^2 = \begin{pmatrix} \alpha^2+2 & 2\alpha+4 \\ \alpha+2 & 6 \end{pmatrix}

Hence,

A24A+3I=(α24α+52α4α21)A^2 - 4A + 3I = \begin{pmatrix} \alpha^2-4\alpha+5 & 2\alpha-4 \\ \alpha-2 & 1 \end{pmatrix}

Compute the right-hand side:

B2=(1+β22ββ+1)B^2 = \begin{pmatrix} 1+\beta & 2 \\ 2\beta & \beta+1 \end{pmatrix}

So,

B2+3B=(β+455ββ+4)B^2 + 3B = \begin{pmatrix} \beta+4 & 5 \\ 5\beta & \beta+4 \end{pmatrix}

Equating corresponding entries gives:

1=β+4β=31 = \beta + 4 \Rightarrow \beta = -3

and

2α4=52α=9α=922\alpha - 4 = 5 \Rightarrow 2\alpha = 9 \Rightarrow \alpha = \frac{9}{2}

Now check the (2,1)(2,1) entry:

α2=922=52\alpha - 2 = \frac{9}{2} - 2 = \frac{5}{2}

while

5β=5(3)=155\beta = 5(-3) = -15

Since

5215\frac{5}{2} \ne -15

the given relation is inconsistent.

The published solution therefore assumes an intended corrected result:

A3B3=15IA^3 - B^3 = 15I

Let

M=A3B3=15I=(150015)M = A^3 - B^3 = 15I = \begin{pmatrix} 15 & 0 \\ 0 & 15 \end{pmatrix}

Then

det(M)=15×15=225\det(M) = 15 \times 15 = 225

For a 2×22 \times 2 matrix,

det(adj(M))=det(M)\det(\operatorname{adj}(M)) = \det(M)

Therefore,

det(adj(M))=225\det(\operatorname{adj}(M)) = 225

and the required value is

(225)2=50625(225)^2 = 50625

So, using the assumed intended correction from the published solution, the answer is 5062550625.

Why the Published Answer Requires an Assumption

Given: the extracted solution explicitly notes that the original question statement is contradictory.

Find: why the answer 5062550625 cannot be obtained directly from the printed relation.

The source question text has

A24A+2IB23B+I=OA^2-4A+2I-B^2-3B+I=O

which simplifies to

A24A+3I=B2+3BA^2-4A+3I=B^2+3B

Using the displayed forms of AA and BB, entrywise comparison gives incompatible values. That means there is no real pair α,β\alpha, \beta satisfying all entries simultaneously.

Because of this contradiction, the numerical result 5062550625 is not derivable from the printed equation alone. The extracted solution works backward from the official answer and assumes the intended outcome was

A3B3=15IA^3-B^3=15I

Under that assumption,

det(A3B3)=det(15I)=152=225\det(A^3-B^3)=\det(15I)=15^2=225

and for a 2×22 \times 2 matrix,

det(adj(M))=det(M)\det(\operatorname{adj}(M))=\det(M)

Hence,

(det(adj(A3B3)))2=(225)2=50625(\det(\operatorname{adj}(A^3-B^3)))^2=(225)^2=50625

Common mistakes

  • Using the printed matrix equation without checking consistency is a mistake. Entrywise comparison gives contradictory conditions on α\alpha and β\beta. Always verify all four entries before proceeding to determinant calculations.

  • Applying det(adj(M))=(detM)n1\det(\operatorname{adj}(M)) = (\det M)^{n-1} incorrectly is a common error. For a 2×22 \times 2 matrix, this becomes det(adj(M))=det(M)\det(\operatorname{adj}(M)) = \det(M), not det(M)2\det(M)^2.

  • Confusing det(kI)\det(kI) with kk is incorrect. For a 2×22 \times 2 matrix, det(kI)=k2\det(kI)=k^2. So if M=15IM=15I, then det(M)=225\det(M)=225, not 1515.

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