Matrices & DeterminantshardPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Matrix Statements with Cayley–Hamilton: Only (S2) Correct | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Let A=[121α]A=\begin{bmatrix}1&2\\1&\alpha\end{bmatrix} and B=[33β2]B=\begin{bmatrix}3&3\\\beta&2\end{bmatrix}. If A2−4A+I=OA^2-4A+I=O and B2−5B−6I=OB^2-5B-6I=O, then among the two statements: (S1): [(B−A)(B+A)]T=[1315710]\big[(B-A)(B+A)\big]^{T}=\begin{bmatrix}13&15\\7&10\end{bmatrix}, and (S2): det⁡(adj(A+B))=−5\det\big(\mathrm{adj}(A+B)\big)=-5,
Aonly (S1) is correct
Bonly (S2) is correctcorrect
Cboth (S1) and (S2) are correct
Dboth (S1) and (S2) are wrong
Solution
Step 1: By Cayley–Hamilton, A2−(tr A)A+(det⁡A)I=OA^2-(\mathrm{tr}\,A)A+(\det A)I=O. Comparing with A2−4A+I=OA^2-4A+I=O: tr A=4\mathrm{tr}\,A=4:
1+α=4 ⇒ α=3.1+\alpha=4\ \Rightarrow\ \alpha=3.
 Step 2: Similarly for BB: B2−(tr B)B+(det⁡B)I=OB^2-(\mathrm{tr}\,B)B+(\det B)I=O compared with B2−5B−6I=OB^2-5B-6I=O gives det⁡B=−6\det B=-6:
6−3β=−6 ⇒ β=4.6-3\beta=-6\ \Rightarrow\ \beta=4.
 Step 3: Now A=[1213]A=\begin{bmatrix}1&2\\1&3\end{bmatrix}, B=[3342]B=\begin{bmatrix}3&3\\4&2\end{bmatrix}. Then
B−A=[213−1],B+A=[4555].B-A=\begin{bmatrix}2&1\\3&-1\end{bmatrix},\qquad B+A=\begin{bmatrix}4&5\\5&5\end{bmatrix}.
(B−A)(B+A)=[1315710].(B-A)(B+A)=\begin{bmatrix}13&15\\7&10\end{bmatrix}.
 Step 4: Check (S1): its transpose is
[(B−A)(B+A)]T=[1371510]≠[1315710].\big[(B-A)(B+A)\big]^{T}=\begin{bmatrix}13&7\\15&10\end{bmatrix}\ne\begin{bmatrix}13&15\\7&10\end{bmatrix}.
 So (S1) is FALSE. Step 5: Check (S2): A+B=[4555]A+B=\begin{bmatrix}4&5\\5&5\end{bmatrix}, so
adj(A+B)=[5−5−54],det⁡(adj(A+B))=(5)(4)−(−5)(−5)=20−25=−5.\mathrm{adj}(A+B)=\begin{bmatrix}5&-5\\-5&4\end{bmatrix},\qquad \det\big(\mathrm{adj}(A+B)\big)=(5)(4)-(-5)(-5)=20-25=-5.
 So (S2) is TRUE. Step 6: Hence only (S2) is correct. Correct answer: (2)
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