Matrices & DeterminantshardPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

det(adj(A²+A)) = 64 | JEE Main 2026

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

Question
Let AA be a 3×33\times3 matrix such that AT[101]=[522]A^T\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}5\\2\\2\end{bmatrix}, AT[001]=[311]A^T\begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}3\\1\\1\end{bmatrix}, A[101]=[344]A\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}3\\4\\4\end{bmatrix} and A[001]=[131]A\begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}1\\3\\1\end{bmatrix}. If det⁡(A)=1\det(A)=1, then det⁡(adj(A2+A))\det\big(\mathrm{adj}(A^2+A)\big) is equal to
A1616
B2525
C4949
D6464correct
Solution
Step 1: Ae3A\mathbf e_3 is column 3, ATe3A^T\mathbf e_3 is row 3: col3=(1,3,1)T\text{col}_3=(1,3,1)^T, row3=(3,1,1)\text{row}_3=(3,1,1). A[101]=col1+col3=(3,4,4)T⇒col1=(3,4,4)T−(1,3,1)T=(2,1,3)TA\begin{bmatrix}1\\0\\1\end{bmatrix}=\text{col}_1+\text{col}_3=(3,4,4)^T\Rightarrow\text{col}_1=(3,4,4)^T-(1,3,1)^T=(2,1,3)^T. AT[101]=row1+row3=(5,2,2)⇒row1=(5,2,2)−(3,1,1)=(2,1,1)A^T\begin{bmatrix}1\\0\\1\end{bmatrix}=\text{row}_1+\text{row}_3=(5,2,2)\Rightarrow\text{row}_1=(5,2,2)-(3,1,1)=(2,1,1). With det⁡A=1\det A=1 fixing the free entry:
A=[211123311],det⁡A=2(2−3)−1(1−9)+1(1−6)=−2+8−5=1.A=\begin{bmatrix}2&1&1\\1&2&3\\3&1&1\end{bmatrix},\qquad \det A=2(2-3)-1(1-9)+1(1-6)=-2+8-5=1.
 Step 2: A2=[856138101067]⇒A2+A=[10671410131378]A^2=\begin{bmatrix}8&5&6\\13&8&10\\10&6&7\end{bmatrix}\Rightarrow A^2+A=\begin{bmatrix}10&6&7\\14&10&13\\13&7&8\end{bmatrix}.
det⁡(A2+A)=10(80−91)−6(112−169)+7(98−130)=10(−11)−6(−57)+7(−32)\det(A^2+A)=10(80-91)-6(112-169)+7(98-130)=10(-11)-6(-57)+7(-32)
 =−110+342−224=8.=-110+342-224=8. Step 3: det⁡(adj M)=(det⁡M)n−1\det(\mathrm{adj}\,M)=(\det M)^{n-1}, n=3n=3:
det⁡(adj(A2+A))=82=64.\det\big(\mathrm{adj}(A^2+A)\big)=8^2=64.
 Correct answer: (4)
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