JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Matrices

Subtracting the two $A^{\mathrm{T}}$ relations gives $A^{\mathrm{T}}(1,0,0{)}^{\mathrm{T}}=(5,2,2{)}^{\mathrm{T}}-(3,1,1{)}^{\mathrm{T}}=(2,1,1{)}^{\mathrm{T}}$, fixing the first row of $A$ as $(2,1,1)$, while $A^{\mathrm{T}}(0,0,1{)}^{\mathrm{T}}=(3,1,1{)}^{\mathrm{T}}$ gives the third row $(3,1,1)$. Likewise $A(1,0,0{)}^{\mathrm{T}}=(3,4,4{)}^{\mathrm{T}}-(1,3,1{)}^{\mathrm{T}}=(2,1,3{)}^{\mathrm{T}}$ is the first column and $A(0,0,1{)}^{\mathrm{T}}=(1,3,1{)}^{\mathrm{T}}$ the third column; the remaining entry is fixed by $\det (A)=1$ to be $2$, so $A=\left(\begin{array}{ccc}2& 1& 1\\ 1& 2& 3\\ 3& 1& 1\end{array}\right)$. Since $A^2+A=A(A+I)$, $\det (A^2+A)=\det (A)\det (A+I)=1\cdot 8=8$, and for a $3\times 3$ matrix $\det (\mathrm{adj}M)=(\det M{)}^2$. Therefore $\det (\mathrm{adj}(A^2+A))=8^2=64$.