JEE Main 2026 — Mathematics Algebra

Let $C_1, C_2, C_3$C1​,C2​,C3​ be the columns of $A$A and $R_1, R_2, R_3$R1​,R2​,R3​ be the rows of $A$A.

From the given equations, we have:

$$A \begin{bmatrix}0\\0\\1\end{bmatrix} = C_3 = \begin{bmatrix}1\\3\\1\end{bmatrix}$$A​001​​=C3​=​131​​

$$A \begin{bmatrix}1\\0\\1\end{bmatrix} = C_1 + C_3 = \begin{bmatrix}3\\4\\4\end{bmatrix} \Rightarrow C_1 = \begin{bmatrix}2\\1\\3\end{bmatrix}$$A​101​​=C1​+C3​=​344​​⇒C1​=​213​​

Similarly, for the transpose $A^T$AT, the columns correspond to the rows of $A$A:

$$A^T \begin{bmatrix}0\\0\\1\end{bmatrix} = R_3^T = \begin{bmatrix}3\\1\\1\end{bmatrix} \Rightarrow R_3 = \begin{bmatrix}3 & 1 & 1\end{bmatrix}$$AT​001​​=R3T​=​311​​⇒R3​=[3​1​1​]

$$A^T \begin{bmatrix}1\\0\\1\end{bmatrix} = R_1^T + R_3^T = \begin{bmatrix}5\\2\\2\end{bmatrix} \Rightarrow R_1^T = \begin{bmatrix}2\\1\\1\end{bmatrix} \Rightarrow R_1 = \begin{bmatrix}2 & 1 & 1\end{bmatrix}$$AT​101​​=R1T​+R3T​=​522​​⇒R1T​=​211​​⇒R1​=[2​1​1​]

Using the elements of $C_1, C_3, R_1,$C1​,C3​,R1​, and $R_3$R3​, we can construct the matrix $A$A with an unknown central element $b$b:

$$A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & b & 3 \\ 3 & 1 & 1 \end{bmatrix}$$A=​213​1b1​131​​

We are given that $\det(A) = 1$det(A)=1. Expanding along the first row:

$$\det(A) = 2(b - 3) - 1(1 - 9) + 1(1 - 3b) = 1$$det(A)=2(b−3)−1(1−9)+1(1−3b)=1

$$2b - 6 + 8 + 1 - 3b = 1$$2b−6+8+1−3b=1

$$3 - b = 1 \Rightarrow b = 2$$3−b=1⇒b=2

Thus, the matrix $A$A is:

$$A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$$A=​213​121​131​​

We need to find $\det(\operatorname{adj}(A^2 + A))$det(adj(A2+A)). Using the properties of determinants and adjoints:

$$\det(\operatorname{adj}(A^2 + A)) = (\det(A^2 + A))^2 = (\det(A) \cdot \det(A + I))^2$$det(adj(A2+A))=(det(A2+A))2=(det(A)⋅det(A+I))2

First, find $A + I$A+I:

$$A + I = \begin{bmatrix} 3 & 1 & 1 \\ 1 & 3 & 3 \\ 3 & 1 & 2 \end{bmatrix}$$A+I=​313​131​132​​

Now, calculate $\det(A + I)$det(A+I):

$$\det(A + I) = 3(6 - 3) - 1(2 - 9) + 1(1 - 9)$$det(A+I)=3(6−3)−1(2−9)+1(1−9)

$$\det(A + I) = 3(3) - 1(-7) + 1(-8) = 9 + 7 - 8 = 8$$det(A+I)=3(3)−1(−7)+1(−8)=9+7−8=8

Since $\det(A) = 1$det(A)=1, we have:

$$\det(A^2 + A) = 1 \times 8 = 8$$det(A2+A)=1×8=8

Finally, for a $3 \times 3$3×3 matrix:

$$\det(\operatorname{adj}(A^2 + A)) = 8^{3-1} = 8^2 = 64$$det(adj(A2+A))=83−1=82=64

Answer: $64$64