JEE Main 2026 — Matrices Question with Solution

Let $C_1,C_2,C_3$ be the columns of $A$ and $R_1,R_2,R_3$ be the rows of $A$.

From the given equations, we have:
$A\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=C_3=\left[\begin{array}{c}1\\ 3\\ 1\end{array}\right]$

$A\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]=C_1+C_3=\left[\begin{array}{c}3\\ 4\\ 4\end{array}\right]\Rightarrow C_1=\left[\begin{array}{c}2\\ 1\\ 3\end{array}\right]$

Similarly, for the transpose $A^T$, the columns correspond to the rows of $A$:
$A^T\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=R_3^T=\left[\begin{array}{c}3\\ 1\\ 1\end{array}\right]\Rightarrow R_3=\left[\begin{array}{ccc}3& 1& 1\end{array}\right]$

$A^T\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]=R_1^T+R_3^T=\left[\begin{array}{c}5\\ 2\\ 2\end{array}\right]\Rightarrow R_1^T=\left[\begin{array}{c}2\\ 1\\ 1\end{array}\right]\Rightarrow R_1=\left[\begin{array}{ccc}2& 1& 1\end{array}\right]$

Using the elements of $C_1,C_3,R_1,$ and $R_3$, we can construct the matrix $A$ with an unknown central element $b$:
$A=\left[\begin{array}{ccc}2& 1& 1\\ 1& b& 3\\ 3& 1& 1\end{array}\right]$

We are given that $\det (A)=1$. Expanding along the first row:
$\det (A)=2(b-3)-1(1-9)+1(1-3b)=1$

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

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

Thus, the matrix $A$ is:
$A=\left[\begin{array}{ccc}2& 1& 1\\ 1& 2& 3\\ 3& 1& 1\end{array}\right]$

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

First, find $A+I$:
$A+I=\left[\begin{array}{ccc}3& 1& 1\\ 1& 3& 3\\ 3& 1& 2\end{array}\right]$

Now, calculate $\det (A+I)$:
$\det (A+I)=3(6-3)-1(2-9)+1(1-9)$

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

Since $\det (A)=1$, we have:
$\det (A^2+A)=1\times 8=8$

Finally, for a $3\times 3$ matrix:
$\det (\mathrm{adj}(A^2+A))=8^{3-1}=8^2=64$

Answer: $64$