JEE Main 2026 — Matrices Question with Solution

The determinant of matrix $A$ is given by:

$|A|=1(0-3)-1(-10-1)+2(-6-0)$

$|A|=-3+11-12=-4$

We know the property $A(\text{adj}A)=|A|I$, which gives $\text{adj}A=|A|A^{-1}$.

Taking the inverse on both sides:

$(\text{adj}A{)}^{-1}=\frac{A}{|A|}=\frac{A}{-4}$

Let $B=2(\text{adj}A{)}^{-1}$. Substituting the above expression, we get:

$B=2\left(\frac{A}{-4}\right)=-\frac{1}{2}A$

We need to find the matrix $\text{adj}(\text{adj}B)$. For a square matrix of order $n$, we have $\text{adj}(\text{adj}B)=|B{|}^{n-2}B$. Since the order is $n=3$:

$\text{adj}(\text{adj}B)=|B{|}^{3-2}B=|B|B$

Now, we calculate the determinant of $B$:

$|B|=\left|-\frac{1}{2}A\right|={\left(-\frac{1}{2}\right)}^3|A|=-\frac{1}{8}(-4)=\frac{1}{2}$

Substituting $|B|$ back into the expression for $\text{adj}(\text{adj}B)$:

$\text{adj}(\text{adj}B)=\frac{1}{2}B=\frac{1}{2}\left(-\frac{1}{2}A\right)=-\frac{1}{4}A$

The sum of all elements of matrix $A$ is:

$1+1+2-2+0+1+1+3+5=12$

Therefore, the sum of all elements of the matrix $-\frac{1}{4}A$ is:

$-\frac{1}{4}\times 12=-3$

Answer: $-3$