JEE Main 2026 — Mathematics Algebra

The determinant of matrix $A$A is given by:

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

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

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

Taking the inverse on both sides:

$$(\text{adj}A)^{-1} = \dfrac{A}{|A|} = \dfrac{A}{-4}$$(adjA)−1=∣A∣A​=−4A​

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

$$B = 2\left(\dfrac{A}{-4}\right) = -\dfrac{1}{2}A$$B=2(−4A​)=−21​A

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

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

Now, we calculate the determinant of $B$B:

$$|B| = \left|-\dfrac{1}{2}A\right| = \left(-\dfrac{1}{2}\right)^3|A| = -\dfrac{1}{8}(-4) = \dfrac{1}{2}$$∣B∣=​−21​A​=(−21​)3∣A∣=−81​(−4)=21​

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

$$\text{adj}(\text{adj}B) = \dfrac{1}{2}B = \dfrac{1}{2}\left(-\dfrac{1}{2}A\right) = -\dfrac{1}{4}A$$adj(adjB)=21​B=21​(−21​A)=−41​A

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

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

Therefore, the sum of all elements of the matrix $-\dfrac{1}{4}A$−41​A is:

$$-\dfrac{1}{4} \times 12 = -3$$−41​×12=−3

Answer: $-3$−3