Maths Q6 – JEE-MAIN 2026

Let

A = [\begin{matrix} 1 & 1 & 2 \\ - 2 & 0 & 1 \\ 1 & 3 & 5 \end{matrix}]

. Then the sum of all elements of the matrix

a d j (a d j (2 (a d j A^{- 1})))

is equal to:

Simplify the given expression using properties of adjoint and inverse of a matrix.

● Video Walkthrough

Step 1: Simplify the matrix expression using adjoint properties✦ Active

Let the given expression be $P = a d j (a d j (2 (a d j A^{- 1})))$ . For a $3 \times 3$ matrix ( $n = 3$ ), we use the following properties:

a d j (a d j (M)) = | M |^{n - 2} M = | M | M

| k M | = k^{n} | M |

a d j (M^{- 1}) = (a d j M)^{- 1}

(a d j M)^{- 1} = \frac{1}{| M |} M

| a d j M | = | M |^{n - 1} = | M |^{2}

Applying these properties sequentially to $P$ :

P = | 2 (a d j A^{- 1}) | \cdot (2 (a d j A^{- 1}))

P = (2^{3} | a d j A^{- 1} |) \cdot (2 (a d j A^{- 1})) = 16 | a d j A^{- 1} | a d j A^{- 1}

P = 16 | (a d j A)^{- 1} | (a d j A)^{- 1}

P = 16 \frac{1}{| a d j A |} (a d j A)^{- 1}

P = 16 \frac{1}{| a d j A |} (\frac{1}{| A |} A)

P = 16 \frac{1}{| A |^{2}} \frac{1}{| A |} A = 16 \frac{1}{| A |^{3}} A

Step 2: Calculate the determinant of matrix A○ Expand

Given $A = [\begin{matrix} 1 & 1 & 2 \\ - 2 & 0 & 1 \\ 1 & 3 & 5 \end{matrix}]$ . The determinant $| A |$ is calculated as:

| A | = 1 (0 \cdot 5 - 1 \cdot 3) - 1 (- 2 \cdot 5 - 1 \cdot 1) + 2 (- 2 \cdot 3 - 0 \cdot 1)

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

| A | = - 3 - 1 (- 11) + 2 (- 6)

| A | = - 3 + 11 - 12 = 8 - 12 = - 4

💡 Teacher's Secret Hint

Double-check determinant calculations to avoid arithmetic errors.

Step 3: Substitute the determinant value and find the sum of elements○ Expand

Substitute $| A | = - 4$ into the simplified expression for $P$ from Step 1:

P = 16 \frac{1}{(- 4)^{3}} A = 16 \frac{1}{- 64} A = - \frac{1}{4} A

Now, find the sum of all elements of matrix $A$ :

Sum of elements of A = (1 + 1 + 2) + (- 2 + 0 + 1) + (1 + 3 + 5)

= 4 + (- 1) + 9 = 3 + 9 = 12

The sum of all elements of matrix $P$ is then:

Sum of elements of P = - \frac{1}{4} \times (Sum of elements of A)

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

💡 Teacher's Secret Hint

Ensure all elements, including negative ones, are correctly summed for matrix A.

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Maths Question 6 – JEE-MAIN 2026