JEE Mathematics 2025 Question with Solution

Given: $A$ is a square matrix of order $3$, $\det(A) = -2$, and

$$\det(3 \cdot \operatorname{adj}(-6 \cdot \operatorname{adj}(3A))) = 2^{m+n} \cdot 3^{mn}$$

Find: $4m + 2n$

Use the facts for $3 \times 3$ matrices:

• $\det(kM) = k^3 \det(M)$
• $\det(\operatorname{adj} M) = \det(M)^2$
• $\operatorname{adj}(kM) = k^2 \operatorname{adj}(M)$

Start with $3A$:

$$\det(3A) = 3^3 \det(A) = 27 \cdot (-2) = -54$$

Now,

$$\det(\operatorname{adj}(3A)) = \det(3A)^2 = (-54)^2 = 54^2 = 2916$$

Let

$$D = -6 \cdot \operatorname{adj}(3A)$$

Then

$$\det(D) = (-6)^3 \det(\operatorname{adj}(3A)) = (-216) \cdot 2916$$

Write this in prime powers:

$$-216 = -2^3 3^3, \qquad 2916 = 2^2 3^6$$

So,

$$\det(D) = -2^{3+2} 3^{3+6} = -2^5 3^9$$

Now take adjugate again:

$$\det(\operatorname{adj}(D)) = \det(D)^2 = (2^5 3^9)^2 = 2^{10} 3^{18}$$

Finally,

$$\det(3 \cdot \operatorname{adj}(D)) = 3^3 \det(\operatorname{adj}(D)) = 3^3 \cdot 2^{10} 3^{18} = 2^{10} 3^{21}$$

Comparing with

$$2^{m+n} \cdot 3^{mn}$$

we get

$$m + n = 10, \qquad mn = 21$$

Since $m > n$, the pair is

$$m = 7, \qquad n = 3$$

Therefore,

$$4m + 2n = 4 \cdot 7 + 2 \cdot 3 = 28 + 6 = 34$$

So the required numerical value is $34$.

Given: $\det(A) = -2$ for a matrix of order $3$.

Find: Evaluate $4m+2n$ from

$$\det(3 \cdot \operatorname{adj}(-6 \cdot \operatorname{adj}(3A))) = 2^{m+n} \cdot 3^{mn}$$

For an $n \times n$ matrix with $n=3$:

$$\det(\operatorname{adj}(M)) = \det(M)^{n-1} = \det(M)^2$$

and for scalar multiplication,

$$\det(kM) = k^3 \det(M)$$

First,

$$\det(3A) = 3^3 \det(A) = 27(-2) = -54$$

Hence,

$$\det(\operatorname{adj}(3A)) = (-54)^2$$

Now multiply by $-6$:

$$\det(-6 \cdot \operatorname{adj}(3A)) = (-6)^3 \cdot (-54)^2$$

Factor each term:

$$(-6)^3 = -2^3 3^3, \qquad (-54)^2 = (2 \cdot 3^3)^2 = 2^2 3^6$$

Therefore,

$$\det(-6 \cdot \operatorname{adj}(3A)) = -2^5 3^9$$

Taking adjugate,

$$\det(\operatorname{adj}(-6 \cdot \operatorname{adj}(3A))) = (-2^5 3^9)^2 = 2^{10} 3^{18}$$

Multiplying the matrix by $3$ outside gives

$$\det(3 \cdot \operatorname{adj}(-6 \cdot \operatorname{adj}(3A))) = 3^3 \cdot 2^{10} 3^{18} = 2^{10} 3^{21}$$

Now compare exponents with

$$2^{m+n} 3^{mn}$$

So,

$$m+n=10, \qquad mn=21$$

The integers satisfying these are $7$ and $3$, and since $m>n$,

$$m=7, \qquad n=3$$

Thus,

$$4m+2n = 4(7)+2(3)=34$$

Therefore, the answer is $34$.