Let A be a matrix of order 3 3 and |A| = 5. If |2 adj(3A adj(2A))| = 2^ 3^ 5^,,, N then + + is equal to

Given: A is a matrix of order 3 3 and |A| = 5. Find: The value of + + if |2 adj(3A adj(2A))| = 2^ 3^ 5^ Use the identities | adj(M)| = |M|^n-1 for an n n matrix, and |kM| = k^n |M| for scalar k. Since the matrices are of order 3 3, |2 adj(3A adj(2A))| = 2^3 | adj(3A adj(2A))| Also, | adj(3A adj(2A))| = |3A adj(2A)|^2 Therefore, |2 adj(3A adj(2A))| = 2^3 |3A adj(2A)|^2 Now simplify the inner determinant: |3A adj(2A)| = |3A| | adj(2A)| For a 3 3 matrix, |3A| = 3^3 |A| and | adj(2A)| = |2A|^2 Also, |2A| = 2^3 |A| Hence, | adj(2A)| = (2^3 |A|)^2 = 2^6 |A|^2 Substitute into the expression: |3A adj(2A)| = 3^3 |A| 2^6 |A|^2 = 2^6 3^3 |A|^3 So, |2 adj(3A adj(2A))| = 2^3 (2^6 3^3 |A|^3)^2 = 2^3 2^12 3^6 |A|^6 = 2^15 3^6 |A|^6 Using |A| = 5, |2 adj(3A adj(2A))| = 2^15 3^6 5^6 Thus, = 15, = 6, = 6 Therefore, + + = 15 + 6 + 6 = 27 So, the correct option is C.

Let A be a matrix of order 3 3 and |A| = 5. If |2 adj(3A adj(2A))… — Solution | JEE Mathematics

MCQMediumJEE 2025Inverse & Adjoint of a Matrix

Let $A$ be a matrix of order $3 \times 3$ and $|A| = 5$ . If $|2 \, \text{adj}(3A \, \text{adj}(2A))| = 2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma}, \alpha, \beta, \gamma \in \mathbb{N}$ then $\alpha + \beta + \gamma$ is equal to

A
$25$
B
$26$
C
$27$
D
$28$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: $A$ is a matrix of order $3 \times 3$ and $|A| = 5$ .

Find: The value of $\alpha + \beta + \gamma$ if

|2 \, \text{adj}(3A \, \text{adj}(2A))| = 2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma}

Use the identities

|\text{adj}(M)| = |M|^{n-1}

for an $n \times n$ matrix, and

|kM| = k^n |M|

for scalar $k$ .

Since the matrices are of order $3 \times 3$ ,

|2 \, \text{adj}(3A \, \text{adj}(2A))| = 2^3 \cdot |\text{adj}(3A \, \text{adj}(2A))|

Also,

|\text{adj}(3A \, \text{adj}(2A))| = |3A \, \text{adj}(2A)|^2

Therefore,

|2 \, \text{adj}(3A \, \text{adj}(2A))| = 2^3 \cdot |3A \, \text{adj}(2A)|^2

Step-by-step Expansion

Now simplify the inner determinant:

|3A \, \text{adj}(2A)| = |3A| \cdot |\text{adj}(2A)|

For a $3 \times 3$ matrix,

|3A| = 3^3 |A|

and

|\text{adj}(2A)| = |2A|^2

Also,

|2A| = 2^3 |A|

Hence,

|\text{adj}(2A)| = (2^3 |A|)^2 = 2^6 |A|^2

Direct Power Counting

Substitute into the expression:

|3A \, \text{adj}(2A)| = 3^3 |A| \cdot 2^6 |A|^2 = 2^6 \cdot 3^3 \cdot |A|^3

So,

|2 \, \text{adj}(3A \, \text{adj}(2A))| = 2^3 \cdot \left(2^6 \cdot 3^3 \cdot |A|^3\right)^2

= 2^3 \cdot 2^{12} \cdot 3^6 \cdot |A|^6 = 2^{15} \cdot 3^6 \cdot |A|^6

Using $|A| = 5$ ,

|2 \, \text{adj}(3A \, \text{adj}(2A))| = 2^{15} \cdot 3^6 \cdot 5^6

Thus,

\alpha = 15, \quad \beta = 6, \quad \gamma = 6

Therefore,

\alpha + \beta + \gamma = 15 + 6 + 6 = 27

So, the correct option is C.

Common mistakes

Using $|\text{adj}(M)| = |M|$ is incorrect. For an $n \times n$ matrix, the correct relation is $|\text{adj}(M)| = |M|^{n-1}$ . Here $n = 3$ , so the power must be $2$ .
Forgetting that scalar multiplication affects the determinant by the order of the matrix is wrong. Since the matrix is $3 \times 3$ , $|2M| = 2^3 |M|$ and $|3A| = 3^3 |A|$ , not just $2|M|$ or $3|A|$ .
Computing $|\text{adj}(2A)|$ as $2^3 |A|^2$ is incorrect because $|2A|$ must be found first. The correct sequence is $|2A| = 2^3|A|$ and then $|\text{adj}(2A)| = |2A|^2 = 2^6 |A|^2$ .

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