Let A be a matrix of order 3 x 3 and det(A) = 2. Then det(det(A) adj(5 adj(A^3))) is equal to

Q: Wait, why does the scalar $k$ become $k^3$ when it comes out? Shouldn't it just be $k$?

This is the most common trap in JEE matrix problems! Remember, the determinant is a linear operator on each row. If you multiply a matrix $M$ of order $3 \times 3$ by a scalar $k$, you are multiplying *every* row by $k$. Since there are $3$ rows, you factor out $k$ three times. Thus, $\det(kM) = k \cdot k \cdot k \cdot \det(M) = k^3 \det(M)$. Always check the order of the matrix!

Q: Why can't I just find the matrix $A$ first? Is there a way to solve for the elements?

That is a path to madness! The problem gives you $\det(A) = 2$, but it never gives you the matrix $A$ itself. There are infinite matrices with a determinant of $2$. This problem is designed to test your mastery of *properties*, not your ability to perform matrix multiplication. Trust the properties; they are the shortcuts to the answer.

Q: I keep getting confused with the powers. How do I keep track of the exponents when applying the adjoint property multiple times?

Think of it as a bookkeeping exercise. Every time you apply $\det(\text{adj}(M)) = (\det(M))^{n-1}$, you are essentially raising the determinant to the power of $n-1$. If you apply it twice, you are raising it to $(n-1)^2$. Keep your scratchpad organized, write down the power at each step, and never try to do it all in your head.

Question

Let $A$ be a matrix of order $3 	imes 3$ and $\det(A) = 2$. Then $\det(\det(A) 	ext{adj}(5 	ext{adj}(A^3)))$ is equal to

Options:

(A) $512 	imes 10^6$

(B) $256 	imes 10^6$

(C) $1024 	imes 10^6$

(D) $256 	imes 10^{11}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you look at an expression like $\det(\det(A) 	ext{adj}(5 	ext{adj}(A^3)))$, it is natural to feel a surge of intimidation. It looks like a labyrinth of nested operators.

But here is the secret: in JEE Advanced, complexity is often a mask for elegance. This problem is not about brute force; it is about the surgical application of properties.

Think of this expression as a Russian Matryoshka doll. We must open it one layer at a time, from the outside in. We are given that $A$ is a $3 	imes 3$ matrix, so $n=3$, and we are given $\det(A) = 2$.

### Phase 1: The Outer Shell

We begin with the outermost determinant. The expression is $\det(|A| 	ext{adj}(5 	ext{adj}(A^3)))$.

Here, $|A|$ is just a scalar constant. We invoke the property $\det(kM) = k^n \det(M)$. Since our matrix order $n$ is $3$, the scalar $|A|$ must emerge as $|A|^3$.

This gives us:
$$|A|^3 \det(	ext{adj}(5 	ext{adj}(A^3)))$$

### Phase 2: The Adjoint Barrier

Now, we face the adjoint operator. The property $\det(	ext{adj}(M)) = (\det(M))^{n-1}$ is our best friend.

With $n=3$, the exponent becomes $n-1 = 2$. Applying this to our current expression, the determinant of the adjoint of the inner term becomes the square of the determinant of that inner term:

$$|A|^3 (\det(5 	ext{adj}(A^3)))^2$$

### Phase 3: The Scalar Trap

Now, look inside the bracket: $\det(5 	ext{adj}(A^3))$. This is where many students stumble.

We need to pull the scalar $5$ out of the determinant. Again, because the matrix is $3 	imes 3$, the $5$ comes out as $5^3$.

Because the entire term is being squared from the previous step, we have $(5^3 \det(	ext{adj}(A^3)))^2$. When we distribute that square, $5^3$ becomes $5^6$. Our expression now stands at:

$$|A|^3 \cdot 5^6 \cdot (\det(	ext{adj}(A^3)))^2$$

### Phase 4: The Final Simplification

We have $\det(	ext{adj}(A^3))$ inside the square. Applying the adjoint property again, $\det(	ext{adj}(A^3)) = (\det(A^3))^{n-1} = (\det(A^3))^2$.

Substituting this back, we get:
$$|A|^3 \cdot 5^6 \cdot ((\det(A^3))^2)^2 = |A|^3 \cdot 5^6 \cdot (\det(A^3))^4$$

Using the property $\det(A^k) = (\det(A))^k$, we know $\det(A^3) = (|A|)^3$. Therefore, $(\det(A^3))^4 = (|A|^3)^4 = |A|^{12}$.

Combining everything, we get:
$$|A|^3 \cdot 5^6 \cdot |A|^{12} = 5^6 |A|^{15}$$

### The Grand Finale

Now, the moment of truth. We substitute $|A| = 2$. The expression becomes $5^6 \cdot 2^{15}$.

To make this calculation elegant, we split $2^{15}$ into $2^6 \cdot 2^9$. This allows us to group the terms:

$$(5 \cdot 2)^6 \cdot 2^9 = 10^6 \cdot 512$$

The final answer is **$512 	imes 10^6$**. By respecting the properties and peeling the layers methodically, the beast was tamed.

Animated Solution for Mathematics - Matrices and Determinants: Let $A$ be a matrix of order $3 \times 3$ and $det (A) = 2$ . Then $det (det (A) adj (5 adj (A^{3})))$ is equal to

Select Answer:

Visualized Solution

Given Information and Goal

Applying Scalar Property: $det (k M) = k^{n} det (M)$

Adjoint Property: $det (adj (M)) = (det (M))^{n - 1}$

Handling the Scalar Constant $5$

Second Adjoint Application

Simplifying the Power of $∣ A ∣$

Final Substitution and Calculation

The Sigma Insight: Properties of Determinants

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

The Grand Finale

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A be a matrix of order 3×3 and det(A)=2. Then det(det(A)adj(5adj(A3))) is equal to

Select Answer:

Visualized Solution

Given Information and Goal

Applying Scalar Property: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(kM)=kndet(M)

Adjoint Property: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(adj(M))=(det(M))n−1

Handling the Scalar Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5