Let A be a non-singular matrix of order 3. If det(3 adj(2 adj((det A)A))) = 3⁻¹³ · 2⁻¹⁰ and det(3 ad

Q: Wait, why does the constant $k$ come out as $k^3$ instead of just $k$?

That is a classic trap! Remember that for a matrix $M$ of order $n$, the determinant $\det(kM)$ is equal to $k^n \det(M)$. Since our matrix $A$ is of order 3, $n=3$. Therefore, every time you pull a scalar constant out of a determinant, it must be raised to the power of 3. It is not just a simple multiplication; it is a scaling of the volume in 3D space!

Q: Can I just calculate $\det(A)$ first and then plug it into the expression?

You could, but that would be incredibly difficult because you do not know the elements of $A$. The beauty of this problem lies in using the properties of determinants to manipulate the expression symbolically. By treating $\det(A)$ as a variable (let's call it $|A|$), we can solve for it algebraically without ever needing to know the individual entries of the matrix.

Q: Why is $\det(\text{adj}(M)) = (\det M)^2$?

Great question! The general property is $\det(\text{adj}(M)) = (\det M)^{n-1}$. Since our matrix $A$ (and consequently any matrix derived from it in this problem) is of order 3, $n=3$. Thus, $n-1 = 2$. This is why the adjoint operator effectively squares the determinant of the matrix inside it. It is a powerful simplification tool!

Question

Let $A$ be a non-singular matrix of order 3. If $\det(3 	ext{adj}(2 	ext{adj}((\det A)A))) = 3^{-13} \cdot 2^{-10}$ and $\det(3 	ext{adj}(2 A)) = 2^m \cdot 3^n$, then $|3m + 2n|$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler in the world of mathematics. Today, we are not just solving a problem; we are embarking on a journey through the elegant, nested structure of matrix algebra.

When you first look at an expression like $\det(3 	ext{adj}(2 	ext{adj}((\det A)A)))$, it is natural to feel a surge of intimidation. It looks like a labyrinth of operators, constants, and nested brackets.

But here is the secret: this is not a labyrinth; it is an onion. And like any onion, the only way to reach the core without tears is to peel it, one layer at a time.

### The Toolkit: Our Mathematical Foundation

Before we touch the expression, let us ensure our toolkit is sharp. We are dealing with a non-singular matrix $A$ of order $n=3$. This means $\det(A) 
eq 0$.

To navigate this problem, we rely on two fundamental properties of determinants:

1. The scalar multiplication property: $\det(kM) = k^n \det(M)$. Since our order $n=3$, every time we pull a constant $k$ out of a determinant, it emerges as $k^3$.

2. The adjoint property: $\det(	ext{adj}(M)) = (\det M)^{n-1}$. With $n=3$, this simplifies to $\det(	ext{adj}(M)) = (\det M)^2$.

These two properties are our keys to unlocking the structure.

### Peeling the Onion: Layer by Layer

Let us start with our first expression: $\det(3 	ext{adj}(2 	ext{adj}(|A|A)))$. We look at the outermost layer.

Using our first property, we pull the $3$ out of the determinant. It emerges as $3^3$. Now, our expression is:

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

Next, we face the adjoint operator. We use our second property: $\det(	ext{adj}(M)) = (\det M)^2$. Here, $M = 2 	ext{adj}(|A|A)$.

So, the expression becomes:

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

Now, look inside the square brackets: $\det(2 	ext{adj}(|A|A))$. Again, we have a constant $2$ inside a determinant. We pull it out as $2^3$.

Because this $2^3$ is trapped inside the square bracket, which is raised to the power of $2$, it becomes $(2^3)^2 = 2^6$. Our expression is now:

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

### Reaching the Core

We are closing in on the center. We have the inner adjoint operator: $\det(	ext{adj}(|A|A))$. Applying the adjoint property again, this becomes $(\det(|A|A))^2$.

Since this was already squared, we now have $(\det(|A|A))^4$. Finally, we tackle the innermost term: $\det(|A|A)$.

Here, $|A|$ is just a scalar. Pulling it out gives $|A|^3 \det(A)$, which is $|A|^3 \cdot |A| = |A|^4$. Raising this to the power of $4$, we get:

$$(|A|^4)^4 = |A|^{16}$$

Putting it all together, our massive expression simplifies to the elegant:

$$3^3 \cdot 2^6 \cdot |A|^{16}$$

### The Final Reveal

We are given that this expression equals $3^{-13} \cdot 2^{-10}$. So, we set up the equation:

$$3^3 \cdot 2^6 \cdot |A|^{16} = 3^{-13} \cdot 2^{-10}$$

Dividing both sides by $3^3 \cdot 2^6$, we find:

$$|A|^{16} = 3^{-16} \cdot 2^{-16}$$

Taking the sixteenth root, we find $|A| = 3^{-1} \cdot 2^{-1} = \frac{1}{6}$.

With $|A|$ in hand, the second expression, $\det(3 	ext{adj}(2A))$, becomes a simple exercise. Following the same logic, it simplifies to:

$$3^3 \cdot 2^6 \cdot |A|^2$$

Substituting $|A| = 3^{-1} \cdot 2^{-1}$, we get:

$$3^3 \cdot 2^6 \cdot (3^{-1} \cdot 2^{-1})^2 = 3^3 \cdot 2^6 \cdot 3^{-2} \cdot 2^{-2} = 3^1 \cdot 2^4$$

Comparing this to $2^m \cdot 3^n$, we find $m=4$ and $n=1$. The final calculation is:

$$|3m + 2n| = |3(4) + 2(1)| = \mathbf{14}$$

You have mastered the onion!

Animated Solution for Mathematics - Matrices and Determinants: Let $A$ be a non-singular matrix of order 3. If $det (3 adj (2 adj ((det A) A))) = 3^{- 13} \cdot 2^{- 10}$ and $det (3 adj (2 A)) = 2^{m} \cdot 3^{n}$ , then $∣3 m + 2 n ∣$ is equal to

Enter Numerical Value:

Visualized Solution

Problem Overview & Matrix Order

Property: $det (k A) = k^{n} det A$

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

Extracting the Outer Constant $3$

Resolving the Outer Adjoint

Extracting the Constant $2$

Resolving the Inner Adjoint

Simplifying $det (∣ A ∣ A)$

Finding the Value of $∣ A ∣$

Simplifying the Second Expression

Substitute $∣ A ∣$ to Find $m, n$

Final Calculation of $∣3 m + 2 n ∣$

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Toolkit

Peeling the Onion

Reaching the Core

The Final Reveal

Similar Questions

Let A be a square matrix of order 3 such that $det (A) = - 2$ and $det (3 a d j (- 6 a d j (3 A))) = 2^{m + n} \cdot 3^{mn}$ $m > n$ . Then $4 m + 2 n$ is equal to

If $A$ is a square matrix of order 3 such that $det (A) = 3$ and $det (adj (- 4 adj (- 3 adj (3 adj ((2 A)^{- 1}))))) = 2^{m} 3^{n}$ , then $m + 2 n$ is equal to:

If $A$ is a $3 \times 3$ matrix and $∣ A ∣ = 2$ , then $∣3 adj (∣3 A ∣ A^{2}) ∣$ is equal to

Let $A = 1 α 1 231312$ for $α \in R$ . If $∣ A ∣ = 2$ . If $∣2 adj (2 adj (2 A)) ∣ = 3 2^{n}$ , then $3 n + α$ is equal to

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

Let $A$ and $B$ be two square matrices of order 3 such that $∣ A ∣ = 3$ and $∣ B ∣ = 2$ . Then $∣ A^{T} A (adj (2 A))^{- 1} (adj (4 B)) (adj (A B))^{- 1} A A^{T} ∣$ is equal to:

Let $P = [a_{ij}]$ be a $3 \times 3$ matrix and let $Q = [b_{ij}]$ , where $b_{ij} = 2^{i + j} a_{ij}$ for $1 \leq i, j \leq 3$ . If the determinant of $P$ is 2, then the determinant of the matrix $Q$ is

If $A = [2335]$ , then the determinant of the matrix $(A^{2025} - 3 A^{2024} + A^{2023})$ is

Let $A$ be a $n \times n$ matrix such that $∣ A ∣ = 2$ . If the determinant of the matrix $adj (2 adj (2 A^{- 1}))$ is $2^{84}$ , then $n$ is equal to ______.

Let $A$ be a $3 \times 3$ matrix of non-negative real elements such that $A 111 = 3111$ . Then the maximum value of $det (A)$ is ______

Enter Numerical Value:

Visualized Solution

Problem Overview & Matrix Order

Property: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(kA)=kndetA

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

Extracting the Outer Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3

Resolving the Outer Adjoint

Extracting the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2

Resolving the Inner Adjoint

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(∣A∣A)

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣A∣

Simplifying the Second Expression

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣A∣ to Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m,n

Final Calculation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣3m+2n∣