Let A be a matrix of order 3 × 3 and |A| = 5. If |2 adj(3 A adj(2 A))| = 2^α · 3^β · 5^γ, α, β, γ ∈

Q: Wait, why does the scalar 2 become $2^3$ when it comes out of the determinant?

Great question! Remember that for any $n \times n$ matrix $M$, the determinant $|kM| = k^n |M|$. Since our matrix $A$ is of order $3 \times 3$, the scalar $k=2$ must be raised to the power of $n=3$. So, $2^3 = 8$ is the correct factor to pull out.

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

This comes from the fundamental property $|\text{adj}(M)| = |M|^{n-1}$. Since we are dealing with a $3 \times 3$ matrix, $n=3$, so $n-1 = 2$. Therefore, the determinant of the adjoint is simply the square of the determinant of the original matrix.

Q: Can I just distribute the adjoint inside the product, like $\text{adj}(AB) = \text{adj}(A)\text{adj}(B)$?

Be careful! The property is actually $\text{adj}(AB) = \text{adj}(B)\text{adj}(A)$. However, since we are taking the determinant of the entire expression, we use the property $|MN| = |M||N|$. So, $|\text{adj}(AB)| = |\text{adj}(B)| |\text{adj}(A)| = |B|^{n-1} |A|^{n-1}$. Always stick to the determinant properties to avoid order-of-multiplication traps!

Question

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

Options:

(A) 25

(B) 26

(C) 27

(D) 28

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a $3 	imes 3$ matrix $A$ with determinant $|A| = 5$. We aim to evaluate the expression:
$$|2 	ext{ adj}(3 A 	ext{ adj}(2 A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$$

We will utilize the following fundamental properties of determinants for an $n 	imes n$ matrix:
1. $|kM| = k^n |M|$
2. $|	ext{adj}(M)| = |M|^{n-1}$
3. $|MN| = |M||N|$

### Layer 1: The Outer Scalar

Our first task is to extract the scalar $2$ from the determinant. Since the matrix is of order $n=3$, pulling the scalar $2$ out yields $2^3$.

The expression simplifies to:
$$2^3 |	ext{adj}(3 A 	ext{ adj}(2 A))|$$

### Layer 2: The Adjoint Property

Next, we apply the property $|	ext{adj}(M)| = |M|^{n-1}$. With $n=3$, this becomes $|	ext{adj}(M)| = |M|^2$.

Applying this to our expression, the outer adjoint is removed, and the inner term is squared:
$$2^3 |3 A 	ext{ adj}(2 A)|^2$$

### Layer 3: The Inner Scalar

Inside the square, we have the scalar $3$. Extracting this from the $3 	imes 3$ determinant results in $3^3$.

Because this term is subject to the square from the previous step, the expression becomes:
$$2^3 (3^3 |A 	ext{ adj}(2 A)|)^2$$

### Layer 4: The Product Property

We now apply the product property $|MN| = |M||N|$ to separate the determinant of the product into the product of the determinants.

The expression is now:
$$2^3 (3^3 |A| |	ext{adj}(2 A)|)^2$$

### Layer 5: The Recursive Adjoint

We resolve the final adjoint term $|	ext{adj}(2 A)|$. Using the property $|	ext{adj}(M)| = |M|^2$, we have $|	ext{adj}(2 A)| = |2 A|^2$.

Since $|2 A| = 2^3 |A|$, we substitute this back:
$$|	ext{adj}(2 A)| = (2^3 |A|)^2 = 2^6 |A|^2$$

### Final Calculation

Substituting $|	ext{adj}(2 A)| = 2^6 |A|^2$ into our expression from Layer 4:
$$2^3 (3^3 \cdot |A| \cdot 2^6 \cdot |A|^2)^2$$

Combining the powers of $|A|$ and simplifying:
$$2^3 (3^3 \cdot 2^6 \cdot |A|^3)^2 = 2^3 \cdot 3^6 \cdot 2^{12} \cdot |A|^6$$
$$= 2^{15} \cdot 3^6 \cdot |A|^6$$

Given $|A| = 5$, the final result is:
$$2^{15} \cdot 3^6 \cdot 5^6$$

Comparing this to $2^\alpha \cdot 3^\beta \cdot 5^\gamma$, we identify $\alpha = 15$, $\beta = 6$, and $\gamma = 6$. The final sum is:
$$\alpha + \beta + \gamma = 15 + 6 + 6 = \mathbf{27}$$

Animated Solution for Mathematics - Matrices and Determinants: Let $A$ be a matrix of order $3 \times 3$ and $∣ A ∣ = 5$ . If $∣2 adj (3 A adj (2 A)) ∣ = 2^{α} \cdot 3^{β} \cdot 5^{γ}, α, β, γ \in N$ then $α + β + γ$ is equal to

Select Answer:

Visualized Solution

Problem Setup

Extracting the Outermost Scalar

Applying Adjoint Determinant Property

Extracting Scalar from Inner Determinant

Product Property of Determinants

Evaluating the Inner Adjoint

Extracting Scalar from $2 A$

Substituting Back and Simplifying

Expanding the Outer Square

Final Substitution and Comparison

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

Layer 1

Layer 2

Layer 3

Layer 4

Layer 5

Final Calculation

Similar Questions

Let A be a $3 \times 3$ matrix such that $A + A^{T} = O$ . If $A 1 - 1 0 = 332, A^{2} 1 - 1 0 = - 3 19 - 24$ and $det (adj (2 adj (A + I))) = (2)^{α} \cdot (3)^{β} \cdot (11)^{γ}, α, β, γ$ are non-negative integers, then $α + β + γ$ is equal to .........

Let $A = 246 2 + p 6 + 2 p 12 + 3 p 2 + p + q 8 + 3 p + 2 q 20 + 6 p + 3 q$ . If $det (adj (adj (3 A))) = 2^{m} \cdot 3^{n}, m, n \in N$ , then $m + n$ is equal to

Let $∣ A ∣ = 6$ , where A is a $3 \times 3$ matrix. If $∣ adj (3 adj (A^{2} \cdot adj (2 A))) ∣ = 2^{m} \cdot 3^{n}, m, n \in N$ , then $m + n$ is equal to .........

Let A be a $3 \times 3$ matrix such that $X^{T} A X = 0$ for all nonzero $3 \times 1$ matrices $X = x y z$ . If $A 111 = 14 - 5$ and $A 121 = 04 - 8$ , and $det (a d j (2 (A + I))) = 2^{α} 3^{β} 5^{γ}$ , $α, β, γ \in N,$ then $α^{2} + β^{2} + γ^{2}$ is

For a $3 \times 3$ matrix M, let trace(M) denote the sum of all the diagonal elements of M. Let A be a $3 \times 3$ matrix such that $∣ A ∣ = \frac{1}{2}$ and trace $(A) = 3$ . If $B = a d j (a d j (2 A))$ , then the value of $∣ B ∣ +$ trace $(B)$ equals :

Let the determinant of a square matrix $A$ of order $m$ be $m - n$ , where $m$ and $n$ satisfy $4 m + n = 22$ and $17 m + 4 n = 93$ . If $d e t (n adj (a d j (m A))) = 3^{a} 5^{b} 6^{c}$ , then $a + b + c$ is equal to

Let $A$ be a $3 \times 3$ real matrix. If $det (2 Adj (2 Adj (Adj (2 A)))) = 2^{41}$ , then the value of $det (A^{2})$ equal .

Let $a \in R$ and $A$ be a matrix of order $3 \times 3$ such that $det (A) = - 4$ and $A + I = 12 a a 11 102$ , where $I$ is the identity matrix of order $3 \times 3$ . If $det ((a + 1) adj ((a - 1) A))$ is $2^{m} 3^{n}$ , $m, n \in {0, 1, 2, \dots, 20}$ , then $m + n$ is equal to :

Let $P = 323 - 1 0 - 5 - 2 α 0$ , where $α \in R$ . Suppose $Q = [q_{ij}]$ is a matrix such that $P Q = k I$ , where $k \in R, k \neq = 0$ and $I$ is the identity matrix of order $3$ . If $q_{23} = - k /8$ and $det (Q) = k^{2} /2$ , then

If $A = \frac{1}{5 ! 6 ! 7 !} 5! 6! 7! 6! 7! 8! 7! 8! 9!$ , then $∣ a d j (a d j (2 A)) ∣$ is equal to