Let A = [aij] and B = [bij] be two 3 × 3 real matrices such that bij = (3)^(i+j-2) aji, where i, j =

Q: Wait, why does $a_{ji}$ imply the transpose? I thought it was just a different element.

Great question! In matrix notation, $A = [a_{ij}]$. If we define a new matrix $C$ where $c_{ij} = a_{ji}$, then $C$ is by definition the transpose of $A$, denoted as $A^T$. The indices $i$ and $j$ are just labels; swapping them is the geometric definition of reflecting the matrix across its main diagonal. Recognizing this symmetry is the key to unlocking the problem.

Q: Why can't I just calculate the determinant directly? It seems like a lot of algebra.

You could, but that is the trap! JEE Advanced problems are designed to test your conceptual grasp, not your brute-force calculation speed. By using properties of determinants, we avoid messy arithmetic and reduce the chance of a silly calculation error. Always look for the structure first; the elegance of the solution is usually hidden in the properties, not the arithmetic.

Q: Does $|A^T| = |A|$ hold for any size matrix?

Yes, absolutely! The determinant represents the volume scaling factor of the linear transformation. Transposing a matrix reflects it across the main diagonal, which changes the orientation but preserves the volume. Thus, the determinant remains invariant for any square matrix of size $n \times n$.

Question

Let $A = [a_{ij}]$ and $B = [b_{ij}]$ be two $3 	imes 3$ real matrices such that $b_{ij} = (3)^{(i+j-2)} a_{ji}$, where $i, j = 1, 2, 3$. If the determinant of $B$ is 81, then the determinant of $A$ is :

Options:

(A) $\frac{1}{9}$

(B) $\frac{1}{81}$

(C) $\frac{1}{3}$

(D) 3

Sigma Percentile · Accepted Answer

### The Matrix Detective: Unlocking the Hidden Symmetry

Welcome, future engineer. Today, we are not just solving a matrix problem; we are performing an act of mathematical archaeology. We have been handed a relationship between two matrices, $A$ and $B$, defined by the equation $b_{ij} = 3^{i+j-2} a_{ji}$.

At first glance, this looks like a chaotic mess of indices and powers. But in the world of JEE Advanced, chaos is just order waiting to be discovered.

### Phase 1: Decoding the Cipher

The first step in any great investigation is to look at the details that others ignore. The relation $b_{ij} = 3^{i+j-2} a_{ji}$ is our map.

Notice the indices: $a_{ji}$. If this were $a_{ij}$, we would be looking at a scaled version of $A$. Because the indices are swapped, we are looking at the transpose of $A$, denoted as $A^T$.

To make this manageable, let us decompose that exponent. We can rewrite the term $3^{i+j-2}$ as $3^{i-1} \cdot 3^{j-1}$.

Now, the expression becomes $b_{ij} = 3^{i-1} \cdot 3^{j-1} \cdot a_{ji}$. This separation is crucial, as it allows us to treat the row-dependent factors and the column-dependent factors as separate entities.

### Phase 2: The Art of Factoring

Imagine the determinant of $B$ as a $3 	imes 3$ grid. When we write out the elements using our new formula, we see a pattern emerging.

For the first row ($i=1$), every element is multiplied by $3^{1-1} = 3^0 = 1$. For the second row ($i=2$), every element is multiplied by $3^{2-1} = 3^1 = 3$. For the third row ($i=3$), every element is multiplied by $3^{3-1} = 3^2 = 9$.

Determinant properties allow us to factor out a constant from an entire row. So, we pull $3^0$ from the first row, $3^1$ from the second, and $3^2$ from the third. Our determinant now looks like this:

$$|B| = (3^0 \cdot 3^1 \cdot 3^2) \cdot \begin{vmatrix} a_{11} & a_{21} & a_{31} \ a_{12} & a_{22} & a_{32} \ a_{13} & a_{23} & a_{33} \end{vmatrix}$$

We still have the column factors $3^{j-1}$ lurking inside. We repeat the process for the columns, pulling $3^0$ from the first column, $3^1$ from the second, and $3^2$ from the third.

### Phase 3: The Transpose Revelation

After extracting all these factors, we are left with a beautiful, clean matrix. Look closely at the remaining grid:

$$\begin{vmatrix} a_{11} & a_{21} & a_{31} \ a_{12} & a_{22} & a_{32} \ a_{13} & a_{23} & a_{33} \end{vmatrix}$$

This is exactly the transpose of matrix $A$, or $A^T$. As we know, the determinant of a matrix is identical to the determinant of its transpose: $|A^T| = |A|$.

Our total multiplier is the product of the row factors and the column factors. We extracted $(3^0 \cdot 3^1 \cdot 3^2) = 3^3 = 27$ from the rows, and the same from the columns. Multiplying these together, we get $3^6 = 729$.

### Phase 4: The Final Calculation

We have arrived at the final, elegant equation:

$$|B| = 3^6 \cdot |A|$$

We are given that $|B| = 81$. Substituting this into our equation, we get:

$$81 = 729 \cdot |A|$$

Solving for $|A|$, we find:

$$|A| = \frac{81}{729} = \frac{1}{9}$$

And there it is. The complexity dissolves, leaving behind a simple, clean fraction. You didn't need to calculate nine separate determinants or perform grueling arithmetic.

You used the properties of the system to reveal the truth. Keep this mindset—look for the structure, trust the properties, and the answer will always reveal itself. The final answer is **$|A| = 1/9$**.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [a_{ij}]$ and $B = [b_{ij}]$ be two $3 \times 3$ real matrices such that $b_{ij} = (3)^{(i + j - 2)} a_{j i}$ , where $i, j = 1, 2, 3$ . If the determinant of $B$ is 81, then the determinant of $A$ is :

Select Answer:

Visualized Solution

The Setup & The Trap

Splitting the Powers

Constructing the Determinant

Factoring from Rows

Factoring from Columns

The Transpose Matrix

Determinant Property

The Final Equation

Final Calculation

The Sigma Insight: Properties of Determinants

The Matrix Detective

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let $A = [a_{ij}]$ and $B = [b_{ij}]$ be two $3 \times 3$ real matrices such that $b_{ij} = (3)^{(i + j - 2)} a_{j i}$ , where $i, j = 1, 2, 3$ . If the determinant of $B$ is $81$ , then the determinant of $A$ is:

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 $A$ 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 $A$ be a $3 \times 3$ matrix with $det (A) = 4$ . Let $R_{i}$ denote the $i^{t h}$ row of $A$ . If a matrix $B$ is obtained by performing the operation $R_{2} \to 2 R_{2} + 5 R_{3}$ on $2 A$ , then $det (B)$ is equal to:

Let $P$ and $Q$ be $3 \times 3$ matrices $P \neq = Q$ . If $P^{3} = Q^{3}$ and $P^{2} Q = Q^{2} P$ then determinant of $(P^{2} + Q^{2})$ is equal to:

The trace of a square matrix is defined to be the sum of its diagonal entries. If $A$ is a $2 \times 2$ matrix such that the trace of $A$ is 3 and the trace of $A^{3}$ is $- 18$ , then the value of the determinant of $A$ is ____.

The value of the determinant $111 a b c a^{2} - b c b^{2} - c a c^{2} - ab$ is .........

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