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 do we reject the negative root here? Or is it just $3^{-2}$?

Great question! We aren't rejecting a root; we are solving an equation. We have $3^4 = 3^6 |A|$. Dividing both sides by $3^6$ gives $|A| = 3^{4-6} = 3^{-2}$. This is simply $\frac{1}{3^2} = \frac{1}{9}$. There is no negative root involved because the determinant of a matrix is a single value, not a variable with multiple solutions.

Q: Why can't we just use energy conservation? (Wait, wrong topic... why can't we just use $|B| = |A| imes |C|$?)

You are thinking of the property $|AB| = |A||B|$. That is a powerful tool! However, matrix $B$ is not a simple product of $A$ and another matrix $C$. It is a transformation of the elements of $A$. The factoring method is the most direct and elegant way to handle this specific structure.

Q: Why is $|A^T| = |A|$? Does the transpose change the determinant?

It absolutely does not! The determinant is a measure of the 'volume' spanned by the row vectors (or column vectors) of a matrix. Transposing a matrix swaps rows and columns, but the set of vectors remains the same, just viewed differently. Thus, the volume—and the determinant—remains unchanged. It is one of the most beautiful and useful properties in linear algebra!

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) $3$

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

(D) $\frac{1}{27}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear student. Today, we are going to embark on a journey through the elegant world of linear algebra. We have a problem that, at first glance, might look like a tedious exercise in calculation, but I promise you, it is a masterclass in pattern recognition.

We are given two $3 	imes 3$ matrices, $A$ and $B$, connected by a beautiful, rhythmic relationship: $b_{ij} = 3^{i+j-2} a_{ji}$. Our mission is to uncover the determinant of $A$, given that the determinant of $B$ is $81$.

### Decoding the Matrix

Imagine you are standing before the matrix $B$. It is not just a grid of numbers; it is a structure governed by a rule. If we write out the elements of $B$ using the formula $b_{ij} = 3^{i+j-2} a_{ji}$, we begin to see the hidden architecture.

For the first row, where $i=1$, the powers of $3$ are $3^{1+j-2} = 3^{j-1}$. So, the first row becomes $3^0 a_{11}, 3^1 a_{21}, 3^2 a_{31}$.

Notice something? The indices of $A$ are $a_{11}, a_{21}, a_{31}$. The column index of $A$ has become the row index of $B$. This is the first trap! Many students rush and assume $b_{ij}$ is just a scaled version of $a_{ij}$. But no, the indices are swapped; we are looking at the transpose of $A$ in disguise.

### The Art of Factoring

Now, let us construct the full matrix $B$:

$$B = \begin{pmatrix} 3^0 a_{11} & 3^1 a_{21} & 3^2 a_{31} \ 3^1 a_{12} & 3^2 a_{22} & 3^3 a_{32} \ 3^2 a_{13} & 3^3 a_{23} & 3^4 a_{33} \end{pmatrix}$$

This looks intimidating, doesn't it? But remember, in the world of determinants, we have a superpower: we can factor out common terms from any row or column.

Let us look at the columns. From the first column, we can factor out $3^0$. From the second, we can factor out $3^1$. From the third, we can factor out $3^2$. This gives us a factor of $3^0 \cdot 3^1 \cdot 3^2 = 3^3$ outside the determinant.

But we are not done! Look at the rows of the remaining matrix. The first row has $3^0$ common, the second has $3^1$ common, and the third has $3^2$ common. Factoring these out gives us another $3^3$. The total factor is $3^3 \cdot 3^3 = 3^6$.

### The Transpose Revelation

After we have extracted all these powers of $3$, what remains inside the determinant? We are left with:

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

This is the determinant of the transpose of $A$, denoted as $|A^T|$. And here is the moment of truth: the property $|A^T| = |A|$ is our key to the kingdom.

The determinant of a matrix is invariant under transposition. So, our equation simplifies beautifully to $|B| = 3^6 |A|$.

### Final Calculation

We know $|B| = 81$, which is $3^4$. So, we have the simple algebraic equation:

$$3^4 = 3^6 |A|$$

Dividing both sides by $3^6$, we get:

$$|A| = \frac{3^4}{3^6} = 3^{-2} = \frac{1}{9}$$

And there it is! The complexity melts away when you understand the underlying structure. You didn't need to calculate a single $3 	imes 3$ determinant; you just needed to see the pattern, respect the properties of determinants, and trust the math.

The final answer is **$|A| = \frac{1}{9}$**.

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 Matrix Relationship

Expanding Matrix $B$

Completing the Matrix

Factoring Powers from Columns

Factoring Powers from Rows

Recognizing the Remaining Determinant

Solving for $∣ A ∣$

Final Answer

The Sigma Insight: Properties of Determinants

Analyzing the Setup

Decoding the Matrix

The Art of Factoring

The Transpose Revelation

Final Calculation

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