Let P = [pij] and Q = [qij] be two square matrices of order 3 such that qij = 2^(i+j-1)pij and det(Q

Q: Wait, why can't I just say $\det(Q) = 2^{i+j-1} \det(P)$? Why is this wrong?

This is a classic trap! You cannot pull $2^{i+j-1}$ out of the determinant because it is not a constant scalar. It is a variable expression that changes for every single element in the matrix. You can only factor out a constant that is common to an entire row or an entire column. Always treat the determinant as a structure, not a simple algebraic product.

Q: Why did we factor out from rows first and then columns? Does the order matter?

The order does not matter at all! The determinant is a property of the matrix itself, and row/column operations are just tools to reveal its value. We chose rows first to organize our thoughts, but you could start with columns and arrive at the exact same result. The crucial step is to ensure you account for *all* powers of $2$ present in the matrix, regardless of the order of operations.

Q: Is the formula $\det(\text{adj}(\text{adj } P)) = |P|^{(n-1)^2}$ always true?

Yes, it is a powerful identity derived from the fundamental property $\text{adj}(A) = |A|A^{-1}$. For any non-singular square matrix of order $n$, this formula holds. It is a lifesaver in JEE Advanced exams because it saves you from calculating the adjoint matrix explicitly, which is a massive time-sink.

Question

Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i+j-1)}p_{ij}$ and $\det(Q) = 2^{10}$. Then the value of $\det(	ext{adj}(	ext{adj } P))$ is:

Options:

(A) 81

(B) 16

(C) 124

(D) 32

Sigma Percentile · Accepted Answer

### The Matrix Landscape: Unveiling the Hidden Structure

Welcome, future engineers! Today, we are going to peel back the layers of a matrix problem that, at first glance, looks like a simple exercise in algebra but is actually a beautiful lesson in the properties of determinants.

We are given two matrices, $P$ and $Q$, related by the expression $q_{ij} = 2^{(i+j-1)}p_{ij}$. Our mission is to find the value of $\det(	ext{adj}(	ext{adj } P))$, given that $\det(Q) = 2^{10}$.

### Phase 1: Decoding the Matrix $Q$

Imagine you are standing before the matrix $Q$. It looks intimidating because every entry is scaled by a different power of $2$. Let us write it out explicitly to see the pattern:

$$
Q = \begin{pmatrix} 2^1 p_{11} & 2^2 p_{12} & 2^3 p_{13} \ 2^2 p_{21} & 2^3 p_{22} & 2^4 p_{23} \ 2^3 p_{31} & 2^4 p_{32} & 2^5 p_{33} \end{pmatrix}
$$

Do you see the rhythm? As we move across the rows and down the columns, the exponent of $2$ grows. This is not a uniform scaling.

If we were to try to pull a single factor out of the entire matrix, we would fail. But remember the golden rule of determinants: we can factor out a constant from a *single* row or a *single* column at a time. This is our key to unlocking the problem.

### Phase 2: The Art of Factoring (The JEE Trap)

Let us tackle this row by row. Look at the first row: $2^1 p_{11}, 2^2 p_{12}, 2^3 p_{13}$. We can pull out $2^1$.

Now look at the second row: $2^2 p_{21}, 2^3 p_{22}, 2^4 p_{23}$. We can pull out $2^2$. Finally, the third row: $2^3 p_{31}, 2^4 p_{32}, 2^5 p_{33}$. We can pull out $2^3$.

After factoring these out, our determinant looks like this:

$$
\det(Q) = (2^1 \cdot 2^2 \cdot 2^3) \det \begin{pmatrix} p_{11} & 2^1 p_{12} & 2^2 p_{13} \ p_{21} & 2^1 p_{22} & 2^2 p_{23} \ p_{31} & 2^1 p_{32} & 2^2 p_{33} \end{pmatrix}
$$

Many students stop here, thinking they are done. But look closely at the columns of the remaining matrix! The second column still has a common factor of $2^1$, and the third column has a common factor of $2^2$.

If you miss this, you miss the entire solution. We must factor these out as well:

$$
\det(Q) = (2^1 \cdot 2^2 \cdot 2^3) \cdot (2^1 \cdot 2^2) \det(P)
$$

Calculating the total power of $2$: $1 + 2 + 3 + 1 + 2 = 9$. Thus, we arrive at the elegant relationship: $\det(Q) = 2^9 \det(P)$.

### Phase 3: The Adjoint Mystery

With $\det(Q) = 2^{10}$, we solve for $\det(P)$:

$$
2^{10} = 2^9 \det(P) \
\Rightarrow \det(P) = \frac{2^{10}}{2^9} = 2
$$

Now, we face the final hurdle: $\det(	ext{adj}(	ext{adj } P))$. Do not panic! We do not need to calculate the adjoint matrix.

We have a powerful theorem for this. For any square matrix $A$ of order $n$, the determinant of the double adjoint is given by:

$$
\det(	ext{adj}(	ext{adj } A)) = |A|^{(n-1)^2}
$$

Since our matrix $P$ is of order $n=3$, the exponent becomes $(3-1)^2 = 2^2 = 4$. Therefore, we simply need to calculate $|P|^4$.

Substituting our value of $|P| = 2$, we get $2^4 = 16$.

### Conclusion

And there you have it! By carefully decomposing the matrix and applying the properties of determinants, we turned a complex-looking expression into a simple calculation.

Remember, in JEE Advanced, the complexity is often just a mask for a fundamental property waiting to be applied. The final answer is **16**.

Animated Solution for Mathematics - Matrices and Determinants: Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $det (Q) = 2^{10}$ . Then the value of $det (adj (adj P))$ is:

Select Answer:

Visualized Solution

Understanding the Relation $q_{ij}$

Expanding Matrix $Q$

Factoring from Rows

Factoring from Columns

Relating Determinants

Solving for $det (P)$

The Double Adjoint Property

Final Calculation

The Sigma Insight: Adjoint and Inverse of a Matrix

The Matrix Landscape

Phase 1

Phase 2

Phase 3

Conclusion

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