Let a ∈ R and A be a matrix of order 3 × 3 such that det(A) = -4 and A + I = [[1, a, 1], [2, 1, 0],

Q: Wait, why does the scalar $k$ come out as $k^3$ when I pull it out of the determinant?

That is a fantastic question! Remember that a determinant represents the volume scaling factor of a linear transformation. In a $3 \times 3$ matrix, you are scaling the volume in three dimensions (length, width, and height). If you multiply every element of the matrix by a scalar $k$, you are effectively scaling each of the three dimensions by $k$. Therefore, the volume scales by $k \times k \times k = k^3$. This is why for an $n \times n$ matrix, the scalar comes out as $k^n$.

Q: Can I just calculate the adjoint matrix directly instead of using the property?

Technically, yes, but I strongly advise against it! Calculating the adjoint of a $3 \times 3$ matrix involves finding nine cofactors and then transposing them. It is a tedious process prone to arithmetic errors. In JEE Advanced, time is your most precious resource. Using the property $\det(\text{adj}(M)) = (\det(M))^{n-1}$ is elegant, faster, and significantly reduces the chance of making a silly mistake.

Q: What happens to the negative sign when I square $\det(2A)$?

Great catch! When you calculate $\det(2A)$, you get $-32$. When you square this value, the negative sign disappears because $(-32)^2 = (-1)^2 \times (32)^2 = 32^2$. Always be careful with signs when dealing with powers. It is often safer to convert everything to prime factors (like powers of 2) early on to avoid these sign traps.

Question

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

Options:

(A) 14

(B) 17

(C) 15

(D) 16

Sigma Percentile · Accepted Answer

### The Matrix Mystery: Unmasking the Unknown

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to peel back the layers of a matrix problem that, at first glance, might look like a chaotic mess of symbols.

Beneath the surface lies a beautiful, logical structure waiting to be revealed. Let us embark on this journey together.

### Phase 1: The Art of Isolation

We start with the equation:

$$A + I = \begin{bmatrix} 1 & a & 1 \ 2 & 1 & 0 \ a & 1 & 2 \end{bmatrix}$$

Many students freeze here, wondering how to handle the matrix $A$ when it is trapped inside this sum. Remember, the identity matrix $I$ is the most humble of all matrices; it is the 'do-nothing' operator.

To isolate $A$, we perform the matrix equivalent of subtraction: $A = (A + I) - I$. By subtracting the identity matrix—which has ones on the main diagonal and zeros elsewhere—we find our matrix $A$:

$$A = \begin{bmatrix} 0 & a & 1 \ 2 & 0 & 0 \ a & 1 & 1 \end{bmatrix}$$

### Phase 2: The Determinant Hunt

Now that we have $A$, we need to find the value of $a$. We are given that $\det(A) = -4$.

Let us expand the determinant along the first row. Expanding along the first row, we get:

$$\det(A) = 0(0 - 0) - a(2 - 0) + 1(2 - 0)$$

Simplifying this, we arrive at $\det(A) = -2a + 2$. Equating this to our given value of $-4$:

$$-2a + 2 = -4 \Rightarrow -2a = -6 \Rightarrow a = 3$$

We have successfully unmasked the variable!

### Phase 3: The Property Playground

Now, we face the target expression: $\det((a + 1) 	ext{adj}((a - 1)A))$. With $a = 3$, this becomes $\det(4 	ext{adj}(2A))$.

We must use the properties of determinants to dismantle this expression. First, the scalar property: $\det(kM) = k^n \det(M)$. Since our matrix is $3 	imes 3$, $n = 3$.

Pulling out the scalar $4$, we get:

$$4^3 \det(	ext{adj}(2A)) = 64 \det(	ext{adj}(2A))$$

Next, we tackle the adjoint. The property $\det(	ext{adj}(M)) = (\det(M))^{n-1}$ is our best friend here. With $n = 3$, the exponent is $2$.

Thus, $\det(	ext{adj}(2A)) = (\det(2A))^2$. Our expression now looks like $64 (\det(2A))^2$.

### Phase 4: The Final Calculation

We need $\det(2A)$. Using the scalar property again:

$$\det(2A) = 2^3 \det(A) = 8 	imes (-4) = -32$$

Now, substitute this back into our expression: $64 	imes (-32)^2$. To make the final step easy, let us work with powers of $2$.

$$64 = 2^6 \quad 	ext{and} \quad (-32)^2 = (2^5)^2 = 2^{10}$$

Multiplying these, we get $2^6 	imes 2^{10} = 2^{16}$.

Comparing this to the form $2^m 3^n$, we see that $m = 16$ and $n = 0$. The final result is:

**$m + n = 16$**

Select Answer:

Visualized Solution

Problem Overview

Finding Matrix $A$

Calculating $det (A)$

Solving for $a$

Target Expression Setup

Applying Scalar Property

Applying Adjoint Property

Evaluating $det (2 A)$

Final Power Calculation

Conclusion

The Sigma Insight: Adjoint and Inverse of a Matrix

The Matrix Mystery

Phase 1

Phase 2

Phase 3

Phase 4

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