Let A be a 3 × 3 matrix such that A + Aᵀ = O. If A [1, -1, 0]ᵀ = [3, 3, 2]ᵀ, A² [1, -1, 0]ᵀ = [-3, 1

Q: Wait, why are the diagonal elements of a skew-symmetric matrix always zero?

Great question! By definition, a skew-symmetric matrix satisfies $A^T = -A$. If we look at the diagonal elements, where $i = j$, the condition becomes $a_{ii} = -a_{ii}$. The only number that is equal to its own negative is zero. So, $2a_{ii} = 0$, which forces $a_{ii} = 0$ for all $i$.

Q: Why can't I just calculate $A^2$ directly? It seems straightforward.

While you *could* calculate $A^2$ directly, it is a massive time-sink and prone to algebraic errors. In JEE Advanced, time is your most precious resource. By using the property $A^2 \mathbf{v} = A(A \mathbf{v})$, you leverage the work you already did in the first step. Always look for ways to reuse your previous results!

Q: I'm confused about the scalar property inside the determinant. Does it just come out as $k$?

Not quite! When you pull a scalar $k$ out of a determinant of an $n \times n$ matrix, it comes out as $k^n$. Since we are dealing with a $3 \times 3$ matrix, $n=3$, so the scalar $k$ comes out as $k^3$. Always remember to raise the scalar to the power of the matrix's order!

Question

Let A be a $3 	imes 3$ matrix such that $A + A^T = O$. If $A \begin{bmatrix} 1 \ -1 \ 0 \end{bmatrix} = \begin{bmatrix} 3 \ 3 \ 2 \end{bmatrix}, A^2 \begin{bmatrix} 1 \ -1 \ 0 \end{bmatrix} = \begin{bmatrix} -3 \ 19 \ -24 \end{bmatrix}$ and $\det(	ext{adj}(2 	ext{ adj}( A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma, \alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a $3 	imes 3$ matrix $A$. The problem provides a secret key: $A + A^T = O$.

This is not just an equation; it is a structural blueprint. It tells us that $A$ is a **skew-symmetric matrix**.

In the world of linear algebra, this is a beautiful constraint. It forces the diagonal elements to be zero and ensures that the elements mirrored across the diagonal are perfect negatives of each other. We can immediately write the general form:

$$A = \begin{bmatrix} 0 & -c & b \ c & 0 & -a \ -b & a & 0 \end{bmatrix}$$

This is our starting point, our canvas.

### Decoding the Unknowns

We are given two conditions involving a vector $\mathbf{v} = \begin{bmatrix} 1 \ -1 \ 0 \end{bmatrix}$. The first condition, $A \mathbf{v} = \begin{bmatrix} 3 \ 3 \ 2 \end{bmatrix}$, is our first clue.

By performing the matrix multiplication, we get a new vector:

$$\begin{bmatrix} c \ c \ -b-a \end{bmatrix} = \begin{bmatrix} 3 \ 3 \ 2 \end{bmatrix}$$

Equating these, we instantly find $c = 3$ and $a + b = -2$.

Now, for the second condition: $A^2 \mathbf{v} = \begin{bmatrix} -3 \ 19 \ -24 \end{bmatrix}$. Here is where the master's intuition kicks in. Do not calculate $A^2$; instead, recognize that $A^2 \mathbf{v} = A(A \mathbf{v})$.

Since we already know $A \mathbf{v} = \begin{bmatrix} 3 \ 3 \ 2 \end{bmatrix}$, we simply multiply our matrix $A$ (with $c=3$) by this vector:

$$A(A\mathbf{v}) = \begin{bmatrix} 0 & -3 & b \ 3 & 0 & -a \ -b & a & 0 \end{bmatrix} \begin{bmatrix} 3 \ 3 \ 2 \end{bmatrix} = \begin{bmatrix} -9 + 2b \ 9 - 2a \ -3b + 3a \end{bmatrix}$$

Comparing this to $\begin{bmatrix} -3 \ 19 \ -24 \end{bmatrix}$, we solve for $a$ and $b$. We find $b = 3$ and $a = -5$. A quick sanity check confirms our work: $-3(3) + 3(-5) = -9 - 15 = -24$.

### The Determinant Dance

With $a = -5, b = 3, c = 3$, our matrix $A$ is fully revealed. The problem asks us to work with $A + I$. Adding the identity matrix $I$ is like adding a gentle shift to the diagonal:

$$A+I = \begin{bmatrix} 1 & -3 & 3 \ 3 & 1 & 5 \ -3 & -5 & 1 \end{bmatrix}$$

Calculating the determinant of this matrix gives us $44$. Now, we address the grand finale: $\det(	ext{adj}(2 	ext{ adj}(A+I)))$.

Let $M = A+I$. We need to evaluate $\det(	ext{adj}(2 	ext{ adj } M))$. We use the property $\det(	ext{adj } B) = (\det B)^{n-1}$. With $n=3$, this becomes $(\det B)^2$.

Here, $B = 2 	ext{ adj } M$. So, the expression is $(\det(2 	ext{ adj } M))^2$. Using the scalar property $\det(kB) = k^n \det B$, we pull out the $2$ as $2^3 = 8$:

$$\det(2 	ext{ adj } M) = 8 \det(	ext{adj } M)$$

Since $\det(	ext{adj } M) = (\det M)^2$, the inner part becomes $8 (\det M)^2$. Squaring this, we get $64 (\det M)^4$.

### The Final Tally

We know $\det M = 44 = 2^2 \cdot 11$. Substituting this into our expression:

$$64 \cdot (2^2 \cdot 11)^4 = 2^6 \cdot (2^8 \cdot 11^4) = 2^{14} \cdot 11^4$$

Comparing this to the form $2^\alpha \cdot 3^\beta \cdot 11^\gamma$, we find $\alpha = 14, \beta = 0, \gamma = 4$.

The sum $\alpha + \beta + \gamma = 14 + 0 + 4 = \mathbf{18}$.

Enter Numerical Value:

Visualized Solution

Identify Skew-Symmetry

Apply First Condition

Solve for $c$ and $a + b$

Analyze Second Condition

Solve for $a$ and $b$

Construct Matrix $A + I$

Calculate $det (A + I)$

Determinant Property: Adjoint

Determinant Property: Scalar Multiplication

Combine and Simplify

Prime Factorization

Final Answer

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

Decoding the Unknowns

The Determinant Dance

The Final Tally

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