Key Concepts and Formulas

1. Quadratic Forms and Skew-Symmetric Matrices: For a real square matrix $A$, if $X^T A X = O$ for all non-zero (or all) column vectors $X$, then $A$ must be a skew-symmetric matrix. A matrix $A$ is skew-symmetric if its transpose is its negative, i.e., $A^T = -A$.

   • Reasoning: Any square matrix $A$ can be uniquely decomposed into a symmetric part $A_{sym} = \frac{A+A^T}{2}$ and a skew-symmetric part $A_{skew} = \frac{A-A^T}{2}$. For any skew-symmetric matrix $M$, $X^T M X = 0$ because $X^T M X$ is a scalar, so $(X^T M X)^T = X^T M^T X = X^T (-M) X = -X^T M X$, which implies $2(X^T M X) = 0$, so $X^T M X = 0$. Thus, $X^T A X = X^T (A_{sym} + A_{skew}) X = X^T A_{sym} X + X^T A_{skew} X = X^T A_{sym} X$. Since $X^T A X = O$, we have $X^T A_{sym} X = O$ for all $X$. A fundamental result states that if a symmetric matrix $S$ satisfies $X^T S X = O$ for all $X$, then $S$ must be the zero matrix. Hence, $A_{sym} = O$, which implies $A = A_{skew}$, meaning $A$ is skew-symmetric.

2. General Form of a $3 \times 3$ Skew-Symmetric Matrix: For a $3 \times 3$ skew-symmetric matrix, the diagonal elements are zero, and $a_{ij} = -a_{ji}$. $A = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix}$

3. Determinant Properties:

   • For an $n \times n$ matrix $M$ and a scalar $k$, $\det(kM) = k^n \det(M)$.
   • For an $n \times n$ matrix $M$, $\det(\operatorname{adj}(M)) = (\det(M))^{n-1}$. This formula is valid for any invertible matrix $M$. If $M$ is singular (i.e., $\det(M)=0$), then $\det(\operatorname{adj}(M))=0$ for $n \ge 2$.

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Step-by-Step Solution

Step 1: Determine the nature of matrix A. The problem states that $X^T A X = O$ for all non-zero $3 \times 1$ matrices $X$. As established in Key Concept 1, this condition implies that $A$ is a skew-symmetric matrix. For a $3 \times 3$ skew-symmetric matrix, its general form is: $A = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix}$ where $a, b, c$ are real numbers.

Step 2: Use the given matrix equations to find the elements of A. We are provided with two matrix equations:

1. $A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \\ -5 \end{bmatrix}$
2. $A \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 4 \\ -8 \end{bmatrix}$

Let's substitute the general form of $A$ into these equations:

From the first equation: $\begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0(1)+a(1)+b(1) \\ -a(1)+0(1)+c(1) \\ -b(1)-c(1)+0(1) \end{bmatrix} = \begin{bmatrix} a+b \\ -a+c \\ -b-c \end{bmatrix}$ Equating this result to $\begin{bmatrix} 1 \\ 4 \\ -5 \end{bmatrix}$, we obtain a system of linear equations: (1) $a+b = 1$ (2) $-a+c = 4$ (3) $-b-c = -5 \implies b+c = 5$

From the second equation: $\begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix} \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 0(1)+a(2)+b(1) \\ -a(1)+0(2)+c(1) \\ -b(1)-c(2)+0(1) \end{bmatrix} = \begin{bmatrix} 2a+b \\ -a+c \\ -b-2c \end{bmatrix}$ Equating this result to $\begin{bmatrix} 0 \\ 4 \\ -8 \end{bmatrix}$, we get another system of equations: (4) $2a+b = 0$ (5) $-a+c = 4$ (This equation is identical to (2)) (6) $-b-2c = -8 \implies b+2c = 8$

Now, we solve the system of independent equations (1), (2), (4), and (6) for $a, b, c$: Subtract equation (1) from equation (4): $(2a+b) - (a+b) = 0 - 1 \implies a = -1$. Substitute $a=-1$ into equation (1): $(-1)+b = 1 \implies b = 2$. Substitute $a=-1$ into equation (2): $-(-1)+c = 4 \implies 1+c = 4 \implies c = 3$.

We can verify these values using equations (3) and (6): For (3): $b+c = 2+3 = 5$. This is consistent. For (6): $b+2c = 2+2(3) = 2+6 = 8$. This is also consistent.

Thus, the values for the elements are $a=-1, b=2, c=3$. The matrix $A$ is: $A = \begin{pmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{pmatrix}$

Step 3: Calculate $A+I$ and its determinant. First, we find the matrix $A+I$: $A+I = \begin{pmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{pmatrix} + \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -1 & 2 \\ 1 & 1 & 3 \\ -2 & -3 & 1 \end{pmatrix}$ Next, we calculate $\det(A+I)$ using cofactor expansion along the first row: $\det(A+I) = 1 \cdot \det \begin{pmatrix} 1 & 3 \\ -3 & 1 \end{pmatrix} - (-1) \cdot \det \begin{pmatrix} 1 & 3 \\ -2 & 1 \end{pmatrix} + 2 \cdot \det \begin{pmatrix} 1 & 1 \\ -2 & -3 \end{pmatrix}$ $= 1 \cdot ( (1)(1) - (3)(-3) ) + 1 \cdot ( (1)(1) - (3)(-2) ) + 2 \cdot ( (1)(-3) - (1)(-2) )$ $= 1 \cdot (1+9) + 1 \cdot (1+6) + 2 \cdot (-3+2)$ $= 1 \cdot 10 + 1 \cdot 7 + 2 \cdot (-1)$ $= 10 + 7 - 2 = 15$ So, $\det(A+I) = 15$.

Step 4: Calculate $\det(2(A+I))$. Using the determinant property $\det(kM) = k^n \det(M)$ with $k=2$ and $n=3$ (since $A$ is a $3 \times 3$ matrix): $\det(2(A+I)) = 2^3 \det(A+I) = 8 \times 15 = 120$

Step 5: Calculate $\det(\operatorname{adj}(2(A+I)))$. Using the determinant property $\det(\operatorname{adj}(M)) = (\det(M))^{n-1}$ with $M = 2(A+I)$ and $n=3$: $\det(\operatorname{adj}(2(A+I))) = (\det(2(A+I)))^{3-1} = (\det(2(A+I)))^2$ Substitute the value from Step 4: $= (120)^2$ To find the prime factorization, we factorize 120: $120 = 12 \times 10 = (2^2 \times 3) \times (2 \times 5) = 2^3 \times 3^1 \times 5^1$. So, $(120)^2 = (2^3 \times 3^1 \times 5^1)^2 = (2^3)^2 \times (3^1)^2 \times (5^1)^2 = 2^6 \times 3^2 \times 5^2$

Step 6: Determine $\alpha, \beta, \gamma$ and calculate $\alpha^2+\beta^2+\gamma^2$. We are given that $\det(\operatorname{adj}(2(A+I))) = 2^\alpha 3^\beta 5^\gamma$. Comparing this with our calculated result $2^6 \times 3^2 \times 5^2$: $\alpha = 6$ $\beta = 2$ $\gamma = 2$

The problem states that $\alpha, \beta, \gamma \in N$. In the context of JEE, $N$ typically refers to the set of positive integers $\{1, 2, 3, \ldots\}$. Our values $6, 2, 2$ are all positive integers, satisfying this condition.

Finally, we calculate $\alpha^2+\beta^2+\gamma^2$: $\alpha^2+\beta^2+\gamma^2 = 6^2 + 2^2 + 2^2 = 36 + 4 + 4 = 44$

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Common Mistakes & Tips

• Identifying Skew-Symmetry: A common pitfall is not recognizing that $X^T A X = O$ for all $X$ implies $A$ is skew-symmetric. This is a crucial first step.
• System of Linear Equations: Errors in solving the system of equations for $a, b, c$ can propagate throughout the solution. Double-check your arithmetic and substitutions.
• Determinant Calculation: Be meticulous with signs during cofactor expansion. A single sign error can lead to an incorrect determinant value.
• Properties of Adjoint and Determinants: Ensure you correctly apply the formulas $\det(kM) = k^n \det(M)$ and $\det(\operatorname{adj}(M)) = (\det(M))^{n-1}$. Misapplying the exponent for $k$ or $n-1$ is a frequent mistake.
• Prime Factorization: Carefully perform prime factorization of the final determinant value to correctly identify $\alpha, \beta, \gamma$.

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Summary

This problem is a comprehensive test of matrix properties, starting with the identification of a skew-symmetric matrix from a quadratic form, then solving a system of linear equations to determine its elements, and finally applying fundamental determinant and adjoint properties. The matrix $A$ is found to be skew-symmetric, leading to a system of equations whose solution determines $A$. Subsequently, $A+I$ is calculated, its determinant is found, and then the properties of determinants and adjoints are used to find the prime factorization of $\det(\operatorname{adj}(2(A+I)))$. From this, the exponents $\alpha, \beta, \gamma$ are determined, and their squared sum is computed.

The final answer is $\boxed{44}$.