Let A be a 3x3 matrix such that X^T A X = 0 for all nonzero 3x1 matrices X = [x, y, z]^T. If A [1, 1

Q: Why does $X^{T}AX=0$ imply skew-symmetry? Can't it be something else?

That is a brilliant question! If $X^{T}AX=0$, then $X^{T}AX = (X^{T}AX)^{T} = X^{T}A^{T}X$. This implies $X^{T}(A+A^{T})X = 0$. For this to hold for all $X$, the matrix $(A+A^{T})$ must be the zero matrix, meaning $A = -A^{T}$. This is the very definition of a skew-symmetric matrix. It is not just a coincidence; it is a fundamental property of quadratic forms!

Q: Do I really need to find the full matrix $A$ to solve this?

Yes, you do. While the properties of determinants are powerful, they only work on the specific matrix $A+I$. You cannot bypass the calculation of the elements $a, b,$ and $c$ because the determinant depends on the specific values within the matrix. Think of it as building the foundation before you can build the roof.

Q: Why is the determinant of $2(A+I)$ not just $2 \times \det(A+I)$?

This is a classic JEE trap! Remember, when you factor out a scalar $k$ from a determinant of an $n \times n$ matrix, it comes out as $k^{n}$. Since our matrix is $3 \times 3$, you must pull out $2^{3}$, which is $8$, not just $2$. Always check the dimension of the matrix before pulling out constants!

Question

Let A be a $3	imes3$ matrix such that $X^{T}AX=0$ for all nonzero $3	imes1$ matrices $X=\begin{bmatrix}x\ y\ z\end{bmatrix}$. If $A\begin{bmatrix}1\ 1\ 1\end{bmatrix}=\begin{bmatrix}1\ 4\ -5\end{bmatrix}$ and $A\begin{bmatrix}1\ 2\ 1\end{bmatrix}=\begin{bmatrix}0\ 4\ -8\end{bmatrix}$, and $\det(adj(2(A+I)))=2^{\alpha}3^{\beta}5^{\gamma}$, $\alpha, \beta, \gamma\in N,$ then $\alpha^{2}+\beta^{2}+\gamma^{2}$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you encounter a condition like $X^{T}AX=0$ for all non-zero $X$, recognize it as the hallmark of a **skew-symmetric matrix**. In such a matrix, the diagonal elements are zero, and the off-diagonal elements are negatives of their counterparts across the diagonal.

We define the matrix $A$ as:

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

By identifying this structure, we reduce the problem to finding the three variables $a$, $b$, and $c$.

### The Hunt for the Variables

We utilize the provided transformations to generate a system of linear equations. Given $A\begin{bmatrix}1\ 1\ 1\end{bmatrix}=\begin{bmatrix}1\ 4\ -5\end{bmatrix}$, we perform matrix multiplication to obtain:

$$a + b = 1$$
$$-a + c = 4$$
$$-b - c = -5$$

Using the second transformation $A\begin{bmatrix}1\ 2\ 1\end{bmatrix}=\begin{bmatrix}0\ 4\ -8\end{bmatrix}$, we obtain:

$$2a + b = 0$$

Solving this system, we subtract the first equation ($a+b=1$) from the fourth ($2a+b=0$) to find $a = -1$. Substituting $a$ back, we find $b = 2$ and $c = 3$. Thus, the matrix is:

$$A = \begin{bmatrix} 0 & -1 & 2 \ 1 & 0 & 3 \ -2 & -3 & 0 \end{bmatrix}$$

### The Determinant Dance

Next, we calculate the determinant of $A+I$. Adding the identity matrix $I$ to $A$ yields:

$$A+I = \begin{bmatrix} 1 & -1 & 2 \ 1 & 1 & 3 \ -2 & -3 & 1 \end{bmatrix}$$

Expanding the determinant along the first row:

$$\det(A+I) = 1(1 - (-9)) - (-1)(1 - (-6)) + 2(-3 - (-2))$$
$$\det(A+I) = 10 + 7 - 2 = 15$$

We now evaluate $\det(	ext{adj}(2(A+I)))$. Using the property $\det(kA) = k^{n}\det(A)$ for a $3 	imes 3$ matrix:

$$\det(2(A+I)) = 2^{3} 	imes \det(A+I) = 8 	imes 15 = 120$$

Applying the adjoint property $\det(	ext{adj}(M)) = (\det M)^{n-1}$, where $n-1 = 2$:

$$\det(	ext{adj}(2(A+I))) = 120^{2}$$

### The Final Victory

We express $120^{2}$ in terms of its prime factors:

$$120^{2} = (2^{3} 	imes 3^{1} 	imes 5^{1})^{2} = 2^{6} 	imes 3^{2} 	imes 5^{2}$$

Comparing this to $2^{\alpha}3^{\beta}5^{\gamma}$, we identify $\alpha = 6$, $\beta = 2$, and $\gamma = 2$. The final calculation is:

$$\alpha^{2} + \beta^{2} + \gamma^{2} = 6^{2} + 2^{2} + 2^{2} = 36 + 4 + 4 = 44$$

The final answer is **44**.

Enter Numerical Value:

Visualized Solution

Identify Matrix Type

Apply First Vector Product

Apply Second Vector Product

Solve for Variables $a, b, c$

Form Matrix $A + I$

Calculate $det (A + I)$

Determinant of $2 (A + I)$

Determinant of Adjoint

Prime Factorization

Final Calculation

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Hunt for the Variables

The Determinant Dance

The Final Victory

Similar Questions

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Enter Numerical Value:

Visualized Solution

Identify Matrix Type

Apply First Vector Product

Apply Second Vector Product

Solve for Variables .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a,b,c

Form Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A+I

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(A+I)

Determinant of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2(A+I)