A homogeneous polynomial of the second degree in `n` variables i.e., the expression
`phi=sum_(i=1)^(n)sum_(j=1)^(n)a_(ij)x_(i)x_(j)` where `a_(ij)=a_(ji)` is called a quadratic form in `n` variables `x_(1),x_(2)`….`x_(n)` if `A=[a_(ij)]_(nxn)` is a symmetric matrix and `x=[{:(x_(1)),(x_(2)),(x_(n)):}]` then
`X^(T)AX=[X_(1)X_(2)X_(3) . . . . .X_(n)][{:(a_(11),a_(12) ....a_(1n)),(a_(21),a_(22)....a_(2n)),(a_(n1),a_(n2)....a_(n n)):}][{:(x_(1)),(x_(2)),(x_(n)):}] =sum_(i=1)^(n)sum_(j=1)^(n)a_(ij)x_(i)x_(j)=phi`
Matrix A is called matrix of quadratic form `phi`.
Q. The quadratic form of matrix `A[{:(0,2,1),(2,3,-5),(1,-5,8):}]` is

Options

A. `3x_(2)^(2)+8x_(3)^(2)+2x_(1)x_(2)+x_(1)x_(3)-5x_(2)x_(3)`
B. `3x_(2)^(2)+8x_(3)^(2)+4x_(1)x_(2)+2x_(1)x_(3)-10x_(3)x_(2)`
C. `x_(1)^(2)+2x_(2)^(2)+x_(3)^(2)+3x_(1)x_(2)-5x_(2)x_(3)+8x_(1)x_(2)`
D. `3x_(1)^(2)+8x_(2)^(2)+4x_(1)x_(2)+2x_(1)x_(3)+10x_(3)x_(2)`

Explanation

To find the quadratic form of the given matrix \( A = \begin{bmatrix} 0 & 2 & 1 \\ 2 & 3 & -5 \\ 1 & -5 & 8 \end{bmatrix} \), we will follow these steps:

### Step 1: Define the vector \( x \)
We define the vector \( x \) as:
\[ x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \]

### Step 2: Compute \( x^T A \)
We need to compute the product \( x^T A \). The transpose of \( x \) is:
\[ x^T = \begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix} \]
Now, we multiply \( x^T \) by \( A \):
\[ x^T A = \begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix} \begin{bmatrix} 0 & 2 & 1 \\ 2 & 3 & -5 \\ 1 & -5 & 8 \end{bmatrix} \]

Calculating this product:
- The first element:
\[ 0 \cdot x_1 + 2 \cdot x_2 + 1 \cdot x_3 = 2x_2 + x_3 \]
- The second element:
\[ 2 \cdot x_1 + 3 \cdot x_2 - 5 \cdot x_3 = 2x_1 + 3x_2 - 5x_3 \]
- The third element:
\[ 1 \cdot x_1 - 5 \cdot x_2 + 8 \cdot x_3 = x_1 - 5x_2 + 8x_3 \]

Thus, we have:
\[ x^T A = \begin{bmatrix} 2x_2 + x_3 \\ 2x_1 + 3x_2 - 5x_3 \\ x_1 - 5x_2 + 8x_3 \end{bmatrix} \]

### Step 3: Compute \( x^T A x \)
Now, we need to compute \( x^T A x \):
\[ x^T A x = \begin{bmatrix} 2x_2 + x_3 \\ 2x_1 + 3x_2 - 5x_3 \\ x_1 - 5x_2 + 8x_3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \]

Calculating this product:
- The first term:
\[ (2x_2 + x_3) x_1 = 2x_1 x_2 + x_1 x_3 \]
- The second term:
\[ (2x_1 + 3x_2 - 5x_3) x_2 = 2x_1 x_2 + 3x_2^2 - 5x_2 x_3 \]
- The third term:
\[ (x_1 - 5x_2 + 8x_3) x_3 = x_1 x_3 - 5x_2 x_3 + 8x_3^2 \]

Combining all these terms, we get:
\[ x^T A x = (2x_1 x_2 + x_1 x_3) + (2x_1 x_2 + 3x_2^2 - 5x_2 x_3) + (x_1 x_3 - 5x_2 x_3 + 8x_3^2) \]

### Step 4: Simplify the expression
Combining like terms:
- Coefficient of \( x_1^2 \): 0
- Coefficient of \( x_2^2 \): \( 3x_2^2 \)
- Coefficient of \( x_3^2 \): \( 8x_3^2 \)
- Coefficient of \( x_1 x_2 \): \( 4x_1 x_2 \)
- Coefficient of \( x_1 x_3 \): \( 2x_1 x_3 \)
- Coefficient of \( x_2 x_3 \): \( -10x_2 x_3 \)

Thus, the quadratic form is:
\[ x^T A x = 3x_2^2 + 8x_3^2 + 4x_1 x_2 + 2x_1 x_3 - 10x_2 x_3 \]

### Final Answer
The quadratic form of the matrix \( A \) is:
\[ 3x_2^2 + 8x_3^2 + 4x_1 x_2 + 2x_1 x_3 - 10x_2 x_3 \]