​​$$\text{For } a, b \in \mathbb{R}, \text{ let} \\p(x, y) = a^2x_1y_1 + abx_2y_1 + abx_1y_2 + b^2x_2y_2, \quad x = (x_1, x_2), \; y = (y_1, y_2) \in \mathbb{R}^2. \\\text{For what values of } a \text{ and } b \text{ does } p: \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R} \text{ define an inner product?}$$​​

​​​

Correct option is D

Result used : 

​$$\text{The function } p(x, y) \text{ is given as:} \\p(x, y) = a^2x_1y_1 + abx_2y_1 + abx_1y_2 + b^2x_2y_2 \\\text{This can be represented as a quadratic form:} \\p(x, y) = \begin{bmatrix}x_1 & x_2\end{bmatrix}\begin{bmatrix}a^2 & ab \\ab & b^2\end{bmatrix}\begin{bmatrix}y_1 \\y_2\end{bmatrix} \\\text{Here, the matrix associated with the bilinear form is:} \\M =\begin{bmatrix}a^2 & ab \\ab & b^2\end{bmatrix} \\\text{Step 1: Symmetry} \\\text{The matrix } M \text{ is symmetric since } M = M^T. \text{ Hence, } p(x, y) \text{ satisfies the symmetry condition.} \\\text{Step 2: Positivity} \\\text{To ensure } p(x, y) \text{ defines an inner product, the matrix } M \text{ must be positive definite.} \\\text{A matrix is positive definite if:} \\\text{All leading principal minors are positive.}$$​

Solution: 

​$$p(x, y) = a^2x_1y_1 + abx_2y_1 + abx_1y_2 + b^2x_2y_2 \\\text{coefficient matrix:} \quad \begin{bmatrix}a^2 & ab \\ab & b^2\end{bmatrix}, \quad \text{where } a_{ij} = \text{coefficient of } x_i y_j. \\\text{Now, if this matrix is positive definite, then } p(x, y) \text{ will define an inner product.} \\\text{Let us check the positivity of the leading principal minors.} \\\Delta_1 = [a^2] > 0 \; \text{(always)} \\\Delta_2 = \begin{vmatrix}a^2 & ab \\ab & b^2\end{vmatrix}= a^2b^2 - a^2b^2 = 0 \; \text{(always)} \\\text{Since } \Delta_2 \text{ is never positive, matrix will never be positive definite }\\ p(x, y) \text{ cannot define an inner product.} \\\text{Thus, there are no values of } a, b \text{ such that } p(x, y) \text{ defines an inner product.}$$​​