Question

If $x, y \in R, X = [x \quad y \quad 1]^T, A = \begin{bmatrix} 5 & 3 & -1 \\ 3 & 2 & -1 \\ -1 & -1 & 1 \end{bmatrix}$ and $X^TAX = 0$ then $2y + x = \_\_\_$

Answer 1

3

Answer 2

The given equation is $X^TAX = 0$. $X^TAX = \begin{bmatrix} x & y & 1 \end{bmatrix} \begin{bmatrix} 5 & 3 & -1 \\ 3 & 2 & -1 \\ -1 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ 1 \end{bmatrix}$ $X^TAX = \begin{bmatrix} x & y & 1 \end{bmatrix} \begin{bmatrix} 5x + 3y - 1 \\ 3x + 2y - 1 \\ -x - y + 1 \end{bmatrix}$ $X^TAX = x(5x + 3y - 1) + y(3x + 2y - 1) + 1(-x - y + 1)$ $X^TAX = 5x^2 + 3xy - x + 3xy + 2y^2 - y - x - y + 1$ $X^TAX = 5x^2 + 6xy + 2y^2 - 2x - 2y + 1$ Given $X^TAX = 0$, so $5x^2 + 6xy + 2y^2 - 2x - 2y + 1 = 0$. Let $k = 2y + x$. Then $x = k - 2y$. Substituting $x$ in the equation: $5(k - 2y)^2 + 6(k - 2y)y + 2y^2 - 2(k - 2y) - 2y + 1 = 0$ $5(k^2 - 4ky + 4y^2) + 6ky - 12y^2 + 2y^2 - 2k + 4y - 2y + 1 = 0$ $5k^2 - 20ky + 20y^2 + 6ky - 12y^2 + 2y^2 - 2k + 2y + 1 = 0$ $(20 - 12 + 2)y^2 + (-20k + 6k + 2)y + (5k^2 - 2k + 1) = 0$ $10y^2 + (-14k + 2)y + (5k^2 - 2k + 1) = 0$ For real solutions of $y$, the discriminant must be non-negative. $\Delta = (-14k + 2)^2 - 4(10)(5k^2 - 2k + 1) \ge 0$ $4(1 - 7k)^2 - 40(5k^2 - 2k + 1) \ge 0$ $4(1 - 14k + 49k^2) - 200k^2 + 80k - 40 \ge 0$ $4 - 56k + 196k^2 - 200k^2 + 80k - 40 \ge 0$ $-4k^2 + 24k - 36 \ge 0$ $k^2 - 6k + 9 \le 0$ $(k - 3)^2 \le 0$ Since $(k-3)^2$ cannot be negative, $(k-3)^2 = 0$, which implies $k = 3$. Therefore, $2y + x = 3$.