The equation of the path traversed by the ball headed by the footballer is $y = ax^2 + bx + c$; (where $0 \leq x \leq 14$ and $a, b, c \in \mathbb{R}$ and $a \neq 0$) with respect to a XY-coordinate system in the vertical plane. The ball passes through the points $(2, 15), (4, 25)$ and $(14, 15)$. Determine the values of $a, b$ and $c$ by solving the system of linear equations in $a, b$ and $c$, using matrix method. Also find the equation of the path traversed by the ball.

$a = -0.5, b = 8, c = 1$; Eq: $y = -0.5x^2 + 8x + 1$

The correct answer is Option (1) → $a = -0.5, b = 8, c = 1$; Eq: $y = -0.5x^2 + 8x + 1$ ##

$y = ax^2 + bx + c$

$15 = 4a + 2b + c$

$25 = 16a + 4b + c$

$15 = 196a + 14b + c$

The set of equations can be represented in the matrix form as $AX = B$, where

$A = \begin{bmatrix} 4 & 2 & 1 \\ 16 & 4 & 1 \\ 196 & 14 & 1 \end{bmatrix}, X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \text{ and } B = \begin{bmatrix} 15 \\ 25 \\ 15 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 4 & 2 & 1 \\ 16 & 4 & 1 \\ 196 & 14 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 15 \\ 25 \\ 15 \end{bmatrix}$

$|A| = 4(4 - 14) - 2(16 - 196) + 1(224 - 784)$

$= -40 + 360 - 560 = -240 \neq 0$

Hence $A^{-1}$ exists.

Now, $\text{adj}(A) = \begin{bmatrix} -10 & 180 & -560 \\ 12 & -192 & 336 \\ -2 & 12 & -16 \end{bmatrix}^T = \begin{bmatrix} -10 & 12 & -2 \\ 180 & -192 & 12 \\ -560 & 336 & -16 \end{bmatrix}$

$\begin{bmatrix} a \\ b \\ c \end{bmatrix} = -\frac{1}{240} \begin{bmatrix} -10 & 12 & -2 \\ 180 & -192 & 12 \\ -560 & 336 & -16 \end{bmatrix} \begin{bmatrix} 15 \\ 25 \\ 15 \end{bmatrix}$

$= -\frac{1}{240} \begin{bmatrix} 120 \\ -1920 \\ -240 \end{bmatrix}$

$∴a = -\frac{1}{2}, b = 8, c = 1$

So, the equation becomes

$y = -\frac{1}{2}x^2 + 8x + 1$