The given equation of the conicoid is: $$ 5x^2 + 4y^2 - 4yz + 2xz + 2x - 4y - 8z + 2 = 0 $$ A conicoid in general form is central if it can be written in the quadratic form: $$ \mathbf{x}^T A \mathbf{x} + 2\mathbf{B}^T \mathbf{x} + C = 0 $$ where $\mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, $A$ is the symmetric matrix of quadratic coefficients, $\mathbf{B}$ contains the linear coefficients. A conicoid is central if $A$ is non-singular (determinant ≠ 0) and the centre $\mathbf{x}_0 = -A^{-1}\mathbf{B}$ exists. ### Step 1: Identify the quadratic matrix $A$ and linear vector $\mathbf{B}$ Quadratic terms: $5x^2 + 4y^2 - 4yz + 2xz$ Write as symmetric matrix $A$: $$ A = \begin{bmatrix} 5 & 0 & 1 \\ 0 & 4 & -2 \\ 1 & -2 & 0 \end{bmatrix} $$ because the cross-terms are split: * $2xz \Rightarrow A_{13} = A_{31} = 1$ * $-4yz \Rightarrow A_{23} = A_{32} = -2$ Linear terms: $2x - 4y - 8z \Rightarrow \mathbf{B} = \begin{bmatrix} 1 \\ -2 \\ -4 \end{bmatrix}$ (half of the coefficients as per $2\mathbf{B}^T \mathbf{x}$) Constant: $C = 2$ ### Step 2: Check if the conicoid is central For a central conicoid, $\det A \neq 0$. Compute determinant of $A$: $$ \det A = \begin{vmatrix} 5 & 0 & 1 \\ 0 & 4 & -2 \\ 1 & -2 & 0 \end{vmatrix} = 5(4 \cdot