Determinant of the matrix associated with the quadratic form If A is the matrix associated with the quadratic form $4x^2+9y^2+2z^2+8yz+6zx+6xy$ then what is the determinant of A? I don't know how to solve quadratic form of a matrix pls help me 

Consider the matrix $$A= \begin{pmatrix}
4 & 3 & 3 \\
3 & 9 & 4  \\
3 & 4 & 2  \\
\end{pmatrix}.$$
Note that the coefficients relating to $x^2,y^2$ and $z^2$ lie on its diagonal. The other entries correspond to half of the coefficients of the interaction terms. Multiplying this matrix with $\begin{pmatrix}
x&y&z \\
\end{pmatrix}^T$ and $\begin{pmatrix}
x\\y\\z \\
\end{pmatrix}$ yields
\begin{align} \begin{pmatrix}
x&y&z \\
\end{pmatrix}^T A \begin{pmatrix}
x\\y\\z \\
\end{pmatrix} &= \begin{pmatrix}
x&y&z \\
\end{pmatrix}^T  \begin{pmatrix}
4 & 3 & 3 \\
3 & 9 & 4  \\
3 & 4 & 2  \\
\end{pmatrix}\begin{pmatrix}
x\\y\\z \\
\end{pmatrix}\\
&=\begin{pmatrix}
x&y&z \\
\end{pmatrix}^T  \begin{pmatrix}
4x + 3y + 3z \\
3x + 9y + 4z  \\
3x + 4y + 2z  \\
\end{pmatrix}\\
&= 4x^2 +3xy + 3xz +3xy +9y^2+4yz+3xz+4yz +2z^2\\
&= 4x^2 +9y^2 +2z^2 +6xy + 6xz + 8yz.
\end{align}
Hence, $A$ is the matrix we are looking for which produces the desired function.
However, we are interested in the determinant of $A$. Luckily, $A$ is a $3 \times 3$ matrix. Hence, we can use the rule of Sarrus to calculate the determinant:
\begin{align}
\det(A)= A&= \begin{vmatrix}
4 & 3 & 3 \\
3 & 9 & 4  \\
3 & 4 & 2  \\
\end{vmatrix}\\
 &= 4\cdot 9 \cdot 2 +3\cdot4\cdot3 +3\cdot4\cdot3 - 3\cdot9\cdot3 -3\cdot3\cdot2 -4\cdot4\cdot4\\
&= 72 +36+36 - 81-18 -64\\
&=-19.
\end{align}
Hence, det$(A)=-19$.
