# Determinant of the matrix associated with the quadratic form [closed]

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

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• Can you at least form $A$? – amd May 17 '18 at 7:04

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$(A)=-19$.