Question about quadratic form

Question

Suppose that $\mathbf{x}^{\prime}=(x, y, z)$ and $$\mathbf{A}=\left(\begin{array}{lll} a & d & e \\ d & b & f \\ e & f & c \end{array}\right)$$ Then find $\mathbf{x}^{\prime} \mathbf{A} \mathbf{x}$, which is called a "quadratic form."

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What I have tried, but I am doubtful about its accuracy: The equation $\mathbf{x}^{\prime} \mathbf{A} \mathbf{x}$ is a type of quadratic equation where the variables are represented by matrices and the coefficients are derived from a given matrix $\mathbf{A}$. In order to find $\mathbf{x}^{\prime} \mathbf{A} \mathbf{x}$ in this case, we must first calculate $\mathbf{A}^{\prime}$. After we have the value of $\mathbf{A}^{\prime}$, we can calculate $\mathbf{x}^{\prime} \mathbf{A} \mathbf{x}$ by $\mathbf{x}^{\prime} \mathbf{A} \mathbf{x} = \mathbf{x} \mathbf{A}^{\prime}$.

Proof: We can do this because they are both equal to the trace of the matrix $\mathbf{A}$. The trace of any matrix is the sum of the elements in the main diagonal. As such, $\mathbf{x}^{\prime} \mathbf{A} \mathbf{x}$ is equal to the sum of the elements in the main diagonal of the matrix $\mathbf{A}$, and $\mathbf{x} \mathbf{A}^{\prime}$ is equal to the same sum. Consequently, $\mathbf{x}^{\prime} \mathbf{A} \mathbf{x} = \mathbf{x} \mathbf{A}^{\prime}$.

Thanks for the help!

Answer 1

$\mathbf{x}^{\prime}$ is a $1 \times 3$ matrix; $\mathbf{x}$ is a $3 \times 1$ matrix. Multiply the matrices $\mathbf{x}^{\prime}A\mathbf{x}$. Your answer is the $1 \times 1$matrix whose only element is $$ax^2+by^2+cz^2+2dxy+2exz+2fyz$$

Answer 2

You can also use trace to obtain this result. $$ s= \mathbf{x}^T \mathbf{A} \mathbf{x} = \mathrm{tr} \left( \mathbf{x}^T \mathbf{A} \mathbf{x} \right) = \mathrm{tr} \left( \mathbf{A} \mathbf{x}\mathbf{x}^T \right) = \mathbf{A} : \mathbf{x}\mathbf{x}^T $$ where we use the colon operator : to denote the Frobenius inner product and the symmetry of $\mathbf{A}$.

Using your notations $$ s= \begin{pmatrix} a & d & e \\ d & b & f \\ e & f & c \end{pmatrix} : \begin{pmatrix} x^2 & xy & xz \\ yx & y^2 & yz \\ zx & zy & z^2 \end{pmatrix} $$ Since $\mathbf{A}:\mathbf{B} = \sum_{ij} A_{ij}B_{ij} $, the final result is easily found