Word Count : 870

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We are to reduce the quadratic form to its orthogonal canonical form and find its principal axes. The quadratic form is: $$ Q(x,y,z) = -x^2 + y^2 + z^2 + 4xy + 4xz $$ --- ### Step 1: Write the quadratic form in matrix form The general quadratic form is $$ Q = [x\ y\ z] \begin{bmatrix} a_{11} & \frac{a_{12}}{2} & \frac{a_{13}}{2} \\ \frac{a_{12}}{2} & a_{22} & \frac{a_{23}}{2} \\ \frac{a_{13}}{2} & \frac{a_{23}}{2} & a_{33} \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} $$ Given: $$ Q(x,y,z) = -x^2 + y^2 + z^2 + 4xy + 4xz $$ Comparing: * $x^2$ coefficient: $-1$ * $y^2$ coefficient: $1$ * $z^2$ coefficient: $1$ * $2a_{12}xy = 4xy \Rightarrow a_{12} = 2$ * $2a_{13}xz = 4xz \Rightarrow a_{13} = 2$ * $yz$ term absent $\Rightarrow a_{23} = 0$ So the symmetric matrix is: $$ A = \begin{bmatrix} -1 & 2 & 2 \\ 2 & 1 & 0 \\ 2 & 0 & 1 \end{bmatrix} $$ --- ### Step 2: Find eigenvalues of $A$ Solve $\det(A - \lambda I) = 0$: $$ \begin{vmatrix} -1-\lambda & 2 & 2 \\ 2 & 1-\lambda & 0 \\ 2 & 0 & 1-\lambda \end{vmatrix} = 0 $$ Expand determinant along the first row: $$ (-1-\lambda)\begin{vmatrix} 1-\lambda & 0 \\ 0 & 1-\lambda \end{vmatrix} - ______ _____ _________ ___ ________ ______ __________ __________ _______ ____ ______ ____.
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