Answer to Question #340436 in Linear Algebra for Micky

Question #340436

Determine whether the polynomial x²+2y²+4xy+2yz+6xz is a quadratic form and if so write it in the form X^T AX, where A is a symmetric matrix.

1
Expert's answer
2022-05-16T11:29:39-0400

By definition, quadratic form is a polynomial with terms all of degree two. Thus, "p=x^2+2y^2+4xy+2yz+6xz" is a quadratic form. Denote:"X=\\left(\\begin{array}{ll}x\\\\y\\\\z \\end{array}\\right)". The aim is to find the matrix "A=\\left(\\begin{array}{lll}a_{11}&a_{12}&a_{13}\\\\a_{12}&a_{22}&a_{23}\\\\a_{13}&a_{23}&a_{33}\\end{array}\\right)" satisfying "p=(x\\,\\,y\\,\\,z)\\left(\\begin{array}{lll}a_{11}&a_{12}&a_{13}\\\\a_{12}&a_{22}&a_{23}\\\\a_{13}&a_{23}&a_{33}\\end{array}\\right)\\left(\\begin{array}{ll}x\\\\y\\\\z \\end{array}\\right)".

The right side of the latter equality is: "2xya_{12}+2xza_{13}+2yza_{23}+a_{11}x^2+a_{22}y^2+a_{33}z^2".

By comparing the coefficients of the latter expression and "p", we get: "a_{11}=1", "a_{12}=2", "a_{13}=3", "a_{22}=2", "a_{23}=1", "a_{33}=0." Thus, "A=\\left(\\begin{array}{lll}1&2&3\\\\2&2&1\\\\3&1&0\\end{array}\\right)".

Answer: "p=x^2+2y^2+4xy+2yz+6xz" can be presented as: "p=X^{\\top}AX", where "X=\\left(\\begin{array}{ll}x\\\\y\\\\z \\end{array}\\right)", "A=\\left(\\begin{array}{lll}1&2&3\\\\2&2&1\\\\3&1&0\\end{array}\\right)".


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