Question

I have done all works, please help me to check, the detailed steps have been added in my replys

a) Express the quadratic form
(X1)^2 + (X2)^2 - 4(X3)^2 + 6(X1)(X3) -8(X2)(X3)
in the matrix notation xTAx.

b) Classify the quadratic form as positive definite, negative definite or indefinite.

Answer 1

A)-----

\[x_{1}^{2} + x _{2}^{2} - 4x _{3}^{2} + 6x_{1}x_{3} - 8x_{2}x_{3} \]=
\[x_{1}^{2} + x _{2}^{2} - 4x _{3}^{2} + 0x_{1}x_{2} + 6x_{1}x_{3} - 8x_{2}x_{3} \] so
\[a_{1} = 1, a_{2} = 1, a_{3} = -4, a_{4} = 0, a_{5} = 6, a_{6} = -8\]
As\[x^{T}Ax = \left[ x_{1} x_{2} x_{3}\right] \left[\begin{matrix}a_{1} & a_{4}/2 & a_{5}/2\\ a_{4}/2 & a_{2} & a_{6}/2\\ a_{5}/2 & a_{6}/2 & a_{3}\end{matrix}\right] \left[\begin{matrix} x_{1} \\ x_{2}\\x_{3} \end{matrix}\right]\]so

\[A=\left[\begin{matrix}1 & 0 &3 \\ 0&1&-4\\3 &-4&-4\end{matrix}\right]\]

B)-----
\[A_{1} =1\]\[A_{2} = \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]\]\[A_{3}=\left[\begin{matrix}1 & 0 &3 \\ 0&1&-4\\3 &-4&-4\end{matrix}\right]\]
\[\det[A_{1}]=1\] \[\det[A_{2}]=1\] \[\det[A_{3}]=-29\]

so A is indefinite, is this enough to proof A is indefinite?