91Ó°ÊÓ

First, we need to rewrite the quadratic form q(x, y) in matrix notation. We are given: \[q(x, y)=2 x^{2}-6 x y-3 y^{2}\] Matrix notation for a quadratic form is given as: \[q(x, y) = \begin{bmatrix}x & y\end{bmatrix} A \begin{bmatrix}x\\y\end{bmatrix}\] To match the given quadratic form, we can find the matrix A as follows: \[A = \begin{bmatrix}2 & -3\\ -3 & -3\end{bmatrix}\] Thus, we have rewritten the quadratic form in matrix notation as: \[q(x, y) = \begin{bmatrix}x & y\end{bmatrix} \begin{bmatrix}2 & -3\\ -3 & -3\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}\] Matrix A that represents the quadratic form is: \[A = \begin{bmatrix}2 & -3\\ -3 & -3\end{bmatrix}\]

We are given the linear substitution as: \[x=s+2 t\] \[y=3 s-t\] Matrix notation for the linear substitution is given as: \[\begin{bmatrix}x\\y\end{bmatrix} = P \begin{bmatrix}s\\t\end{bmatrix}\] To match the given linear substitution, we can find the matrix P as follows: \[P = \begin{bmatrix}1 & 2\\ 3 & -1\end{bmatrix}\] Thus, we have rewritten the linear substitution in matrix notation as: \[\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}1 & 2\\ 3 & -1\end{bmatrix} \begin{bmatrix}s\\t\end{bmatrix}\] Matrix P that corresponds to the linear substitution is: \[P = \begin{bmatrix}1 & 2\\ 3 & -1\end{bmatrix}\]

Now we will find the transformed quadratic form q(s, t) using direct substitution. Substitute the given linear substitution into the quadratic form: \[q(s, t) = 2(s+2t)^2 - 6(s+2t)(3s-t) - 3(3s-t)^2\] Now expand and simplify the expression: \[q(s, t) = 2(s^2 + 4st + 4t^2) - 6(s^2 + 2st - 3st + 2t^2) - 3(9s^2 - 6st + t^2)\] \[q(s, t) = 2s^2 + 8st + 8t^2 - 6s^2 - 12st + 36st - 12t^2 - 27s^2 + 18st - 3t^2\] \[q(s, t) = -31s^2 + 50st - 7t^2\] Therefore, the transformed quadratic form q(s, t) is: \[q(s, t) = -31s^2 + 50st - 7t^2\]

Finally, we will find the transformed quadratic form q(s, t) using matrix notation. We know that: \[q(x, y) = \begin{bmatrix}x & y\end{bmatrix} A \begin{bmatrix}x\\y\end{bmatrix}\] And we have found that: \[\begin{bmatrix}x\\y\end{bmatrix} = P \begin{bmatrix}s\\t\end{bmatrix}\] Now substitute this linear substitution into the quadratic form: \[q(s, t) = \begin{bmatrix}s & t\end{bmatrix} P^T A P \begin{bmatrix}s\\t\end{bmatrix}\] We already found: \[P = \begin{bmatrix}1 & 2\\ 3 & -1\end{bmatrix}\] \[A = \begin{bmatrix}2 & -3\\ -3 & -3\end{bmatrix}\] Compute the product \(P^T A P\): \[P^T A P = \begin{bmatrix}1 & 3\\ 2 & -1\end{bmatrix} \begin{bmatrix}2 & -3\\ -3 & -3\end{bmatrix} \begin{bmatrix}1 & 2\\ 3 & -1\end{bmatrix}\] \[P^T A P = \begin{bmatrix}-7 & 6\\ 11 & -5\end{bmatrix} \begin{bmatrix}1 & 2\\ 3 & -1\end{bmatrix}\] \[P^T A P = \begin{bmatrix}-31 & 50\\ 50 & -7\end{bmatrix}\] Thus, the transformed quadratic form q(s, t) using matrix notation is: \[q(s, t) = \begin{bmatrix}s & t\end{bmatrix} \begin{bmatrix}-31 & 50\\ 50 & -7\end{bmatrix} \begin{bmatrix}s\\t\end{bmatrix}\] \[q(s, t) = -31s^2 + 50st - 7t^2\] Both direct substitution and matrix notation methods yield the same result for the transformed quadratic form q(s, t): \[q(s, t) = -31s^2 + 50st - 7t^2\]