Question

Reduce the quadratic form 3x2 + 2y2 + 3z2 – 2xy – 2yz into a canonical
form using an orthogonal transformation.

Answer 1

Answer:

relationship whenever y'all down through

Answer 2

Answer:- The matrix of this quadratic form is (3−10−12−10−13)\begin{pmatrix} 3 & -1 & 0\ -1 & 2 &-1\ 0 & -1 &3 \end{pmatrix}⎝⎛3−10−12−10−13⎠⎞We have to find its eigenvalues and eigenvectors. det⁡∣3−x−10−12−x−10−13−x∣=(3−x)2(2−x)−2(3−x)\det\begin{vmatrix} 3-x & -1 & 0\ -1 & 2-x &-1\ 0 & -1 & 3-x \end{vmatrix}=(3-x)^2(2-x)-2(3-x)det∣∣∣∣∣∣∣3−x−10−12−x−10−13−x∣∣∣∣∣∣∣=(3−x)2(2−x)−2(3−x) =(3−x)((3−x)(2−x)−2)=(3−x)(x2−5x+4)=(3−x)(x−1)(x−4)=(3-x)((3-x)(2-x)-2)=(3-x)(x^2-5x+4)=(3-x)(x-1)(x-4)=(3−x)((3−x)(2−x)−2)=(3−x)(x2−5x+4)=(3−x)(x−1)(x−4) Therefore, the eigenvalues are 1, 3 and 4. 1) Consider the eigenvalue 1: (3−1−10−12−1−10−13−1)(xyz)=(000)\begin{pmatrix} 3-1 & -1 & 0\ -1 & 2-1 &-1\ 0 & -1 & 3-1 \end{pmatrix} \begin{pmatrix} x\ y\ z \end{pmatrix}= \begin{pmatrix} 0\ 0\ 0 \end{pmatrix}⎝⎛3−1−10−12−1−10−13−1⎠⎞⎝⎛xyz⎠⎞=⎝⎛000⎠⎞(2−10−11−10−12)(xyz)=(000)\begin{pmatrix} 2 & -1 & 0\ -1 &1 &-1\ 0 & -1 & 2 \end{pmatrix} \begin{pmatrix} x\ y\ z \end{pmatrix}= \begin{pm…

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