Question

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

Answer 1

Answer:

The given quadratic form is

"3x^2+5y^2+3z^2-2xy-2yz+2xz"

The matrix of the given quadratic form is

"A=\\begin{pmatrix}\n3 & -1 &1 \\\\\n-1& 5 & -1 \\\\\n1 & -1 & 3 \n\\end{pmatrix}"

We write , "A= IAI"

i,e, "\\begin {pmatrix}\n3&-1&1 \\\\\n-1&5&-1 \\\\\n1&-1&3\n\\end{pmatrix}" "=\\begin{pmatrix}\n1&0&0 \\\\\n0&1&0\\\\\n0&0&1\n\\end{pmatrix}" "A\\begin{pmatrix}\n1&0&0 \\\\\n0&1&0 \\\\\n0&0&1\n\\end{pmatrix}"

Now we shall reduce "A" to diagonal form by applying congruence operation on it . Performing "R_2\\rightarrow 3R_2+R_1,R_3\\rightarrow 3R_3-R_1;C_2\\rightarrow C_2+\\frac{1}{3}C_1,"

"C_3\\rightarrow C_3-\\frac{1}{3}C_1;R_3\\rightarrow 7R_3+R_2;C_3\\rightarrow C_3-\\frac{1}{7}C_2"

We get ,

"\\begin{pmatrix}\n3&0&0\\\\\n0&14&0\\\\\n0&0&54\n\\end{pmatrix}" "=" "\\begin{pmatrix}\n1&0&0 \\\\\n1&3&0\\\\\n-6&3&21\n\\end{pmatrix}" "A\\begin{pmatrix}\n1& \\frac{1}{3} & \\frac{-8}{21} \\\\\n0&1&\\frac{-1}{7} \\\\\n0&0&1 \n\\end{pmatrix}"

Step-by-step explanation:

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Answer 2

Answer: The matrix of this quadratic form is (3-10-12-10-13)\begin{pmatrix} 3 & -1 & 0-1 & 2 &-10 & -1&3 \end{pmatrix} (3-10-12-10-13)We have to find its eigenvalues and eigenvectors. detl3-x-10-12-x-10-13-xl-(3-x)2(2-x) -2(3-x)\det\begin{vmatrix) 3-x & -1 & 01-1 & 2-x &-110 & -1 & 3-x \end{vmatrix}=(3 x)^2(2-x)-2(3

x)det3-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 ((3-x)(2-x)-2)=(3-x)(x2-5x+4)=(3-x)(x-1) (x-4) Therefore, the eigenvalues are 1, 3 x)-2)=(3-x)(x^2-5x+4)=(3-x)(x-1)(x-4)=(3-X)

(x-4) Therefore, the eigenvalues are 1, 3

and 4. 1) Consider the eigenvalue 1:

(3-1-10-12-1-10-13-1)(xyz)= \begin{pmatrix} O\ O\ O \end{pmatrix} -1 & 0-1 &1 &-10 & -1 & 2 \end{pmatrix} \begin{pmatrix} xl yl z \end{pmatrix)= \begin{pm...

(000)\begin{pmatrix} 3-1 & -1 & 0\ -1 & 2-1

&-110 & -1 & 3-1 \end{pmatrix}

\begin{pmatrix} x\y\ z \end{pmatrix}=

((3-1-10-12-1-10-13-1)(xyz)-(000)(2-1

0-11-10-12)(xyz)=(000)\begin{pmatrix} 2 & -1 & 01-1&1 &-110 & -1 & 2 \end{pmatrix} \begin{pmatrix} x\y\ z \end{pmatrix}= \bigin{pm...

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