10. Reduce the Quadratic form2x2 + 2y2 + 2z2 - 2xy + 2xz - 2yz into the canonical form byOrthogonal transformation and discuss its nature.
Answers
We know that signature is the difference between the number of positive and negative square terms of a quadratic form. Therefore, we have signature = 2 – 1 = 1
Answer:
The given quadratic form is
3x^2+5y^2+3z^2-2xy-2yz+2xz3x
2
+5y
2
+3z
2
−2xy−2yz+2xz
The matrix of the given quadratic form is
A=\begin{pmatrix} 3 & -1 &1 \\ -1& 5 & -1 \\ 1 & -1 & 3 \end{pmatrix}A=
⎝
⎛
3
−1
1
−1
5
−1
1
−1
3
⎠
⎞
We write , A= IAIA=IAI
i,e, \begin {pmatrix} 3&-1&1 \\ -1&5&-1 \\ 1&-1&3 \end{pmatrix}
⎝
⎛
3
−1
1
−1
5
−1
1
−1
3
⎠
⎞
=\begin{pmatrix} 1&0&0 \\ 0&1&0\\ 0&0&1 \end{pmatrix}=
⎝
⎛
1
0
0
0
1
0
0
0
1
⎠
⎞
A\begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}A
⎝
⎛
1
0
0
0
1
0
0
0
1
⎠
⎞
Now we shall reduce AA 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,R
2
→3R
2
+R
1
,R
3
→3R
3
−R
1
;C
2
→C
2
+
3
1
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_2C
3
→C
3
−
3
1
C
1
;R
3
→7R
3
+R
2
;C
3
→C
3
−
7
1
C
2
We get ,
\begin{pmatrix} 3&0&0\\ 0&14&0\\ 0&0&54 \end{pmatrix}
⎝
⎛
3
0
0
0
14
0
0
0
54
⎠
⎞
== \begin{pmatrix} 1&0&0 \\ 1&3&0\\ -6&3&21 \end{pmatrix}
⎝
⎛
1
1
−6
0
3
3
0
0
21
⎠
⎞
A\begin{pmatrix} 1& \frac{1}{3} & \frac{-8}{21} \\ 0&1&\frac{-1}{7} \\ 0&0&1 \end{pmatrix}A
⎝
⎛
1
0
0
3
1
1
0
21
−8
7
−1
1
⎠
⎞