Answer to Question #232022 in Linear Algebra for Anuj

Question #232022
a. Find the orthogonal and normal canonical forms of 2y^2-2yz+2zx-2xy.


b. The operation,* defined by a*b= sin(ab), is a binary operation on N


True or false with full explanation
1
Expert's answer
2021-11-15T17:47:05-0500

1.f(x;y;z)=2y22yz+2zx2xy=(y22yz+z2)+(y22xy+x2)z2+2xzx2=(yz)2+(yx)2(zx)2.Let’s make the transformation of variables: yz=a;yx=b;zx=c.We have f(a;b;c)=a2+b2c2. It is the orthogonal and normal canonical form.2.For all a,bN there exists the unique value sin(ab).The domain of the function y=sin(x) is R , and by the properties of this function y=sin(x) it is unambiguous. So "" is a binary operation.1.f(x;y;z)=2y^2-2yz+2zx-2xy=(y^2-2yz+z^2)+(y^2-2xy+x^2)-z^2+2xz-x^2=(y-z)^2+ (y-x)^2-(z-x)^2. \\ \text{Let's make the transformation of variables: } y-z=a; y-x=b; z-x=c. \text{We have } f(a;b;c)=a^2+b^2-c^2.\text{ It is the orthogonal and normal canonical form.} \\ 2.\text{For all $a,b\in N$ there exists the unique value $\sin (ab)$} .\\ \text{The domain of the function $y=\sin(x)$ is $\R$ , and by the properties of this function $y=\sin(x)$ it is unambiguous. So $"*"$ is a binary operation.}


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