Question #219092
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
2021-09-14T06:08:51-0400

Answer:-

(a) 2y22yz+2zx2xy(a) \ 2y^2-2yz+2zx-2xy

The matrix of the quadratic equation is;

A=011120102A=\begin{vmatrix} 0 & -1 &1\\ -1 & 2&0\\ 1&0 & -2 \end{vmatrix}


The characteristic roots are;

i.e

A=0λ1112λ0102λ=01112λ0102λA=\begin{vmatrix} 0-\lambda& -1 &1\\ -1 & 2-\lambda&0\\ 1&0 & -2-\lambda \end{vmatrix}=\begin{vmatrix} 0& -1 &1\\ -1 & 2-\lambda&0\\ 1&0 & -2-\lambda \end{vmatrix}


The characteristic vector for λ=0\lambda = 0 is given by;

i.e

1x2+x3=01x1+2x2=0x12x3=0-1x_2 +x_3 = 0\\ -1x_1 +2x_2 = 0\\ x_1 -2x_3 = 0


Solving for first 2,

x12=x2=x3\dfrac{x_1}{2}= -x_2 = x_3


giving the Eignen vector X1=k1(2,1,1)X_1 = k_1(2,-1,1)


When λ=3\lambda = 3, the characteristic vector becomes

4x22x3=04x11x2=02x15x3=0-4x_2 -2x_3 = 0\\ -4x_1 -1x_2 = 0\\ -2x_1 -5x_3 = 0


which can then become

4x2+2x3=04x1+1x2=02x1+5x3=04x_2+2x_3 = 0\\ 4x_1 +1x_2 = 0\\ 2x_1 +5x_3 = 0


solving again we get

x1=x24=x38x_1 = \dfrac{x_2}{4}= \dfrac{x_3}{8}


X2=k2(1,4,8)X_2 = k_2(1,4,8)


Similarly, X3=k(1,8,4)X_3= k(1,8,4)


X1.X2=X1.X3=X2.X3=0X_1.X_2 = X_1.X_3 = X_2.X_3= 0


The normalised vector is

A=1/32/31/34/31/38/31/38/34/3A=\begin{vmatrix} -1/3& 2/3 &1/3\\ 4/3 & 1/3&8/3\\ 1/3&8/3&4/3 \end{vmatrix}


b)

ab=sin(ab)a*b = sin(ab)

Let a be 2 and b be 5

sin(10°)=0.174sin(10) r=0.544sin(10°) = 0.174\\ sin(10)\ r = -0.544


since, sinab \neq N,

the operation is not binary





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