Answer:-
( a )   2 y 2 − 2 y z + 2 z x − 2 x y (a) \ 2y^2-2yz+2zx-2xy ( a )   2 y 2 − 2 yz + 2 z x − 2 x y 
The matrix of the quadratic equation is;
A = ∣ 0 − 1 1 − 1 2 0 1 0 − 2 ∣ A=\begin{vmatrix}
   0 & -1 &1\\
   -1 & 2&0\\
1&0 & -2
\end{vmatrix} A = ∣ ∣  0 − 1 1  − 1 2 0  1 0 − 2  ∣ ∣  
The characteristic roots are;
i.e
A = ∣ 0 − λ − 1 1 − 1 2 − λ 0 1 0 − 2 − λ ∣ = ∣ 0 − 1 1 − 1 2 − λ 0 1 0 − 2 − λ ∣ A=\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} A = ∣ ∣  0 − λ − 1 1  − 1 2 − λ 0  1 0 − 2 − λ  ∣ ∣  = ∣ ∣  0 − 1 1  − 1 2 − λ 0  1 0 − 2 − λ  ∣ ∣  
The characteristic vector for λ = 0 \lambda = 0 λ = 0 
i.e
− 1 x 2 + x 3 = 0 − 1 x 1 + 2 x 2 = 0 x 1 − 2 x 3 = 0 -1x_2 +x_3 = 0\\
-1x_1 +2x_2 = 0\\
x_1 -2x_3 = 0 − 1 x 2  + x 3  = 0 − 1 x 1  + 2 x 2  = 0 x 1  − 2 x 3  = 0 
Solving for first 2,
x 1 2 = − x 2 = x 3 \dfrac{x_1}{2}= -x_2 = x_3 2 x 1   = − x 2  = x 3  
giving the Eignen vector X 1 = k 1 ( 2 , − 1 , 1 ) X_1 = k_1(2,-1,1) X 1  = k 1  ( 2 , − 1 , 1 ) 
When λ = 3 \lambda = 3 λ = 3 
− 4 x 2 − 2 x 3 = 0 − 4 x 1 − 1 x 2 = 0 − 2 x 1 − 5 x 3 = 0 -4x_2 -2x_3 = 0\\
-4x_1 -1x_2 = 0\\
-2x_1 -5x_3 = 0 − 4 x 2  − 2 x 3  = 0 − 4 x 1  − 1 x 2  = 0 − 2 x 1  − 5 x 3  = 0 
which can then become
4 x 2 + 2 x 3 = 0 4 x 1 + 1 x 2 = 0 2 x 1 + 5 x 3 = 0 4x_2+2x_3 = 0\\
4x_1 +1x_2 = 0\\
2x_1 +5x_3 = 0 4 x 2  + 2 x 3  = 0 4 x 1  + 1 x 2  = 0 2 x 1  + 5 x 3  = 0 
solving again we get
x 1 = x 2 4 = x 3 8 x_1 = \dfrac{x_2}{4}= \dfrac{x_3}{8} x 1  = 4 x 2   = 8 x 3   
X 2 = k 2 ( 1 , 4 , 8 ) X_2 = k_2(1,4,8) X 2  = k 2  ( 1 , 4 , 8 ) 
Similarly, X 3 = k ( 1 , 8 , 4 ) X_3= k(1,8,4) X 3  = k ( 1 , 8 , 4 ) 
X 1 . X 2 = X 1 . X 3 = X 2 . X 3 = 0 X_1.X_2 = X_1.X_3 = X_2.X_3= 0 X 1  . X 2  = X 1  . X 3  = X 2  . X 3  = 0 
The normalised vector is
A = ∣ − 1 / 3 2 / 3 1 / 3 4 / 3 1 / 3 8 / 3 1 / 3 8 / 3 4 / 3 ∣ A=\begin{vmatrix}
   -1/3& 2/3 &1/3\\
   4/3 & 1/3&8/3\\
1/3&8/3&4/3
\end{vmatrix} A = ∣ ∣  − 1/3 4/3 1/3  2/3 1/3 8/3  1/3 8/3 4/3  ∣ ∣  
b)
a ∗ b = s i n ( a b ) a*b = sin(ab) a ∗ b = s in ( ab ) 
Let a be 2 and b be 5
s i n ( 10 ° ) = 0.174 s i n ( 10 )   r = − 0.544 sin(10°) = 0.174\\
 sin(10)\ r = -0.544 s in ( 10° ) = 0.174 s in ( 10 )   r = − 0.544 
since, sinab ≠ \neq  = 
the operation is not binary
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