(a) 2y2−2yz+2zx−2xy
The matrix of the quadratic equation is;
A=∣∣0−11−12010−2∣∣
The characteristic roots are;
i.e
A=∣∣0−λ−11−12−λ010−2−λ∣∣=∣∣0−11−12−λ010−2−λ∣∣
The characteristic vector for λ=0 is given by;
i.e
−1x2+x3=0−1x1+2x2=0x1−2x3=0
Solving for first 2,
2x1=−x2=x3
giving the Eignen vector X1=k1(2,−1,1)
When λ=3, the characteristic vector becomes
−4x2−2x3=0−4x1−1x2=0−2x1−5x3=0
which can then become
4x2+2x3=04x1+1x2=02x1+5x3=0
solving again we get
x1=4x2=8x3
X2=k2(1,4,8)
Similarly, X3=k(1,8,4)
X1.X2=X1.X3=X2.X3=0
The normalised vector is
A=∣∣−1/34/31/32/31/38/31/38/34/3∣∣
b)
a∗b=sin(ab)
Let a be 2 and b be 5
sin(10°)=0.174sin(10) r=−0.544
since, sinab = N,
the operation is not binary
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