Answer to Question #234684 in Calculus for Anuj

Question #234684

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
2021-09-13T00:30:10-0400

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

The matrix of the quadratic equation is;

"A=\\begin{vmatrix}\n 0 & -1 &1\\\\\n -1 & 2&0\\\\\n1&0 & -2\n\\end{vmatrix}"


The characteristic roots are;

i.e

"A=\\begin{vmatrix}\n 0-\\lambda& -1 &1\\\\\n -1 & 2-\\lambda&0\\\\\n1&0 & -2-\\lambda\n\\end{vmatrix}=\\begin{vmatrix}\n 0& -1 &1\\\\\n -1 & 2-\\lambda&0\\\\\n1&0 & -2-\\lambda\n\\end{vmatrix}"


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

i.e

"-1x_2 +x_3 = 0\\\\\n-1x_1 +2x_2 = 0\\\\\nx_1 -2x_3 = 0"


Solving for first 2,

"\\dfrac{x_1}{2}= -x_2 = x_3"


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


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

"-4x_2 -2x_3 = 0\\\\\n-4x_1 -1x_2 = 0\\\\\n-2x_1 -5x_3 = 0"


which can then become

"4x_2+2x_3 = 0\\\\\n4x_1 +1x_2 = 0\\\\\n2x_1 +5x_3 = 0"


solving again we get

"x_1 = \\dfrac{x_2}{4}= \\dfrac{x_3}{8}"


"X_2 = k_2(1,4,8)"


Similarly, "X_3= k(1,8,4)"


"X_1.X_2 = X_1.X_3 = X_2.X_3= 0"


The normalised vector is

"A=\\begin{vmatrix}\n -1\/3& 2\/3 &1\/3\\\\\n 4\/3 & 1\/3&8\/3\\\\\n1\/3&8\/3&4\/3\n\\end{vmatrix}"


b)

"a*b = sin(ab)"

Let a be 2 and b be 5

"sin(10\u00b0) = 0.174\\\\\n sin(10)\\ r = -0.544"


since, sinab "\\neq" N,

the operation is not binary



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS