Answer to Question #94958 in Linear Algebra for Chief

Question #94958
Determine if the matrix p=
[√3/3, √6/6, -√2/2 ]
[-√3/3, √6/3, 0 ]
[√3/3, √6/6, √2/2] is othognal.
1
Expert's answer
2019-09-23T06:00:12-0400

To determine, whether the matrix A is orthogonal we should multiply matrix to its transpose, if the result is equal to the identity matrix, then A is orthogonal:

"A=\\begin {pmatrix}\n \\cfrac{1}{\\sqrt{3}} & \\cfrac{1}{\\sqrt{6}} & \\cfrac{-1}{\\sqrt{2}}\\\\\n \\cfrac{-1}{\\sqrt{3}}& \\cfrac{2}{\\sqrt{6}}& 0\\\\\n \\cfrac{1}{\\sqrt{3}} & \\cfrac{1}{\\sqrt{6}}& \\cfrac{1}{\\sqrt{2}}\n\\end{pmatrix}"

"A^T=\\begin{pmatrix}\n \\cfrac{1}{\\sqrt{3}} & \\cfrac{-1}{\\sqrt{3}} & \\cfrac{1}{\\sqrt{3}}\\\\ \n \\cfrac{1}{\\sqrt{6}}& \\cfrac{2}{\\sqrt{6}}& \\cfrac{1}{\\sqrt{6}}\\\\\n \\cfrac{-1}{\\sqrt{2}} & 0& \\cfrac{1}{\\sqrt{2}}\n\\end{pmatrix}"

"AA^T=" "\\begin{pmatrix}\n (\\cfrac{1}{3}+\\cfrac{1}{6}+\\cfrac{1}{2}) & (\\cfrac{-1}{3} + \\cfrac{2}{6})& (\\cfrac{1}{3}+\\cfrac{1}{6}-\\cfrac{1}{2})\\\\ \n (\\cfrac{-1}{3} + \\cfrac{2}{6})& (\\cfrac{1}{3}+\\cfrac{4}{6})& (\\cfrac{-1}{3} + \\cfrac{2}{6})\\\\\n (\\cfrac{1}{3}+\\cfrac{1}{6}-\\cfrac{1}{2}) & (\\cfrac{-1}{3} + \\cfrac{2}{6})& (\\cfrac{1}{3}+\\cfrac{1}{6}+\\cfrac{1}{2})\n\\end{pmatrix}=\n\\begin{pmatrix}\n1&0&0\\\\\n0&1&0\\\\\n0&0&1\n\\end{pmatrix}"

So A is an orthogonal matrix.


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