My Math Forum 3x3 Matrix

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 January 10th, 2016, 03:20 PM #1 Senior Member   Joined: Nov 2013 Posts: 137 Thanks: 1 3x3 Matrix Hi! I have a doubt... Look this matrix equation: $\begin{bmatrix} A\\ B \end{bmatrix} = \begin{bmatrix} +\frac{1}{\sqrt{2}} & +\frac{1}{\sqrt{2}}\\ +\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} X\\ Y \end{bmatrix}$ $\begin{bmatrix} X\\ Y \end{bmatrix} = \begin{bmatrix} +\frac{1}{\sqrt{2}} & +\frac{1}{\sqrt{2}}\\ +\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} A\\ B \end{bmatrix}$ By analogy, should exist a matrix 3x3 such that: $\begin{bmatrix} A\\ B\\ C\\ \end{bmatrix} = \begin{bmatrix} ? & ? & ?\\ ? & ? & ?\\ ? & ? & ?\\ \end{bmatrix} \begin{bmatrix} X\\ Y\\ Z\\ \end{bmatrix}$ $\begin{bmatrix} X\\ Y\\ Z\\ \end{bmatrix} = \begin{bmatrix} ? & ? & ?\\ ? & ? & ?\\ ? & ? & ?\\ \end{bmatrix} \begin{bmatrix} A\\ B\\ C\\ \end{bmatrix}$ So, what values need be replaced in ? for the matrix equation above be right? Thanks, Henry
 January 10th, 2016, 05:34 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,916 Thanks: 2199 $\begin{pmatrix} 1/3 & -2/3 & -2/3\\ -2/3 & 1/3 & -2/3\\ -2/3 & -2/3 & 1/3 \end{pmatrix}$ is a possibility. Thanks from Jhenrique
 January 11th, 2016, 04:06 AM #3 Senior Member   Joined: Nov 2013 Posts: 137 Thanks: 1 But, this is not the answer that I'm looking for... Would be the answer in this below pages? https://en.wikipedia.org/wiki/Quadra...nge_resolvents https://en.wikipedia.org/wiki/Cubic_...nge.27s_method https://en.wikipedia.org/wiki/Quarti...ange_resolvent
 January 11th, 2016, 07:47 AM #4 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 The OP question is: What is the square root of a 3x3 identity matrix? Given X, find M st A=MX and X=MA -> M^2=I OP gives M for 2x2 I, M^2=I checks
 January 11th, 2016, 09:25 AM #5 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 $\displaystyle M=\begin{vmatrix} 1/\sqrt{2} &1/\sqrt{2} &0 \\ 1/\sqrt{2} & -1/\sqrt{2} &0 \\ 0& 0 & 1 \end{vmatrix}\\$ by inspection. ref previous post. This is a square root, not the square root. Last edited by zylo; January 11th, 2016 at 09:29 AM.

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