ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Fri, 20 Apr 2012 03:34:47 +0200Output of solve - what does it mean?https://ask.sagemath.org/question/8909/output-of-solve-what-does-it-mean/After a lengthy bit of coding,
var('r,t')
eq = [corr[1][1].subs({r_11:r,t_12:t,t_13:t,t_21:t,r_22:r,t_23:t,t_31:t,t_32:t,r_33:r})==0]
solve(eq,r,t)
yields
$$ \newcommand{\Bold}[1]{\mathbf{#1}}\left(\left[r = -\sqrt{-\frac{\alpha_{1} t^{2}}{\alpha_{2}} - \frac{\alpha_{3} t^{2}}{\alpha_{2}} + 1}, r = \sqrt{-\frac{\alpha_{1} t^{2}}{\alpha_{2}} - \frac{\alpha_{3} t^{2}}{\alpha_{2}} + 1}, r = 0\right], \left[1, 1, 1\right]\right)$$
Note that the second element of the output is just `[1,1,1]`, instead of `t==1`. Can someone help with interpreting this?Fri, 20 Apr 2012 01:40:53 +0200https://ask.sagemath.org/question/8909/output-of-solve-what-does-it-mean/Answer by Shashank for <p>After a lengthy bit of coding, </p>
<pre><code>var('r,t')
eq = [corr[1][1].subs({r_11:r,t_12:t,t_13:t,t_21:t,r_22:r,t_23:t,t_31:t,t_32:t,r_33:r})==0]
solve(eq,r,t)
</code></pre>
<p>yields
$$ \newcommand{\Bold}[1]{\mathbf{#1}}\left(\left[r = -\sqrt{-\frac{\alpha_{1} t^{2}}{\alpha_{2}} - \frac{\alpha_{3} t^{2}}{\alpha_{2}} + 1}, r = \sqrt{-\frac{\alpha_{1} t^{2}}{\alpha_{2}} - \frac{\alpha_{3} t^{2}}{\alpha_{2}} + 1}, r = 0\right], \left[1, 1, 1\right]\right)$$
Note that the second element of the output is just <code>[1,1,1]</code>, instead of <code>t==1</code>. Can someone help with interpreting this?</p>
https://ask.sagemath.org/question/8909/output-of-solve-what-does-it-mean/?answer=13483#post-id-13483You are trying to solve a system with one equation and two unknowns. This cannot be done. So what sage is doing is assuming t=1 and then solving it. Fri, 20 Apr 2012 03:34:47 +0200https://ask.sagemath.org/question/8909/output-of-solve-what-does-it-mean/?answer=13483#post-id-13483