Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given $$ M = \pmatrix{a &-1 &-1\\-1 &b &-1\\-1 &-1 &c},A = \pmatrix{e\\0\\-1} $$ and $a,b,c,e>0$:

  1. What requirements should $a,b,c$ meet such that: $MX = b$ has a non negative solution (all components of $X$ are nonnegative) , given $B = \pmatrix{b_1&b_2&b_3}^T, b_1\geq 0, b_2\geq ,b_3\geq 0$ and $\max(b_1,b_2,b_3)>0$.

  2. What requirements should $a,b,c,e$ meet such that $Mx = B$ has a solution $x = \pmatrix{x_1&x_2&x_3}^T$ given $B = \pmatrix{b_1&b_2&b_3}^T, b_1\geq 0, b_2\geq ,b_3\geq 0$ and $\max(b_1,b_2,b_3)>0$. And also satisfy that $MA = S = \pmatrix{s_1&s_2&s_3}^T$ if $x_i<0$ then $s_i <0$.

share|cite|improve this question
Smells like homework. If it is, please tag it as such. – ja72 Feb 24 '12 at 0:26

1 Answer 1


$ax_1-x_2-x_3=b_1 \geq 0 \Rightarrow a \geq \frac{x_2+x_3}{x_1}$

$-1x_1+bx_2-x_3=b_2 \geq 0 \Rightarrow b \geq \frac{x_1+x_3}{x_2}$

$-1x_1-1x_2+cx_3=b_3 \geq 0 \Rightarrow c \geq \frac {x_1+x_2}{x_3}$


$ae+1=s_1$ , if $s_1< 0 \Rightarrow a<\frac{-1}{e}$ ,which is not possible since $a> 0$

$-e+1=s_2$ , if $s_2 < 0 \Rightarrow e> 1$

$-e-c=s_3$ , if $s_3< 0 \Rightarrow c>-e$ ,which is always true since $c>0$

Like in the first case requirements that $a,b,c$ should meet are :

$a \geq \frac{x_2+x_3}{x_1} \land b \geq \frac{x_1+x_3}{x_2} \land c \geq \frac {x_1+x_2}{x_3}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.