# Linear Algebra solution space with real symmetric matrix question

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 $s_i <0$.

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$1)$

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

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

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

$2)$

$ae+1=s_1$ , 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 :

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

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