[ Systems of linear equations ]

Solve the systems of equations:

a) $x_{1}+x_{2}+x_{3}=0$

$x_{2}+x_{3}+x_{4}=0$

...

$x_{99}+x_{100}+x_{1}=0$,

$x_{100}+x_{1}+x_{2}=0$

b) $x+y+z=a$,

$y+z+t=b$,

$y+z+t=c$,

$t+x+y=d$

c) $x_{1}+x_{2}+x_{3}+x_{4}=2 a_{1}$,

$x_{1}+x_{2}-x_{3}-x_{4}=2 a_{2}$

$x_{1}-x_{2}+x_{3}-x_{4}=2 a_{3}$

$x_{1}-x_{2}-x_{3}+x_{4}=2 a_{4}$

d) $x_{1}+2 x_{2}+3 x_{3}+\ldots+n x_{n}=a_{1}$

$n x_{1}+x_{2}+2 x_{3}+\ldots+(n-1) n x_{n}=a_{2}$,

$\cdots$

$2 x_{1}+3 x_{2}+4 x_{3}+\ldots+x_{n}=a_{n}$
a) Subtracting the first equation from the second, we get $x_{4}=x_{1}$. Similarly, $x_{5}=x_{2}, x_{6}=x_{3}, \ldots, x_{3}=x_{100}$. Since 100 is not divisible by 3, it follows that

$x_{1}=x_{2}=x_{3}=\ldots=x_{100}$

b) Adding all the equations, we get $x+y+z+t=1 / 3(a+b+c+d)$. Subtracting each of the equations from this result, we get the answer

c) Adding all the equations, we get $4 x_{1}=2\left(a_{1}+a_{2}+a_{3}+a_{4}\right)$. Adding the first two equations and subtracting the last two, we similarly find $x_{2}$. And so on.

d) Let $s=x_{1}+x_{2}+\ldots+x_{n}$. Adding all the equations, we get $1 / 2 n(n+1) s=a_{1}+a_{2}+\ldots+a_{n}$. Subtracting the second equation from the first, we get

$s-n x_{1}=a_{1}-a_{2}$. Hence, $x_{1}=1 / n\left(s-a_{1}+a_{2}\right)$. The other unknowns are found similarly.

## Answer

a) $(0,0, \ldots, 0)$.

b) $x=1 / 3(a+b+c-2 d), y=1 / 3(a+b-2 c+d), z=1 / 3(a-2 b+c+d), t=1 / 3(b+c+d-2 a)$.

c) $x_{1}=1 / 2\left(a_{1}+a_{2}+a_{3}+a_{4}\right)$, $x_{2}=1 / 2\left(a_{1}+a_{2}-a_{3}-a_{4}\right), x_{3}=1 / 2\left(a_{1}-a_{2}+a_{3}-a_{4}\right), x_{4}=1 / 2\left(a_{1}-a_{2}-a_{3}+a_{4}\right)$.

d) $x_{1}=\frac{1}{n}\left(\frac{2\left(a_{1}+\ldots+a_{n}\right)}{n(n+1)}-a_{1}+a_{2}\right), x_{2}=\frac{1}{n}\left(\frac{2\left(a_{1}+\ldots+a_{n}\right)}{n(n+1)}-a_{2}+a_{3}\right)$

$x_{n}=\frac{1}{n}\left(\frac{2\left(a_{1}+\ldots+a_{n}\right)}{n(n+1)}-a_{n}+a_{1}\right)$.