There are $n$ equations and $n$ variables to solve. All the $n$ equations are of the form, for example...
If $n=3$:
$-x+y+z=a$
$x-y+z=b$
$x+y-z=c$
If $n=4:$
$-x+y+z+p=a$
$x-y+z+p=b$
$x+y-z+p=c$
$x+y+z-p=d$
here $a,b,c,d$ are constants.
How to solve this type of equations and calculate value of variables?
$\endgroup$ 14 Answers
$\begingroup$This problem is the starting point of linear algebra. There are different ways to solve such a system, any of which you can Google: Gaussian elimination and Cramer's rule are the two most common.
A good introduction to Linear Algebra is the book by Gilbert Strang.
$\endgroup$ $\begingroup$If n=3
$ [A|B] = \begin{bmatrix} -1 & 1 & 1 &|a\\ 1 & -1 & 1 &|b\\ 1 & 1 & -1 &|c\\ \end{bmatrix} $
Reduce it to Echelon form as
$ [A|B] = \begin{bmatrix} -1 & 1 & 1 &|a\\ 0 & 0 & 2 &|a+b\\ 0 & 2 & 0 &|a+c\\ \end{bmatrix} $
So we get
$-x+y+z=a$
$2z=a+b$
$2y=a+c$
and find value of x,y and z from above equations
If n=4
$ [A|B] = \begin{bmatrix} -1 & 1 & 1 & 1 &|a\\ 1 & -1 & 1 & 1 &|b\\ 1 & 1 & -1 & 1 &|c\\ 1 & 1 & 1 & -1 &|d\\ \end{bmatrix} $
Reduce it to Echelon form as
$ [A|B] = \begin{bmatrix} -1 & 1 & 1 & 1 &|a\\ 0 & 0 & 2 & 2 &|a+b\\ 0 & 2 & 0 & 2 &|a+c\\ 0 & 2 & 2 & 0 &|a+d\\ \end{bmatrix} $
So we get
$-x+y+z+p=a$
$2z+2p=a+b$
$2y+2p=a+c$
$2y+2z=a+d$
and find value of x,y,z and p from above equations
Similarly,do with other values of n
$\endgroup$ $\begingroup$These equations have a simple solution. A first step is to add up all the equations, and get the value of $x+y+z$, or $x+y+z+p$.
$\endgroup$ $\begingroup$Here, your matrix contains -1 in diagonal, and 1 on other locations.
So, solution is as follows,
For n = 3,
Add all the equations, it will give you,
(n-2)(x + y + z) = (a + b + c)(x+y+z) = (a+b+c)/(n-2)Now use value of (x+y+z), solve all variables.
For example,
-x+y+z = a => (x+y+z) - 2x = a => (a+b+c)/(n-2) - 2x = a => x = ( (a+b+c)/(n-2) - a )/2
Similarly you can calculate all variables for n = 4, 5, 6.......
Hope you clear with my explanations. If any doubt ask me.
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