Vtyjkyuk

Consider the equations

$$sum_j=1^n a_ijx_j = b_i, i = 1,cdots, n-1$$

or equivalently, $Ax = b$ with $Ain mathbbR^(n-1)times n, bin
mathbbR^n-1$ and $xin mathbbR^n$ corresponding to $n-1$
linearly independent hyperplanes. The intersection of these
hyperplanes gives a line in $mathbbR^n$ in the form

$$y = x + lambda d$$

with $lambdain mathbbR$ and $x,dinmathbbR^n$. Find how to
chose $x$ and $d$.

I did an example of a system of two equations:

$$2x+3y+5z = 1$$
$$4x+y+8z=3$$

The intersection turned out to be

$$20x+42z=14$$

It's a line in $mathbbR^3$. But I can't see it in the way the exercise asks be to. I understand that I will always get a line, but how do I convert this line equation in the form the exercise asks?

I will work with the example you have given. First, note that your solution is not correct. You can see this easily, as it doesn't describe a line, but a plane(you have freedom in the $y$-direction).

The way of specifying a line as

$$x+lambda d$$

is called parametrization(as all solutions can be obtained using some value for the parameter $lambda$).

Let's first look at the real solution set, the said line, and how to obtain a parameterized solution right away. Taking your system of equations

$$beginpmatrix2 &3 &5\4 &1 &8endpmatrixbeginpmatrixx\y\zendpmatrix=beginpmatrix1\3endpmatrix$$

we may transform it using Gauss-Jordan elimination into

$$beginpmatrix2 &3 &5\0 &-5 &-2endpmatrixbeginpmatrixx\y\zendpmatrix=beginpmatrix1\1endpmatrix$$

We can now use back-substitution to determine the solutions, i.e. setting $z=c$ and determining the values of $x$ and $y$ in correspondence to $c$. You may then expand the solution vector, in the entries depending on $c$, to an equation $x+cy$ for vectors $x,y$ by pulling out the constants using vector addition and extracting $c$ using the laws of scalar multiplication.

Note, that you may convert the plane you've given as a supposed solution into a parametrized equation as well. For this, we just let $y=c,z=c'$ and can then determine $x$ in correspondence to $z$(and $y$ technically). Note, that as it is a plane, we've needed more parameters to describe the corresponding solutions.

In general, for a linear system of equations, for convenience given as $Ax=b$ for a corresponding coefficient matrix $A$ and solution vector $b$, the solution set $S(A,b)$, given a $xin S(A,b)$, can be specified as $x+S(A,mathbf0)=x+ymid yin S(A,mathbf0)$. The solution set $S(A,mathbf0)$ is commonly called the kernel of $A$. In other words, the solution set of an arbitrary system of equations $Ax=b$ can always be obtained by a linear shift of the solution set to the homogeneous system $Ax=mathbf0$ by any already found solution. I don't want to get into this too much, but I've already written an answer to a question asking what the solution space of a linear system is. It has slightly more depth so you might want to check this out.

In your case, the solution set $S(A,mathbf0)$ will always be a line(as the planes are linearly independent and you have $n-1$ of them, check this), i.e. a one-dimensional subspace of the underlying vector space. Now, as above, we can specify the general solution set $S(A,b)$ as $x+S(A,mathbf0)$ for an arbitrary $xin S(A,b)$. If $xneqmathbf0$(which happens iff $bneq 0$, check this as well), the set $S(A,b)$ is called an affine line, a linear shift of a linear subspace, as it does not contain the null vector anymore.

As $S(A,mathbf0)$ is always one dimensional, you may choose a one-element basis $d$ for $S(A,mathbf0)$. We may, given this basis, write

$$S(A,mathbf0)=lambda dmidlambdainmathbbR$$

as every element of $S(A,mathbf0)$ can be obtained as a linear combination of its basis and since it is one-dimensional, this just amounts to scalar multiplication. Thus

$$S(A,b)=x+S(A,mathbf0)=x+lambda dmidlambdainmathbbR$$

i.e. every element of $S(A,b)$ may be written as $x+lambda d$. Thus, to answer your question more generally, $x$ and $d$ in the specification of your line of solutions always correspond to any already found solution and to a non-null solution to the equation $Ax=mathbf0$ respectively. This solution will automatically be a basis as above, as the space is only $1$-dimensional.

For some further investigations, you may try to generalize these results, e.g. can you always describe a solution set to a linear system of equations as such a parametric equation, etc.