Suppose we are given a linear system of equations as:
$$\begin{align} a_{11}x_1+a_{12}x_2+a_{13}x_3+...+a_{1n}x_n&=0 \\ a_{21}x_1+a_{22}x_2+...+a_{2n}x_n&=0 \\ \vdots \\ a_{m1}x_1+a_{m2}x_2+...+a_{mn}x_n&=0 \end{align}$$
$(a)$ Show that it must have at least one solution.

Answer: We can see the truth of the proposition by putting $(x_1,x_2,.....,x_n)=(0,0,.....,0)$.

$(b)$ Show that if it has two distinct solutions then, it has infinitely many solutions.

Answer: If $(x_1,x_2,......,x_n)$ and $(y_1,y_2,....,y_n)$ be distinct solutions of the equation. If it is so,then at least one of them must not be equal to $(0,0,.....,0)$.WLOG,let $(y_1,....,y_n)\neq (0,0,......,0)$.Already,
$a_{i1}y_1+a_{i2}y_2+......+a_{in}y_n=0$ for ($1\leq i \leq n$), and thus if we multiply $u \neq 0$ to both sides($u\neq 1$), we get $(uy_1,uy_2,....,uy_n)$ as the new solution. There can be infinite number of such $u's $ and thus,infinte such solutions.

$(c)$ If the system has exactly one solution,then the more general equation with $b_1,b_2,...,b_n$ as constants not necessarily all zero, will have at most one solution.

Answer: Let us assume the general system of equations to have more than one solutions in the case when the above given system has only one solution. If it is so, then the only solution to the above given system will be $(0,0,0,....,0)$.But if $(y_1,y_2,....,y_n)$ and $(x_1,x_2,.....,x_n)$ are two distinct solutions to the general form then,
we get that $a_{i1}y_1+a_{i2}y_2+......+a_{in}y_n=b_i$ and
$a_{i1}x_1+a_{i2}x_2+......+a_{in}x_n=b_i $(for $1 \leq i \leq n$). Subtracting these two
equations we get another solution $(y_1-x_1,y_2-x_2,......,y_n-x_n)$ to the above given equation. Thus, we get a contradiction. Hence the proposition in $(c)$ must be true.(As pointed out by @dxiv, the solution $(y_1-x_1,y_2-x_2,......,y_n-x_n)$ is distinct form $(0,0,0,...,0)$ because at $x_k \neq y_k$ for at least one $k$ since the solutions $(x_1,x_2,....,x_n)$ and $(y_1,y_2,......,y_n)$ are distinct solutions to more general equation. I am self-studying and I want someone to verify this. Please let me know if it is correct.

  • 2
    $\begingroup$ At $(c)$ you may add that the $x_i,y_j$ solutions being distinct means $y_k \ne x_k \iff y_k-x_k \ne 0$ for at least one $k$, which is necessary for the solution obtained by subtracting to be distinct from the all-$0$ solution. Other than that, it looks good (+1). $\endgroup$
    – dxiv
    Jul 30, 2021 at 2:52

1 Answer 1


Good job. Your working seems fine.

An extension of part b is that any convex combination of solutions of a linear system is a solution. Hence if you have two solutions, any point in between them is also a solution. This hold regardless of whether the system is homogeneous.


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