Question

Is it possible that all solutions of a homogeneous system of ten linear equations in twelve variables are multiples of one fixed nonzero solution?

Discuss.

Answer 1

Discussing the asked concept:

Considering the system of linear equation according to the question

$$\begin{gathered} a_{1,1}{x}_1+a_{1,2}{x}_2+.......+a_{1,12}{x}_{12}=0 \\ {a}_{2,1}{x}_1+a_{2,2}{x}_2+.......+a_{1,2}{x}_{12}=0 \\ ... \\ ... \\ ... \\ {a}_{10,1}{x}_1+a_{10,2}{x}_2+.......+a_{10,12}{x}_{12}=0 \end{gathered}$$

Now we can write it in matrix form

$A=\left[\begin{array}{c}{a}_{1,1}.................a_{1,12}\\ .\\ .\\ .\\ .\\ a_{10,1}.................a_{10,12}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ .\\ .\\ .\\ .\\ x_{12}\end{array}\right]=b=\left[\begin{array}{c}0\\ 0\\ .\\ .\\ .\\ .\\ .\\ .\\ 0\end{array}\right]$

Therefore, $A$ is a $10\times 12$ matrix.

The matrix has $10$ rows and $12$ columns.

So $A$ has at most $10$ pivot positions, then the maximum rank of matrix $A$ is $10$
Now applying Rank Nullity theorem

$$\begin{array}{l}rankA+dim\mathrm{N}\mathrm{u}\mathrm{l}A=n\\ dim\mathrm{N}\mathrm{u}\mathrm{l}A=n-rankA\\ dim\mathrm{N}\mathrm{u}\mathrm{l}A=12-10\\ dim\mathrm{N}\mathrm{u}\mathrm{l}A=2\end{array}$$

Hence, every solution of the system $Ax=0$ is a linear combination of two linearly independent vectors in the null space of $A$ and one solution of $Ax=0$.

Therefore, we can not find a single vector in $NulA$ which spans $NulA$.

Hence, it is impossible that all solutions of a homogeneous system of ten linear equations in twelve variables are multiples of one fixed nonzero solution