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For a nonhomogeneous system of 2012 equations in 1999 unknowns, answer the following three questions:

  • Can the system be inconsistent?
  • Can the system have infinitely many solutions?
  • Can the system have a unique solution?
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    Heck, consider 3 equations in 2 unknowns. But don't try anything simpler than that, unless you're willing to accept $0x=0$ as an equation in one unknown.2012-11-03

1 Answers 1

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There's homogeneous in the title and nonhomogeneous in the text. In either case, we have a $2012 \times 1999$ matrix $A$. We are looking for solutions to $A\mathbf{x}=\mathbf{0}$ (homogeneous) or $A\mathbf{x}=\mathbf{b}$ where $\mathbf{b} \neq \mathbf{0}$ (inhomogeneous).

Homogeneous

We immediately have $A\mathbf{0}=\mathbf{0}$, so $\mathbf{x}=\mathbf{0}$ is a solution, so the system is consistent. (In fact, all homogeneous systems of linear equations are consistent due to the all-$0$ solution.)

By definition, it will have infinitely many solutions if and only if the null space of $A$ is non-trivial, i.e., if and only if the rank of $A$ is less than $1999$ (by the Rank-Nullity Theorem). One example is when $[A|\mathbf{0}]$ corresponds to the system of equations \begin{align*} x_1 &=0 \\ 2x_1 &=0 \\ 3x_1 &=0 \\ \vdots \\ 2012x_1 &=0. \end{align*}

By definition, it will have a unique solution (and this solution must be the zero solution) if and only if the null space of $A$ is trivial, i.e., if and only if the rank of $A$ is equal to $1999$. One example is when $A$ has $1$'s on the main diagonal and $0$'s elsewhere, which has rank $1999$. This is when $[A|\mathbf{0}]$ corresponds to the system of equations \begin{align*} x_1 &=0 \\ x_2 &=0 \\ \vdots \\ x_{1999} &=0 \\ 0 &=0 \\ \vdots \\ 0 &=0. \\ \end{align*}

Inhomogeneous

It can be inconsistent: e.g. when $[A|\mathbf{b}]$ corresponds to the system of equations \begin{align*} x_1+x_2+\cdots+x_{1999} &=1 \\ x_1+x_2+\cdots+x_{1999} &=2 \\ \vdots \\ x_1+x_2+\cdots+x_{1999} &=2012. \\ \end{align*}

It can have infinitely many solutions: e.g. when $[A|\mathbf{b}]$ corresponds to the system of equations \begin{align*} x_1 &=1 \\ 2x_1 &= 2 \\ 3x_1 &= 3 \\ \vdots \\ 2012x_1 &= 2012. \\ \end{align*}

It can have a unique solution: e.g. when $[A|\mathbf{b}]$ corresponds to the system of equations \begin{align*} x_1 &=1 \\ x_2 &=1 \\ \vdots \\ x_{1999} &=1 \\ 0 &=0 \\ \vdots \\ 0 &=0. \\ \end{align*}


Note: The answers to the second part changes if we work over finite fields.