Question

Let A be a 4 × 4 matrix, and let $\overrightarrow{\mathbf{b}}$and$\overrightarrow{\mathbf{c}}$ be two vectors in${\mathbf{ℝ}}^{\mathbf{4}}$ . We are told that the system$\mathit{A}\overrightarrow{\mathbf{x}}\mathbf{=}\overrightarrow{\mathbf{b}}$ is inconsistent. What can you say about the number of solutions of the system$\mathit{A}\overrightarrow{\mathbf{x}}\mathbf{=}\overrightarrow{\mathbf{c}}$ ?

Short answer

The system $A\overrightarrow{\mathrm{x}}=\overrightarrow{\mathrm{c}}$will have infinitely many solutions or none, if the system $A\overrightarrow{\mathrm{x}}=\overrightarrow{b}$is inconsistent.

Step-by-step solution

Consider the system.

The consistent system will have unique solution or infinitely many solutions with at least one free variable.

The inconsistent system will have no solution or infinitely many solutions with no free variables.

A is a 4 × 4 matrix,$\overrightarrow{b}$ and $\overrightarrow{c}$and are two vectors in ${\mathrm{ℝ}}^4$and the system $A\overrightarrow{\mathrm{x}}=\overrightarrow{\mathrm{c}}$is inconsistent.

Verify consistency of the system.

As the system $A\overrightarrow{x}=\overrightarrow{b}$is inconsistent, thus, the system may have infinitely many solutions or no solution.

As $\overrightarrow{b}$and $\overrightarrow{c}$are the vectors in ${\mathrm{ℝ}}^4$, thus, if $A\overrightarrow{x}=\overrightarrow{b}$is inconsistent, then $A\overrightarrow{x}=\overrightarrow{c}$will have infinitely many solutions or none.

Final answer.

As the system $A\overrightarrow{x}=\overrightarrow{b}$is inconsistent, then, the system $A\overrightarrow{x}=\overrightarrow{c}$will have infinitely many solutions or none.