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Let A be 4 X 4 matrix, and let $\vec{b}$ and $\vec{c}$ be two vectors in R^4. We are told that the system $A \overrightarrow{x} = \overrightarrow{b}$ has a unique solution. What can you say about the number of solutions of the system A $\vec{x}$ = $\vec{c}$ ?

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    Well, we know that A is nonsingular. What does that tell you? – 2017-02-03

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Well, since there is a unique solution that means the matrix has full rank. Meaning it has a trivial kernel and is a representation matrix of an automorphism. (A one to one correspondence from a space to itself)

What that means in your case is, if $\vec c$ is $\vec 0$ then so is the solution $\vec x$ here for $\vec c$. (Here we used the trivial kernel.)

If it's not, then solution is unique since its an automorphism. (Here we used bijction.)

Also if $\vec x$ is the same in both equalities (I'm guessing it's not?), that would imply $\vec b= \vec c$.

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Hint: Let $x_b$ be the unique solution to $Ax = b\neq 0$. Suppose \begin{align} Ax = 0 \end{align} then we see that \begin{align} 0=A(x-x_b+x_b) = A(x-x_b)+Ax_b = A(x-x_b)+b \end{align} which means \begin{align} A(x_b-x) = b \end{align} but the uniqueness implies that $x_b-x=x_b$ which means $x=0$. Hence $A$ is injective (Since $A$ is a finite dimensional square matrix, so...).