The relation between $\det(A)$ and $Ax=b$

Question

I wish to understand the the relation between $\det(A)$ and $Ax=b$.

Hence, what are the solutions for $Ax=b$ (homogeneous and non-homogeneous) when $\det(A)=0$ and $\det(A)\ne 0$?

So, all in all there are $4$ cases to cover.

Thanks!

Answer 1

If $\det A\neq 0$, then $A$ is invertible, so $Ax=b$ has the unique solution $x=A^{-1}b$, and that applies in both the homogeneous case, where we get $x=0$, and in the non-homogeneous case.

If $\det A = 0$, then the homogeneous equation has more than just the trivial solution, i.e., there are infinitely many solutions to $Ax=0$. In this case, the non-homogeneous $Ax=b$ can either be inconsistent, or it can have infinitely many solutions. Which of these two happens depends on whether $b$ is in the column-space of $A$. In the invertible case, the column-space of $A$ is all of $\mathbb{R}^n$, so every $b$ is in it.

Answer 2

If $\det A \ne 0$, then $A$ is invertible, and $A\mathbf{x} = \mathbf{b}$ has the unique solution $\mathbf{x} = A^{-1}\mathbf{b}$.

If $\det A = 0$, then the homogeneous equation $A\mathbf{x} = \mathbf{0}$ has an infinite number of solutions (some hyperplane in $\mathbb{R}^n$, or whatever field you are working over), and the solutions to the nonhomogeneous equation $A\mathbf{x} = \mathbf{b}$ are determined by finding any single solution (if there are any) and adding all the solutions of the homogeneous equation.

Answer 3

I assume here that $A$ is a matrix, $b$, is a column vector, and we're solving $Ax = b$ for the column vector $x$.

If $\det(A) \neq 0$, then $Ax = b$ will have a unique solution.

If $\det(A) = 0$, then $Ax = b$ will either have no solution or infinitely many solutions.

In particular, the equation $Az = \vec 0$ will always have a solution since we can set $z = \vec 0$. The solutions to $Ax = b$ can be characterized as $x = x_0 + z$, where $x_0$ is some vector satisfying $Ax_0 = b$ and $z$ is any vector satisfying $Az = \vec 0$.

Answer 4

$$\det(A)\neq 0 \iff \forall b, \ \exists!   x \ :\ Ax=b$$