$A$ is $3 \times 4$-matrix of rank 3. Then the system of equations $Ax=b$ has exactly one solution. True or False?

Question

I think it's false.

because if $b=0$, then as $\dim(\text{kernel})=1$ we get non-zero vectors $x$ with $Ax=0$.

Is above solution correct?

If $b \neq 0$, then what does it mean when I combine it with the fact that $\dim(\text{kernel})=1$?

Answer 1

You are right, the statement is false.

The situation is as follows : Let $A$ be a $m\times n$-matrix and $A'$ the concatenation of the matrix $A$ and the vector $b$ (The vector $b$ is an additional column). Let $rank(A)=u$ and $rank(A')=v$

• If $u\ne v$, $Ax=b$ has no solution. This case cannot occur, when $b=0$.

• If $u=v=n$ , $Ax=b$ has a unique solution. $Ax=0$ has only the trivial solution.

• If $u=v<n$ , $Ax=b$ has infinite many solutions. We can choose $n-u$ variables arbitarily. $Ax=0$ has non-trivial solutions.

The general solution of the equation $Ax=b$ is the general solution of $Ax=0$ added to a special solution of $Ax=b$ (if existent). If $m<n$ and $rank(A)=m$ (full rank, our case), the number of solutions is always infinite.

Answer 2

Since $rank(A)=3$ there exists a vector $a_1$ of dimension 4 that is normal to the rows of $A$ that defines the nullspace of $A$ i.e. $Aa_1=0$. Then the solutions of $Ax=b$ are given by $$x=A^T(AA^T)^{-1}b+ma_1$$ parametrized by $m\in\mathbb{R}$.