Equivalently, the SLE in matrix form is \(A\mathbf x=\boldsymbol0\). We abbreviate "homogeneous system of linear equations" by HSLE.

*Definition*. Let \(A\) be a matrix. Then its *rank* is the number of nonzero rows in the RREF matrix equivalent to \(A\). Note that this is the same as the number of nonzero rows in any echelon form matrix equivalent to \(A\).

*Definition*. Consider a HSLE \(A\mathbf x=\boldsymbol0\). It always has the *trivial solution* \(\mathbf x=\boldsymbol0\). Any other solution is called a *nontrivial solution*.

*Proposition*. Consider a HSLE \(A\mathbf x=\boldsymbol0\). It either has only the trivial solution or infinitely many solutions.

*Proof*. Suppose that \(\mathbf x_1\) is a nontrivial solution: \(\mathbf x_1\ne\boldsymbol0\), and \(A\mathbf x_1=\boldsymbol0\). Then for any scalar \(c\), \(\mathbf x=c\mathbf x_1\) is also a solution: \[
A(c\mathbf x_1)=c(A\mathbf x_1)=c\boldsymbol0=\boldsymbol0.
\]

*Theorem*. Let \(A\) be an \(m\times n\) matrix with rank \(r\). Then the solution set to the HSLE \(A\mathbf x=\boldsymbol0\) is an \((n-r)\)-parameter family of solutions.

at least one of \(b_1,\dotsc,b_m\ne0\). In other words, in the matrix form \(A\mathbf x=\mathbf b\), we have \(\mathbf b\ne\boldsymbol0\). We abbreviate "nonhomogeneous system of linear equations" by NHSLE.

*Proposition*. Consider a NHSLE \(A\mathbf x=\mathbf b\). Suppose that we have a solution \(A\mathbf x_0=\mathbf b\). Then \(\mathbf x_1\) is also a solution precisely when \(\mathbf x_1-\mathbf x_0\) is a solution to the HSLE \(A\mathbf x=\boldsymbol0\).

*Proof*. \(A(\mathbf x_1-\mathbf x_0)=A\mathbf x_1-A\mathbf x_0=A\mathbf x_1-\mathbf b\). This is \(\boldsymbol0\) precisely when \(A\mathbf x_1=\mathbf b\).

*Corollary*. Suppose that \(A\) is an \(m\times n\) matrix. Let \(r\) be the rank of \(A\). Suppose that \(A\mathbf x=\mathbf b\) has at least one solution. Then the solution set has \(n-r\) parameters.

*Theorem*. Consider a NHSLE \(A\mathbf x=\mathbf b\), where \(A\) is an \(m\times n\) matrix. Let \(p\) be the rank of \(A\), and let \(q\) be the rank of the augmented matrix \((A|\mathbf b)\). Then \(A\mathbf x=\mathbf b\) has a solution precisely when \(p=q\).

*Corollary*. Suppose that \(p=m\), that is in an echelon form matrix equivalent to \(A\) all the rows are nonzero. Then any SLE \(A\mathbf x=\mathbf b\) has a solution.

*Example*. Consider \[
(A|\mathbf b)=\left(\begin{array}{cccc|c}
-2 & 0 & 3 & 1 & -2 \\
3 & 1 & -2 & 0 & 1 \\
-1 & 1 & 4 & 2 & 0
\end{array}\right)\sim
\left(\begin{array}{cccc|c}
1 & -1 & -4 & -2 & 0 \\
0 & -2 & -5 & -3 & -2 \\
0 & 4 & 10 & 6 & 1
\end{array}\right)\sim
\left(\begin{array}{cccc|c}
1 & -1 & -4 & -2 & 0 \\
0 & -2 & -5 & -3 & -2 \\
0 & 0 & 0 & 0 & -3
\end{array}\right).
\] We can see that there is a contradiction, therefore \(A\mathbf x=\mathbf b\) has no solution. Note that \(A\) has rank 2, and \((A|\mathbf b)\) has rank 3.

*Theorem*. Let \(A\) be an \(m\times m\) matrix. Then the following statements are equivalent.

\(A\) is invertible.

For every \(m\)-vector \(\mathbf b\), the SLE \(A\mathbf x=\mathbf b\) has exactly one solution.

The HSLE \(A\mathbf x=\boldsymbol0\) has only the trivial solution.

The rank of \(A\) is \(m\).

The RREF matrix equivalent to \(A\) is \(I_m\).