Suppose we have $m$ linear equations in the $n$ unknowns $x_{1}, \ldots, x_{n}$. The equations can conveniently be expressed as a single matrix equation $\mathbf{A} \mathbf{x}=\mathbf{b}$, where $\mathbf{A}$ is the $m \times n$ matrix of coefficients. The equation $\mathbf{A} \mathbf{x}=\mathbf{b}$ is said to be consistent if it has at least one solution; otherwise, it is inconsistent. The equation is homogeneous if $\mathbf{b}=\mathbf{0}$. The set of solutions of the homogeneous equation $\mathbf{A x}=\mathbf{0}$ is clearly the null space of $\mathbf{A}$. If the equation $\mathbf{A x}=\mathbf{b}$ is consistent, then we can write $$ \mathbf{b}=x_{1}^{0} \mathbf{a}{\mathbf{1}}+\cdots+x{n}^{0} \mathbf{a}{\mathbf{n}} $$ for some $x{1}^{0}, \ldots, x_{n}^{0}$, where $\mathbf{a}{1}, \ldots, \mathbf{a}{\mathbf{n}}$ are the columns of $\mathbf{A}$. Thus $\mathbf{b} \in \mathcal{C}(\mathbf{A})$ Conversely, if $\mathbf{b} \in \mathcal{C}(\mathbf{A})$, then $\mathbf{A x}=\mathbf{b}$ must be consistent. If the equation is consistent and if $\mathbf{x}^{0}$ is a solution of the equation, then the set of all solutions of the equation is given by $$ \left{\mathbf{x}^{\mathbf{0}}+\mathbf{x}: \mathbf{x} \in \mathcal{N}(\mathbf{A})\right} $$ Clearly, the equation $\mathbf{A x}=\mathbf{b}$ has either no solution, a unique solution, or infinitely many solutions.

A matrix $\mathbf{A}$ of order $n \times n$ is said to be nonsingular if $R(\mathbf{A})=n$; otherwise, the matrix is singular.


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