Solution of Linear System by Matrix Inversion

Theorem

Let

• $A$ : Invertible $n\times n$ matrix
• $\mathbf{b}$ : Any $n\times 1$ matrix

Then the linear system $A\mathbf{x}=\mathbf{b}$ has exactly one solution, namely,

$$\mathbf{x}=A^{-1}\mathbf{b}$$

Proof

Let

• $A$ : Invertible $n\times n$ matrix
• $\mathbf{b}$ : Any $n\times 1$ matrix

Since $A(A^{-1}\mathbf{b})=(A{A}^{-1})\mathbf{b}=I\mathbf{b}=\mathbf{b}$, by definition of solution of linear system, it follows that $\mathbf{x}=A^{-1}\mathbf{b}$ is a solution of $A\mathbf{x}=\mathbf{b}$.

Let ${\mathbf{x}}_0$ be any solution of $A\mathbf{x}=\mathbf{b}$. Then $A{\mathbf{x}}_0=\mathbf{b}$. Further,

$$\begin{array}{rl}A{\mathbf{x}}_0& =\mathbf{b}\\ A^{-1}A{\mathbf{x}}_0& =A^{-1}\mathbf{b}\\ {\mathbf{x}}_0& =A^{-1}\mathbf{b}\end{array}$$

We have shown that if ${\mathbf{x}}_0$ is an arbitrary solution of $A\mathbf{x}=\mathbf{b}$, then ${\mathbf{x}}_0=A^{-1}\mathbf{b}$. Therefore, $A^{-1}\mathbf{b}$ is the only solution of $A\mathbf{x}=\mathbf{b}$.