Inverse Matrices and Solving Systems

Definition The $n \times n$ identity matrix $I_n$ has $1$ in each entry of the main diagonal and $0$ everywhere else. For every $n \times n$ matrix $A$ and every $\mathbf{x}$ in $\mathbb{R}^n$

Definition Let $A$ be an $n \times n$ matrix. The matrix $A$ is invertible when there is an $n \times n$ matrix $B$ with

Such a $B$ is unique, because if $C$ also works then $B = B(AC) = (BA)C = C$. It is called the inverse of $A$ and written $A^{-1}$. If $A$ and $B$ are invertible $n \times n$ matrices, then $(A^{-1})^{-1} = A$, and $AB$ is invertible with

Definition Only square matrices have inverses. If $A$ is $m \times n$ with $m \ne n$, no matrix $B$ satisfies both $AB = I_m$ and $BA = I_n$. A square matrix $A$ is invertible exactly when it is nonsingular, that is, when $\det A \ne 0$. Equivalently, its reduced row echelon form is the identity matrix. For square matrices one product is enough to check, because if $A$ and $B$ are $n \times n$ and $AB = I_n$, then also $BA = I_n$ and $B = A^{-1}$.

Definition Let $A$ have rows $(a, b)$ and $(c, d)$. If $ad - bc \ne 0$, then

If $ad - bc = 0$, the matrix $A$ has no inverse. The formula works because

Definition Let $A$ be $n \times n$. Row reduce the $n \times 2n$ matrix $[A \mid I_n]$. If the left block becomes $I_n$, the right block is $A^{-1}$,

If instead a row of zeros appears in the left block, then $A$ is singular and has no inverse.

Definition A system of $m$ linear equations in $n$ variables can be written as the single matrix equation

where $A$ is the $m \times n$ coefficient matrix, $\mathbf{x}$ is the $n \times 1$ column of variables and $\mathbf{b}$ is the $m \times 1$ column of constants. The row-column product $A\mathbf{x}$ rebuilds the left sides of the equations, one row at a time.

Definition If $A$ is an invertible $n \times n$ matrix, then for every $\mathbf{b}$ in $\mathbb{R}^n$ the equation $A\mathbf{x} = \mathbf{b}$ has exactly one solution,

Multiplying both sides on the left by $A^{-1}$ shows that any solution must equal $A^{-1}\mathbf{b}$, and $A(A^{-1}\mathbf{b}) = \mathbf{b}$ shows that it is one. The inverse goes on the left, because $\mathbf{b}A^{-1}$ is not even defined for a column $\mathbf{b}$ when $n \ge 2$.

Definition The inverse of an $n \times n$ matrix $A$ is defined by the products $AA^{-1} = A^{-1}A = I_n$, not entry by entry. In general $A^{-1}$ is not the matrix of reciprocals $1/a_{ij}$, and a matrix with zero entries can still be invertible. Reciprocals do work for a diagonal matrix with nonzero diagonal entries $d_1, \ldots, d_n$. Its inverse is the diagonal matrix with diagonal entries $1/d_1, \ldots, 1/d_n$, and its zeros off the diagonal stay zeros. There is also no division of matrices. Write $A^{-1}B$ or $BA^{-1}$, which are usually different.