$求解矩阵方程AX+E=A^2+X,其中A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 1 & 6 & 1 \\ \end{bmatrix}.$
$解：由AX+E=A^2+X,有(A-E)X+A^2-E.$
$又|A-E|=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 6 & 0 \\ \end{bmatrix}=0,所以A-E不可逆.$
$用待定系数法求解,令X=\begin{bmatrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33} \\ \end{bmatrix}.$
$有\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 6 & 0 \\ \end{bmatrix}\begin{bmatrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33} \\ \end{bmatrix}=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 3 & 0 \\ 2 & 18 & 0 \\ \end{bmatrix}.$
$从而得\begin{bmatrix} 0 & 0 & 0 \\ x_{21} & x_{22} & x_{23} \\ x_{11}+6x_{21} & x_{12}+6x_{22} & x_{13}+6x_{23} \\ \end{bmatrix}=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 3 & 0 \\ 2 & 18 & 0 \\ \end{bmatrix}.$
$故$
$\left\{\begin{aligned} x_{21}=x_{23} =0 \\ x_{22}=3 \\ x_{11}+6 x_{21}=2 \\ x_{12}+6 x_{22}=18 \\ x_{13}+6 x_{23}=2 \end{aligned}\right.$
$于是$
$\left\{\begin{aligned} x_{11}=2 \\ x_{12}=0 \\ x_{13}=0 \\ x_{21}=0 \\ x_{22}=3 \\ x_{23}=0 \\ \end{aligned}\right.$
$所以$
$X=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ a & b & c \\ \end{bmatrix},其中a,b,c为任意常数.$