Solution MISLE.C28 Robert Beezer
Employ Theorem CINM, \begin{equation*} \begin{bmatrix} 1 & 1 & 3 & 1 & 1 & 0 & 0 & 0\\ -2 & -1 & -4 & -1 & 0 & 1 & 0 & 0\\ 1 & 4 & 10 & 2 & 0 & 0 & 1 & 0\\ -2 & 0 & -4 & 5 & 0 & 0 & 0 & 1 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 0 & 38 & 18 & -5 & -2\\ 0 & \leading{1} & 0 & 0 & 96 & 47 & -12 & -5\\ 0 & 0 & \leading{1} & 0 & -39 & -19 & 5 & 2\\ 0 & 0 & 0 & \leading{1} & -16 & -8 & 2 & 1 \end{bmatrix} \end{equation*} And therefore we see that $C$ is nonsingular ($C$ row-reduces to the identity matrix, Theorem NMRRI) and by Theorem CINM, \begin{equation*} \inverse{C}= \begin{bmatrix} 38 & 18 & -5 & -2\\ 96 & 47 & -12 & -5\\ -39 & -19 & 5 & 2\\ -16 & -8 & 2 & 1 \end{bmatrix} \end{equation*}