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{\bf Question}

Find the inverses of the following matrices using cofactors:

$$A = \left( \begin{array}{cccc}
 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}
 \right) \hspace{.2in} B = \left( \begin{array}{ccc} 2 & 1 & 3 \\ 1 & 1 & 2 \\ 2 & 1 & 6
\end{array} \right)$$

\vspace{.25in}

{\bf Answer}

Inverse of $A = \left( \begin{array}{cccc} 1 & 0 & 0 & 1 \\ 1 & 0
& 1 & 0\\1 & 1 & 0 & 0\\1 & 0 & 0 & 0 \end{array} \right)$

det(a) = 1

$A^{-1} = \frac {adj(A)}{|A|}$


Matrix of cofactors
\begin{eqnarray*} [A_{ij}] & = & \left[ \begin{array}{cccc}
\left|\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 &0 \\ 0 & 0 & 0
\end{array}\right| &
-\left|\begin{array}{ccc} 1 & 1 & 0\\ 1 & 0 &0 \\ 1 & 0 & 0
\end{array}\right| &
\left|\begin{array}{ccc} 1 & 0 & 0\\ 1 & 1 &0 \\ 1 & 0 & 0
\end{array}\right| &
-\left|\begin{array}{ccc} 1 & 0 & 1\\ 1 & 1 &0 \\ 1 & 0 & 0
\end{array}\right| \\ \\
-\left|\begin{array}{ccc} 0 & 0 & 1\\ 1 & 0 &0 \\ 0 & 0 & 0
\end{array}\right| &
\left|\begin{array}{ccc} 1 & 0 & 1\\ 1 & 0 &0 \\ 1 & 0 & 0
\end{array}\right| &
-\left|\begin{array}{ccc} 1 & 0 & 1\\ 1 & 1 &0 \\ 1 & 0 & 0
\end{array}\right| &
\left|\begin{array}{ccc} 1 & 0 & 0\\ 1 & 1 &0 \\ 1 & 0 & 0
\end{array}\right| \\ \\
\left|\begin{array}{ccc} 0 & 0 & 1\\ 0 & 1 &0 \\ 0 & 0 & 0
\end{array}\right| &
-\left|\begin{array}{ccc} 1 & 0 & 1\\ 1 & 1 &0 \\ 1 & 0 & 0
\end{array}\right| &
\left|\begin{array}{ccc} 1 & 0 & 1\\ 1 & 0 &0 \\ 1 & 0 & 0
\end{array}\right| &
-\left|\begin{array}{ccc} 1 & 0 & 0\\ 1 & 0 &1 \\ 1 & 0 & 0
\end{array}\right| \\ \\
-\left|\begin{array}{ccc} 0 & 0 & 1\\ 0 & 1 &0 \\ 1 & 0 & 0
\end{array}\right| &
\left|\begin{array}{ccc} 1 & 0 & 1\\ 1 & 1 &0 \\ 1 & 0 & 0
\end{array}\right| &
-\left|\begin{array}{ccc} 1 & 0 & 1\\ 1 & 0 &0 \\ 1 & 1 & 0
\end{array}\right| &
\left|\begin{array}{ccc} 1 & 0 & 0\\ 1 & 0 &1 \\ 1 & 1 & 0
\end{array}\right| \\ \\
\end{array}\right]\\ & = & \left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\
 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & -1 & -1 & -1 \end{array} \right) \end{eqnarray*}

Hence $A^{-1} = \left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\
 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & -1 \\ 1 & 0 & 0 & -1 \end{array}
 \right)$

Finding the inverse of $B = \left( \begin{array}{ccc} 2 & 1 & 3
\\ 1 & 1 & 2 \\ 2 & 1 & 6 \end{array} \right)$

Matrix of cofactors
\begin{eqnarray*} [A_{ij}] & = & \left[ \begin{array}{ccc}
\left|\begin{array}{cc} 1 & 2 \\ 1 & 6 \end{array}\right| &
-\left|\begin{array}{cc} 1 & 2 \\ 2 & 6 \end{array}\right| &
\left|\begin{array}{cc} 1 & 1 \\ 2 & 1 \end{array}\right| \\ \\
-\left|\begin{array}{cc} 1 & 3 \\ 1 & 6 \end{array}\right| &
\left|\begin{array}{cc} 2 & 3 \\ 2 & 6 \end{array}\right| &
-\left|\begin{array}{cc} 2 & 1 \\ 2 & 1 \end{array}\right| \\ \\
\left|\begin{array}{cc} 1 & 3 \\ 1 & 2 \end{array}\right| &
-\left|\begin{array}{cc} 2 & 3 \\ 1 & 2 \end{array}\right| &
\left|\begin{array}{cc} 2 & 1 \\ 1 & 1 \end{array}\right| \\
\end{array}\right]\\ & = & \left( \begin{array}{ccc} 4 & -2 & -1 \\
 -3 & 6 & 0 \\ -1 & -1 & 1\end{array} \right) \end{eqnarray*}

Hence $B^{-1} = \frac{1}{3} \left( \begin{array}{ccc} 4 & -3 & -1
\\ -2 & 6 & -1 \\ -1 & 0 & 1 \end{array} \right)$



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