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\begin{document}

{\bf Question}

Find the inverses of the following matrices and verify that they
are correct.

\begin{description}
\item[(i)]
$A=\left(\begin{array}{ccc} 3 & -2 & -1\\ -4 & 1 & -1\\ 2 & 0 & 1
\end{array} \right)$

\item[(ii)]
$B=\left(\begin{array}{ccc} 1 & -1 & 3\\ 2 & 1 & 4\\ 0 & 1 & 1
\end{array} \right)$

\item[(iii)]
$C=\left(\begin{array}{ccc} 3 & 2 & 6\\ 5 & 2 & 11\\ 7 & 4 & 16
\end{array} \right)$
\end{description}

\medskip

{\bf Answer}
\begin{description}
\item[(i)]
$A=\left(\begin{array} {ccc} 3 & -2 & -1\\ -4 & 1 & -1\\ 2 & 0 & 1
\end{array} \right)=\left(\begin{array} {ccc} A_{11} & A_{12} & A_{13}\\
A_{21} & A_{22} & A_{23}\\ A_{31} & A_{32} & A_{33}
\end{array} \right)$

step (i)

cofactor of $A_{11}=+\left|\begin{array} {cc} 1 & -1\\0 & 1
\end{array} \right|=+1$

cofactor of $A_{12}=-\left|\begin{array} {cc} 1 & -1\\0 & 1
\end{array} \right|=+2$

cofactor of $A_{13}=+\left|\begin{array} {cc} -4 & -1\\2 & 1
\end{array} \right|=-2$

cofactor of $A_{21}=-\left|\begin{array} {cc} -2 & -1\\0 & 1
\end{array} \right|=+2$

cofactor of $A_{22}=+\left|\begin{array} {cc} 3 & -1\\2 & 1
\end{array} \right|=+5$

cofactor of $A_{23}=-\left|\begin{array} {cc} 3 & -2\\2 & 0
\end{array} \right|=-4$

cofactor of $A_{31}=+\left|\begin{array} {cc} -2 & -1\\1 & -1
\end{array} \right|=+3$

cofactor of $A_{32}=-\left|\begin{array} {cc} 3 & -1\\-4 & -1
\end{array} \right|=+7$

cofactor of $A_{33}=+\left|\begin{array} {cc} 3 & -2\\-4 & 1
\end{array} \right|=-5$

Matrix of cofactors is thus:

$$\left(\begin{array} {ccc} +1 & +2 & -2\\ +2 & +5 & -4\\ +3 & +7
& -5
\end{array} \right)$$

step (ii)

Transpose to get $adj A$

$$adj\ A=\left(\begin{array} {ccc} 1 & 2 & 3\\ 2 & 5 & 7\\ -2 & -4
& -5 \end{array} \right)$$

step (iii)

\begin{eqnarray*} det\ A & = & \left|\begin{array} {ccc} 3 & -2 &
-1\\ -4 & 1 & -1\\ 2 & 0 & 1 \end{array}\right|\\ & = &
3\left|\begin{array}{cc} 1 & -1\\ 0 & 1 \end{array} \right|-(-2)
\left|\begin{array}{cc} -4 & -1\\ 2 & 1 \end{array} \right|-1
\times \left|\begin{array}{cc} -4 & 1\\ 2 & 0 \end{array}
\right|\\ & = & 3-4+2=1 \end{eqnarray*}

step (iv)

$$A^{-1}=\ds\frac{adj A}{det A}=\left(\begin{array} {ccc} 1 & 2 &
3\\ 2 & 5 & 7\\ -2 & -4 & -5 \end{array} \right)$$

Check (for $A^{-1}A$ only)

$\left(\begin{array} {ccc} 1 & 2 & 3\\ 2 & 5 & 7\\ -2 & -4 & -5
\end{array} \right)\left(\begin{array} {ccc} 3 & -2 & -1\\ -4 & 1 & -1\\ 2 &
0 & 1 \end{array} \right)=\left(\begin{array} {ccc} 1 & 0 & 0\\ &
1 & 0\\ 0 & 0 & 1 \end{array} \right)=I_3\ \surd$

\newpage
\item[(ii)]
$B=\left(\begin{array} {ccc} 1 & -1 & 3\\ 2 & 1 & 4\\ 0 & 1 & 1
\end{array} \right)=\left(\begin{array} {ccc} B_{11} & B_{12} & B_{13}\\
B_{21} & B_{22} & B_{23}\\ B_{31} & B_{32} & B_{33}
\end{array} \right)$

step (i)

cofactor of $B_{11}=+\left|\begin{array} {cc} 1 & 4\\1 & 1
\end{array} \right|=-3$

cofactor of $B_{12}=-\left|\begin{array} {cc} 2 & 4\\0 & 1
\end{array} \right|=-2$

cofactor of $B_{13}=+\left|\begin{array} {cc} 2 & 1\\0 & 1
\end{array} \right|=+2$

cofactor of $B_{21}=-\left|\begin{array} {cc} -1 & 3\\1 & 1
\end{array} \right|=+4$

cofactor of $B_{22}=+\left|\begin{array} {cc} 1 & 3\\0 & 1
\end{array} \right|=+1$

cofactor of $B_{23}=-\left|\begin{array} {cc} 1 & -1\\0 & 1
\end{array} \right|=-1$

cofactor of $B_{31}=+\left|\begin{array} {cc} -1 & 3\\1 & 4
\end{array} \right|=-7$

cofactor of $B_{32}=-\left|\begin{array} {cc} 1 & 3\\2 & 4
\end{array} \right|=+2$

cofactor of $B_{33}=+\left|\begin{array} {cc} 1 & -1\\2 & 1
\end{array} \right|=+3$

Matrix of cofactors is thus:

$$\left(\begin{array} {ccc} -3 & -2 & 2\\ +4 & +1 & -1\\ -7 & +2 &
+3 \end{array} \right)$$

step (ii)

Transpose to get $adj B$

$$adj\ B=\left(\begin{array} {ccc} -3 & 4 & -7\\ -2 & 1 & 2\\ 2 &
-1 & 3 \end{array} \right)$$

\newpage
step (iii)

\begin{eqnarray*} det\ B & = & \left|\begin{array} {ccc} 1 & -1 &
3\\ 2 & 1 & 4\\ 0 & 1 & 1 \end{array}\right|\\ & = &
1\times\left|\begin{array}{cc} 1 & 4\\ 1 & 1 \end{array}
\right|-(-1) \left|\begin{array}{cc} 2 & 4\\ 0 & 1 \end{array}
\right|+3 \times \left|\begin{array}{cc} 2 & 1\\ 0 & 1
\end{array} \right|\\ & = & -3+2+6=5 \end{eqnarray*}

step (iv)

$$B^{-1}=\ds\frac{adj B}{det B}=\left(\begin{array} {ccc}
-\frac{3}{5} & \frac{4}{5} & -\frac{7}{5}
\\ -\frac{2}{5} & \frac{1}{5} & \frac{2}{5}\\ \frac{2}{5} & -\frac{1}{5} & \frac{3}{5}
\end{array} \right)$$

Check (for $B^{-1}B$ only)

$\left(\begin{array} {ccc} -\frac{3}{5} & \frac{4}{5} &
-\frac{7}{5}\\ -\frac{2}{5} & \frac{1}{5} & \frac{2}{5}\\
\frac{2}{5} & -\frac{1}{5} & \frac{3}{5}
\end{array} \right)\left(\begin{array} {ccc} 1 & -1 & 3\\ 2 & 1 & 4\\ 0 &
1 & 1 \end{array} \right)=\left(\begin{array} {ccc} 1 & 0 & 0\\ &
1 & 0\\ 0 & 0 & 1 \end{array} \right)=I_3\ \surd$

\item[(iii)]
$C=\left(\begin{array} {ccc} 3 & 2 & 6\\ 5 & 3 & 11\\ 7 & 4 & 16
\end{array} \right)=\left(\begin{array} {ccc} C_{11} & C_{12} & C_{13}\\
C_{21} & C_{22} & C_{23}\\ C_{31} & C_{32} & C_{33}
\end{array} \right)$

step (i)

cofactor of $C_{11}=+\left|\begin{array} {cc} 3 & 11\\4 & 16
\end{array} \right|=+4$

cofactor of $C_{12}=-\left|\begin{array} {cc} 5 & 11\\7 & 16
\end{array} \right|=-3$

cofactor of $C_{13}=+\left|\begin{array} {cc} 5 & 3\\7 & 4
\end{array} \right|=-1$

cofactor of $C_{21}=-\left|\begin{array} {cc} 2 & 6\\4 & 16
\end{array} \right|=-8$

cofactor of $C_{22}=+\left|\begin{array} {cc} 3 & 6\\7 & 16
\end{array} \right|=+6$

cofactor of $C_{23}=-\left|\begin{array} {cc} 3 & 2\\7 & 4
\end{array} \right|=+2$

cofactor of $C_{31}=+\left|\begin{array} {cc} 2 & 6\\3 & 11
\end{array} \right|=+4$

cofactor of $C_{32}=-\left|\begin{array} {cc} 3 & 6\\5 & 11
\end{array} \right|=-3$

cofactor of $C_{33}=+\left|\begin{array} {cc} 3 & 2\\5 & 3
\end{array} \right|=-1$

Matrix of cofactors is thus:

$$\left(\begin{array} {ccc} 4 & -3 & -1\\ -8 & 6 & 2\\ 4 & -3 & -1
\end{array} \right)$$

step (ii)

Transpose to get $adj C$

$$adj\ C=\left(\begin{array} {ccc} 4 & -8 & 4\\ -3 & 6 & -3\\ -1 &
2 & -1 \end{array} \right)$$

step (iii)

\begin{eqnarray*} det\ C & = & \left|\begin{array} {ccc} 4 & -8 & 4\\ -3 & 6 & -3\\ -1 &
2 & -1 \end{array}\right|\\ & = & 4\times\left|\begin{array}{cc} 6
& -3\\ 2 & -1 \end{array} \right|-(-8) \left|\begin{array}{cc} -3
& -3\\ -1 & -1 \end{array} \right|+4 \times
\left|\begin{array}{cc} -3 & 6\\ -1 & 2
\end{array} \right|\\ & = & 0-0+0=0 \end{eqnarray*}

$det C=0$. Therefore there is \un{no} inverse matrix such that
$C^{-1}C=I_3$.

\end{description}

\end{document}