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\newcommand{\ds}{\displaystyle}
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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)]
$A=\left(\begin{array}{ccc} 3 &amp; -2 &amp; -1\\ -4 &amp; 1 &amp; -1\\ 2 &amp; 0 &amp; 1
\end{array} \right)$

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

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

\medskip

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

step (i)

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

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

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

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

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

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

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

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

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

Matrix of cofactors is thus:

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

step (ii)

Transpose to get $adj A$

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

step (iii)

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

step (iv)

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

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

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

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

step (i)

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

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

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

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

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

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

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

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

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

Matrix of cofactors is thus:

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

step (ii)

Transpose to get $adj B$

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

\newpage
step (iii)

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

step (iv)

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

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

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

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

step (i)

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

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

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

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

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

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

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

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

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

Matrix of cofactors is thus:

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

step (ii)

Transpose to get $adj C$

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

step (iii)

\begin{eqnarray*} det\ C &amp; = &amp; \left|\begin{array} {ccc} 4 &amp; -8 &amp; 4\\ -3 &amp; 6 &amp; -3\\ -1 &amp;
2 &amp; -1 \end{array}\right|\\ &amp; = &amp; 4\times\left|\begin{array}{cc} 6
&amp; -3\\ 2 &amp; -1 \end{array} \right|-(-8) \left|\begin{array}{cc} -3
&amp; -3\\ -1 &amp; -1 \end{array} \right|+4 \times
\left|\begin{array}{cc} -3 &amp; 6\\ -1 &amp; 2
\end{array} \right|\\ &amp; = &amp; 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}
</pre></body></html>