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\begin{document}

{\bf Question}

Consider the system of equations

\begin{eqnarray*} 2x+3y+6z &amp; = &amp; 1\\ 6x+2y-3z &amp; = &amp; 1\\ 3x-6y+2z &amp; = &amp;
2 \end{eqnarray*}

Show by matrix inversion that the solution set is
$x=\ds\frac{2}{7},\ y=-\ds\frac{1}{7},\ z=\ds\frac{1}{7}$ What is
the geometrical significance of this?

{\bf{Note:}} No marks will be given unless a full working of the
calculation of the inverse matrix is shown.
\medskip

{\bf Answer}
\begin{eqnarray*} 2x+3y+6z &amp; = &amp; 1\\ 6x+2y-3z &amp; = &amp; 1\\ 3x-6y+2z &amp; = &amp; 2
\end{eqnarray*}

$\Rightarrow \left(\begin{array}{ccc} 2 &amp; 3 &amp; 6\\ 6 &amp; 2 &amp; -3\\ 3 &amp;
-6 &amp; 2 \end{array} \right) \left(\begin{array}{c}x\\y\\z
\end{array}\right)=\left(\begin{array}{c} 1\\1\\2 \end{array}
\right)$

              ${\bf{A}} \cdot \ \ \ {\bf{X}}\ \ \ = \ \ \ {\bf{K}}$

Require ${\bf{A}}^{-1}$, since ${\bf{X}}={\bf{A}}^{-1}{\bf{K}}$

Step (iii)

\begin{eqnarray*} \bigtriangleup &amp; = &amp; det
A\\ &amp; = &amp; \left|\begin{array}{ccc}2 &amp; 3 &amp; 6\\ 6 &amp; 2 &amp; -3\\ 3 &amp; -6
&amp; 2 \end{array} \right|\\ &amp; = &amp; 2\left|\begin{array}{cc} 2 &amp;
-3\\-6 &amp; 2 \end{array} \right|-3\left|\begin{array}{cc} 6 &amp; -3\\3
&amp; 2 \end{array} \right|+6\left|\begin{array}{cc} 6 &amp; 2\\3 &amp; -6
\end{array} \right|\\ &amp; = &amp; 2\times (4-18)-3 \times (12+9)\\ &amp; = &amp;
-3\times 42\\ &amp; = &amp; -28-63-252\\ &amp; = &amp; -343\\ &amp; \ne &amp; \un{0}
\end{eqnarray*}

so there exists an inverse

Step (i) Cofactors of matrix are given by

$\left(\begin{array} {ccc} A_{11} &amp; A_{12} &amp; A_{13}\\ A_{21} &amp;
A_{22} &amp; A_{23}\\ A_{31} &amp; A_{32} &amp; A_{33}
\end{array} \right)=\left(\begin{array} {ccc} 2 &amp; 3 &amp; 6\\ 6 &amp; 2 &amp; -3\\ 3 &amp; -6 &amp;
2 \end{array} \right)$

cofactor of $A_{11}=+\left|\begin{array} {cc} 2 &amp; -3\\-6 &amp; 2
\end{array} \right|=-14$ Following + - sign pattern

cofactor of $A_{12}=-\left|\begin{array} {cc} 6 &amp; -3\\3 &amp; 2
\end{array} \right|=-21$

cofactor of $A_{13}=+\left|\begin{array} {cc} 6 &amp; 2\\3 &amp; -6
\end{array} \right|=-42$

cofactor of $A_{21}=-\left|\begin{array} {cc} 3 &amp; 6\\-6 &amp; 2
\end{array} \right|=-42$

cofactor of $A_{22}=+\left|\begin{array} {cc} 2 &amp; 6\\3 &amp; 2
\end{array} \right|=-14$

cofactor of $A_{23}=-\left|\begin{array} {cc} 2 &amp; 3\\3 &amp; -6
\end{array} \right|=+21$

cofactor of $A_{31}=+\left|\begin{array} {cc} 3 &amp; 6\\2 &amp; -3
\end{array} \right|=-21$

cofactor of $A_{32}=-\left|\begin{array} {cc} 2 &amp; 6\\6 &amp; -3
\end{array} \right|=+42$

cofactor of $A_{33}=+\left|\begin{array} {cc} 2 &amp; 3\\6 &amp; 2
\end{array} \right|=-14$

Matrix of cofactors is thus:

$$\left(\begin{array} {ccc} -14 &amp; -21 &amp; -42\\ -42 &amp; -14 &amp; +21\\
-21 &amp; +42 &amp; -14 \end{array} \right)$$

Step (ii)

Transpose this to get $adj A$

$$adj\ A=\left(\begin{array} {ccc} -14 &amp; -42 &amp; -21\\ -21 &amp; -14 &amp;
42\\ -42 &amp; 21 &amp; -14 \end{array} \right)$$

Step(iii)

$det\ A = -343$

Step (iv)

\begin{eqnarray*} A^{-1} &amp; = &amp; \ds\frac{adj\ A}{det\
A}=\ds\frac{1}{-343}\left(\begin{array} {ccc} -14 &amp; -42 &amp; -21\\
-21 &amp; -14 &amp; 42\\ -42 &amp; 21 &amp; -14 \end{array} \right)\\ &amp; = &amp;
\ds\frac{1}{49} \left(\begin{array} {ccc} 2 &amp; 6 &amp; 3\\ 3 &amp; 2 &amp; -6\\
6 &amp; -3 &amp; 2 \end{array} \right) \end{eqnarray*}

Hence

\begin{eqnarray*} {\bf{X}} &amp; = &amp; -\ds\frac{1}{49}\left(\begin{array} {ccc} 2 &amp; 6 &amp; 3\\ 3 &amp; 2 &amp; -6\\
6 &amp; -3 &amp; 2  \end{array} \right)\times
\left(\begin{array}{c}1\\1\\2 \end{array} \right)\\ &amp; = &amp;
\ds\frac{1}{49} \times
\left(\begin{array}{c}2+6+6\\3+2-12\\6-3+4\end{array}\right)\\ &amp; =
&amp; \ds\frac{1}{49}\times
\left(\begin{array}{c}14\\-7\\7\end{array}\right)\\ {\bf{X}} &amp; = &amp;
\un{\left(\begin{array}{c}\frac{2}{7}\\-\frac{1}{7}\\ \frac{1}{7}
\end{array} \right)}
\end{eqnarray*}

This the point of intersection of three planes in 3-D.
\end{document}
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