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{\bf Question}

Consider the following set of simultaneous equations

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

If $k=0$, find the solution by matrix inversion.

{\bf{Note:}} if you fail to show detailed working of the matrix
inversion, no marks will be awarded, even if you can write down
the correct answer.
\medskip

{\bf Answer}

$k=0$
\begin{eqnarray*} y-2z &amp; = &amp; 3\\ x-y+2z &amp; = &amp; 1\\ -x+2y+z &amp; = &amp; 2
\end{eqnarray*}

$\Rightarrow \left(\begin{array}{ccc} 0 &amp; 1 &amp; -2\\ 1 &amp; -1 &amp; 2\\ -1
&amp; 2 &amp; 1 \end{array} \right) \left(\begin{array}{c}x\\y\\z
\end{array}\right)=\left(\begin{array}{c} 3\\1\\2 \end{array}
\right)$

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

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

so solution exists

Step (i) Cofactors of matrix are given by

$\left(\begin{array} {ccc} A_{11} &amp; A_{12} &amp; A_{13}\\ A_{21} &amp;
A_{22} &amp; A_{23}\\ A_{31} &amp; A_{32} &amp; A_{33}
\end{array} \right)=\left(\begin{array} {ccc} 0 &amp; 1 &amp; -2\\ 1 &amp; -1 &amp; 2\\ -1 &amp; 2 &amp; 1
\end{array} \right)$

\newpage
cofactor of $A_{11}=+\left|\begin{array} {cc} -1 &amp; 2\\2 &amp; 1
\end{array} \right|=-5$ Following + - sign pattern

cofactor of $A_{12}=-\left|\begin{array} {cc} 1 &amp; 2\\-1 &amp; 1
\end{array} \right|=-3$

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

cofactor of $A_{21}=-\left|\begin{array} {cc} 1 &amp; -2\\2 &amp; 1
\end{array} \right|=-5$

cofactor of $A_{22}=+\left|\begin{array} {cc} 0 &amp; -2\\-1 &amp; 1
\end{array} \right|=-2$

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

cofactor of $A_{31}=+\left|\begin{array} {cc} 1 &amp; -2\\-1 &amp; 2
\end{array} \right|=0$

cofactor of $A_{32}=-\left|\begin{array} {cc} 0 &amp; -2\\1 &amp; 2
\end{array} \right|=-2$

cofactor of $A_{33}=+\left|\begin{array} {cc} 0 &amp; 1\\1 &amp; -1
\end{array} \right|=-1$

Matrix of cofactors is thus:

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

Step (ii)

Transpose this to get $adj A$

$$adj\ A=\left(\begin{array} {ccc} -5 &amp; -3 &amp; 1\\ -5 &amp; -2 &amp; -1\\ 0
&amp; -2 &amp; -1 \end{array} \right)^T=\left(\begin{array} {ccc} -5 &amp; -5
&amp; 0\\ -3 &amp; -2 &amp; -2\\ 1 &amp; -1 &amp; -1 \end{array} \right)$$

Step (iv)

$A^{-1}=\ds\frac{adj\ A}{det\
A}=\ds\frac{1}{-5}\left(\begin{array} {ccc} -5 &amp; -5 &amp; 0\\ -3 &amp; -2
&amp; -2\\ 1 &amp; -1 &amp; -1 \end{array} \right)$

Hence ${\bf{A}}=-\ds\frac{1}{2}\left(\begin{array} {ccc} -5 &amp; -5 &amp;
0\\ -3 &amp; -2 &amp; -2\\ 1 &amp; -1 &amp; -1 \end{array}
\right)\left(\begin{array}{c}3\\1\\2 \end{array} \right)$

$\Rightarrow \un{\left(\begin{array}{c}x\\y\\z\end{array}\right)=
-\ds\frac{1}{5}\left(\begin{array}{c}-20\\-15\\0
\end{array}\right)=\left(\begin{array}{c} 4\\3\\0
\end{array}\right)}$

\end{document}
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