Question A.9: This example illustrates the solution by transforming the co......

can be solved by row manipulations on the augmented matrix as follows:

= \left|\begin{matrix}1 &4& 6 \\ 0& -2 &-9 \\ 0& -17& -29\end{matrix} \right|\left|\begin{matrix}2 \\ -3 \\ -5\end{matrix} \right|→ R_{3} \ – \frac{17}{2}R_{2}=\left|\begin{matrix}1 &4& 6 \\ 0& -2 &-9 \\ 0& 0&47.5 \end{matrix} \right|\left|\begin{matrix}2 \\ -3 \\ 20.5 \end{matrix} \right|

Thus,

47.5x_{3} = 20.5
-2x_{2} – 9x_{3} = -3
x_{1} + 4x_{2} + 6x_{3} = 2

which gives

A set of simultaneous equations can also be solved by partitioning

\left| \begin{array}{c : c} a_{11}, . . . ,a_{1k} &a_{1m}, . . . ,a_{1n} \\ . . . & . . . \\ a_{k1}, . . . ,a_{kk}&a_{km}, . . . ,a_{kn} \\ \hdashline a_{m1}, . . . ,a_{mk} &a_{mm}, . . . ,a_{mn} \\ . . . & . . . \\ a_{n1}, . . . ,a_{nk}&a_{nm}, . . . ,a_{nn} \end{array} \right|\left| \begin{array}{c} x_{1} \\ . \\ x_{k} \\ \hdashline x_{m} \\ . \\ x_{n} \end{array} \right|= \left| \begin{array}{c} b_{1} \\ . \\ b_{k} \\ \hdashline b_{m} \\ . \\ b_{n} \end{array} \right|      (A.68)

Equation A.68 is horizontally partitioned and rewritten as

\left|\begin{matrix} \overline{A}_{1}&\overline{A}_{2} \\  \overline{A}_{3}&\overline{A}_{4}  \end{matrix} \right| = \left|\begin{matrix} \overline{X}_{1} \\ \overline{X}_{2}  \end{matrix} \right|\left|\begin{matrix} \overline{B}_{1} \\ \overline{B}_{2}  \end{matrix} \right|       (A.69)

Vectors \overline{X}_{1}  and \overline{X}_{2}  are given by

\overline{X}_{1}=⌊\overline{A}_{1}-\overline{A}_{2}\overline{A}^{-1}_{4}\overline{A}_{3}⌋^{-1} ⌊\overline{B}_{1}-\overline{A}_{2}\overline{A}^{-1}_{4}\overline{B}_{2}⌋       (A.70)
\overline{X}_{2}=⌊\overline{A}^{-1}_{4}(\overline{B}_{2}-\overline{A}_{3}\overline{X}_{1})⌋       (A.71)