શ્રેણિક પદ્ધતિથી નીચેનાં સુરેખ સમીકરણોની સંહતિનો ઉકેલ મેળવો : $ 3 x-2 y+3 z =8 \,;\,2 x+y-z =1 \,;\,4 x-3 y+2 z =4$
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Solution The system of equations can be written in the form $\mathrm{AX}=\mathrm{B}$, where$A = \left[ {\begin{array}{*{20}{c}} 3&{ - 2}&3 \\ 2&1&{ - 1} \\ 4&{ - 3}&2 \end{array}} \right],X = \left[ {\begin{array}{*{20}{l}} x \\ y \\ z \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{l}} 8 \\ 1 \\ 4 \end{array}} \right]$
We see that
$|A|=3(2-3)+2(4+4)+3(-6-4)=-17 \neq 0$
Hence, $A$ is nonsingular and so its inverse exists.
Now $\begin{gathered} {{\text{A}}_{11}} = - 1,\,\,\,{{\text{A}}_{12}} = - 8,\,\,\,{{\text{A}}_{13}} = - 10 \hfill \\
{{\text{A}}_{21}} = - 5,\,\,\,{{\text{A}}_{22}} = - 6,\,\,\,{{\text{A}}_{23}} = 1 \hfill \\ {{\text{A}}_{31}} = - 1,\,\,\,{{\text{A}}_{32}} = 9,\,\,\,{{\text{A}}_{33}} = 7 \hfill \\ \end{gathered} $
Therefore ${A^{ - 1}} = - \frac{1}{{17}}\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 5}&{ - 1} \\ { - 8}&{ - 6}&9 \\
{ - 10}&1&7 \end{array}} \right]\,$
$So\quad \,X = {A^{ - 1}}B = - \frac{1}{{17}}\left[ {\begin{array}{*{20}{l}} { - 1}&{ - 5}&{ - 1} \\ { - 8}&{ - 6}&9 \\
{ - 10}&1&7 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} 8 \\ 1 \\ 4 \end{array}} \right]$
$i.e.\,\,\,\,\,\left[ {\begin{array}{*{20}{l}} x \\ y \\ z \end{array}} \right] = - \frac{1}{{17}}\left[ {\begin{array}{*{20}{l}} { - 17} \\ { - 34} \\ { - 51} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 1 \\ 2 \\ 3 \end{array}} \right]$
Hence $ x = 1,y = 2$ and $z = 3$
We see that
$|A|=3(2-3)+2(4+4)+3(-6-4)=-17 \neq 0$
Hence, $A$ is nonsingular and so its inverse exists.
Now $\begin{gathered} {{\text{A}}_{11}} = - 1,\,\,\,{{\text{A}}_{12}} = - 8,\,\,\,{{\text{A}}_{13}} = - 10 \hfill \\
{{\text{A}}_{21}} = - 5,\,\,\,{{\text{A}}_{22}} = - 6,\,\,\,{{\text{A}}_{23}} = 1 \hfill \\ {{\text{A}}_{31}} = - 1,\,\,\,{{\text{A}}_{32}} = 9,\,\,\,{{\text{A}}_{33}} = 7 \hfill \\ \end{gathered} $
Therefore ${A^{ - 1}} = - \frac{1}{{17}}\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 5}&{ - 1} \\ { - 8}&{ - 6}&9 \\
{ - 10}&1&7 \end{array}} \right]\,$
$So\quad \,X = {A^{ - 1}}B = - \frac{1}{{17}}\left[ {\begin{array}{*{20}{l}} { - 1}&{ - 5}&{ - 1} \\ { - 8}&{ - 6}&9 \\
{ - 10}&1&7 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} 8 \\ 1 \\ 4 \end{array}} \right]$
$i.e.\,\,\,\,\,\left[ {\begin{array}{*{20}{l}} x \\ y \\ z \end{array}} \right] = - \frac{1}{{17}}\left[ {\begin{array}{*{20}{l}} { - 17} \\ { - 34} \\ { - 51} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 1 \\ 2 \\ 3 \end{array}} \right]$
Hence $ x = 1,y = 2$ and $z = 3$
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