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यदि$\left[ {\begin{array}{*{20}{c}}1&2&3\\3&1&2\\2&3&1\end{array}} \right]\,\left[ \begin{array}{l}x\\y\\z\end{array} \right] = \left[ {\begin{array}{*{20}{c}}4&{ - 2}\\0&{ - 6}\\{ - 1}&2\end{array}} \right]\,\left[ \begin{array}{l}2\\1\end{array} \right]$, तो $(x,y,z)$=
$( - 4,\,2,\,2)$
$(4,\, - 2,\, - 2)$
$(4,\,2,\,2)$
$( - 4,\, - 2,\, - 2)$
Solution
(a) $\left[ {\begin{array}{*{20}{c}}1&2&3\\3&1&2\\2&3&1\end{array}} \right]\,\,\left[ \begin{array}{l}x\\y\\z\end{array} \right] = \left[ {\begin{array}{*{20}{c}}4\\0\\{ – 1}\end{array}\,\,\begin{array}{*{20}{c}}{ – 2}\\{ – 6}\\2\end{array}} \right]\,\left[ \begin{array}{l}2\\1\end{array} \right]\,$
$\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,x + 2y + 3z = 6\\\, \Rightarrow \,\,\,\,\,3x + y + 2z = – \,6\\\,{\rm{ }}2x + 3y + z = 0\end{array}$
समीकरणों को हल करने पर,
$x = – 4,\,y = 2,\,z = 2$.
.