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यदि $\left[ {\begin{array}{*{20}{c}}1&{\,\,1}&{\,\,1}\\1&{ - 2}&{ - 2}\\1&{\,\,3}&{\,\,1}\end{array}} \right]\,\left[ \begin{array}{l}x\\y\\z\end{array} \right] = \left[ \begin{array}{l}0\\3\\4\end{array} \right]$, तो $\left[ \begin{array}{l}x\\y\\z\end{array} \right]$=
$\left[ \begin{array}{l}1\\1\\1\end{array} \right]$
$\left[ \begin{array}{l}\,\,\,1\\ - 2\\\,\,\,3\end{array} \right]$
$\left[ \begin{array}{l}\,\,\,1\\ - 2\\\,\,\,1\end{array} \right]$
$\left[ \begin{array}{l}\,\,\,\,1\\\,\,\,\,2\\ - 3\end{array} \right]$
Solution
(d) हम जानते हैं कि $\left[ {\begin{array}{*{20}{c}}1&1&1\\1&{ – 2}&{ – 2}\\1&3&1\end{array}} \right]\,\,\left[ \begin{array}{l}x\\y\\z\end{array} \right] = \left[ \begin{array}{l}0\\3\\4\end{array} \right]$
$x + y + z = 0$ ……$(i)$
$x – 2y – 2z = 3$ ……$(ii)$
$x + 3y + z = 4$…..$(iii)$
अर्थात् $x = 1,\,y = 2,\,z = – 3$
i.e., $\left[ \begin{array}{l}{\rm{ }}1\\{\rm{ }}2\\ – 3\end{array} \right]$.