3 and 4 .Determinants and Matrices
medium

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 $\left[ \begin{array}{l}x\\y\\z\end{array} \right]$=

A

$\left[ \begin{array}{l}1\\1\\1\end{array} \right]$

B

$\left[ \begin{array}{l}\,\,\,1\\ - 2\\\,\,\,3\end{array} \right]$

C

$\left[ \begin{array}{l}\,\,\,1\\ - 2\\\,\,\,1\end{array} \right]$

D

$\left[ \begin{array}{l}\,\,\,\,1\\\,\,\,\,2\\ - 3\end{array} \right]$

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., $\left[ \begin{array}{l}{\rm{ }}1\\{\rm{ }}2\\ – 3\end{array} \right]$.

Standard 12
Mathematics

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