If [ * 20 c 1& , ,1 & , ,1 1& - 2 & - 2 1& , ,3 & , ,1 ] , [ l x … - Vidyadip

MCQ

If $\left[ {\begin{array}{*{20}{c}}1&{\,\,1}&{\,\,1}\\1&{ - 2}&{ - 2}\\1&{\,\,3}&{\,\,1}\end{array}} \right]\,\left[ \begin{array}{l}x\\y\\z\end{array} \right] = \left[ \begin{array}{l}0\\3\\4\end{array} \right]$, then $\left[ \begin{array}{l}x\\y\\z\end{array} \right]$is equal to

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

✓

Answer

Correct option: D.

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

d
(d) We have, $\left[ {\begin{array}{*{20}{c}}1&1&1\\1&{ - 2}&{ - 2}\\1&3&1\end{array}} \right]\,\,\left[ \begin{array}{l}x\\y\\z\end{array} \right] = \left[ \begin{array}{l}0\\3\\4\end{array} \right]$

$x + y + z = 0$ ......$(i)$

$x - 2y - 2z = 3$ ......$(ii)$

$x + 3y + z = 4$.....$(iii)$

On solving $x = 1,\,y = 2,\,z = - 3$

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

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