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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
The solution of the equation $\left[ {\begin{array}{*{20}{c}}1&0&1\\{ - 1}&1&0\\0&{ - 1}&1\end{array}} \right]\,\left[ \begin{array}{l}x\\y\\z\end{array} \right] = \left[ \begin{array}{l}1\\1\\2\end{array} \right]$ is $(x,y,z)$=
  • A
    $(1,\,\,1\,,\,1)$
  • B
    $(0,\, - 1\,\,2)$
  • C
    $( - 1,\,\,2,\,\,2)$
  • ✓
    $(-1, 0, 2)$
✓

Answer

Correct option: D.
$(-1, 0, 2)$
Let $A = \left[ {\begin{array}{*{20}{c}}1&0&1\\{ - 1}&1&0\\0&{ - 1}&1\end{array}} \right] $
$\Rightarrow \,{A^{ - 1}}\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&{\frac{{ - 1}}{2}}&{ - 1}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{{ - 1}}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\end{array}} \right]$
$\therefore $ $AX = B$
$ \Rightarrow X = {A^{ - 1}}B\,\left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}&{\frac{{ - 1}}{2}}&{ - 1}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{{ - 1}}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}&{\frac{1}{2}}\end{array}} \right]\,\,\left[ \begin{array}{l}1\\1\\2\end{array} \right] = \left[ \begin{array}{l} - 1\\{\rm{ }}0\\{\rm{ }}2\end{array} \right]$.
Aliter : $\left[ {\begin{array}{*{20}{c}}1&0&1\\{ - 1}&1&0\\0&{ - 1}&1\end{array}} \right]\,\,\,\,\left[ \begin{array}{l}x\\y\\z\end{array} \right] = \left[ \begin{array}{l}1\\1\\2\end{array} \right]$
$\Rightarrow\left[ \begin{array}{l}x + 0y + z\\ - x + y + 0z\\0x - y + z\end{array} \right] = \left[ \begin{array}{l}1\\1\\2\end{array} \right]$   
$\begin{array}{l}x + z = 1\\ - x + y = 1\\z - y = 2\end{array}$
$ \Rightarrow (x,\,y,\,z) = ( - 1,\,0,\,2)$.

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