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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
Solve system of linear equations, using matrix method. $2 x+y+z=1 ; x-2 y-z=\frac{3}{2} ; 3 y-5 z=9$
  • A
    $x=2, y=\frac{1}{2},-\frac{3}{2}$
  • ✓
    $x=1, y=\frac{1}{2},-\frac{3}{2}$
  • C
    $x=1, y=\frac{1}{2},-\frac{-3}{2}$
  • D
    $x=1, y=\frac{-1}{2},-\frac{3}{2}$
✓

Answer

Correct option: B.
$x=1, y=\frac{1}{2},-\frac{3}{2}$
The given system of equation can be written in the form of $A X=B,$
where$A = \left[ {\begin{array}{*{20}{c}}  2&1&1 \\ 1&{ - 2}&{ - 1} \\ 0&3&{ - 5} \end{array}} \right],X = \left[ {\begin{array}{*{20}{l}}  x \\  y \\  z  \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{3}{2}} \\ 9  \end{array}} \right]$
Now,
$|A|=2(10+3)-1(-5-3)+0$
$=2(13)-1(-8)=26+8=34 \neq 0$
Thus $A$ is non $-$ singular.
Therefore, its inverse exists.
Now,
$A_{11}=13, A_{12}=5, A_{13}=3$
$A_{21}=8, A_{22}=-10, A_{23}=-6$
$A_{31}=1, A_{32}=3, A_{33}=-5$
$\therefore A^{-1}=\frac{1}{|A|}(\text{adj} A)=\frac{1}{34}\left[\begin{array}{ccc}13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -16 & -5\end{array}\right]$
$\therefore X=A^{-1} B=\frac{1}{34}\left[\begin{array}{ccc}13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5\end{array}\right]\left[\begin{array}{c}1 \\ \frac{3}{2} \\ 9\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{34}\left[\begin{array}{l}13+12+9 \\ 5-15+27 \\ 3-9-45\end{array}\right]$
$=\frac{1}{34}\left[\begin{array}{c}34 \\ 17 \\ -51\end{array}\right]$
$=\left[\begin{array}{c}1 \\ \frac{1}{2} \\ -\frac{3}{2}\end{array}\right]$
Hence, $x=1, y=\frac{1}{2},$ and $z=-\frac{3}{2}$

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