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સુરેખ સમીકરણોની સંહતિનો ઉકેલ શ્રેણિકના ઉપયોગથી મેળવો : $x-y+2 z=7$ ; $3 x+4 y-5 z=-5$ ; $2 x-y+3 z=12$
$x=2, y=1,z=3$
$x=-2, y=-1,z=3$
$x=-2, y=-1,z=3$
$x=2, y=1,z=-3$
Solution
The given system of equation can be written in the form of $A X=B$, where
$A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and
$B=\left[\begin{array}{c}7 \\ -5 \\ 12\end{array}\right]$
Now,
$|A|=1(12-5)+1(9+10)+2(-3-8)=7+19-22=4 \neq 0$
Thus, $A$ is non-singular. Therefore, its inverse exists.
Now,
$A_{11}=7, A_{12}=-19, A_{13}=11$
$A_{11}=1, A_{22}=-1, A_{23}=-1$
$A_{31}=-3, A_{32}=11, A_{33}=7$
$\therefore A^{-1}=|A|^{(a d j A)}=\frac{1}{4}\left[\begin{array}{ccc}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{array}\right]$
$\therefore X=A^{-1} B=\frac{1}{4}\left[\begin{array}{ccc}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{array}\right]\left[\begin{array}{c}7 \\ -5 \\ 12\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}49-5-36 \\ -133+5+132 \\ -77+5+84\end{array}\right]$
$=\frac{1}{4}\left[\begin{array}{l}8 \\ 4 \\ 12\end{array}\right]=\left[\begin{array}{l}2 \\ 1 \\ 3\end{array}\right]$
Hence, $x=2, y=1$ and $z=3$
