Question

The values of $x,y$ and $z$ for the system of equations $x+2y+3z=6$, $3x-2y+z=2$ and $4x+2y+z=7$ are respectively

• A. $1,1,1$
• B. $1,2,3$
• C. $1,3,2$
• D. $2,3,1$

Answer 1

The correct option is A $1,1,1$

The given system of equations is
$x+2y+3z=6$ ...(i)
$3x-2y+z=2$ ...(ii)
and $4x+2y+z=7$ ...(iii)
Here, $A=\left[\begin{array}{ccc}1& 2& 3\\ 3& -2& 1\\ 4& 2& 1\end{array}\right]$, $B=\left[\begin{array}{c}6\\ 2\\ 7\end{array}\right]$ and $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$
$\therefore \left|A\right|=\left|\begin{array}{ccc}1& 2& 3\\ 3& -2& 1\\ 4& 2& 1\end{array}\right|$
$=1(-2-2)-2(3-4)+3(6+8)$
$=-4+2+42=40$
Now, $adjA=\left[\begin{array}{ccc}-4& 4& 8\\ 1& -11& 8\\ 14& 6& -8\end{array}\right]$
$\therefore A^{-1}=\frac{1}{\left|A\right|}adjA=\frac{1}{40}\left[\begin{array}{ccc}-4& 4& 8\\ 1& -11& 8\\ 14& 6& -8\end{array}\right]$
Now, $X=A^{-1}B=\frac{1}{40}\left[\begin{array}{ccc}-4& 4& 8\\ 1& -11& 8\\ 14& 6& -8\end{array}\right]\left[\begin{array}{c}6\\ 2\\ 7\end{array}\right]$
$=\frac{1}{40}\left[\begin{array}{c}-24+8+56\\ 6-22+56\\ 84+12-56\end{array}\right]=\frac{1}{40}\left[\begin{array}{c}40\\ 40\\ 40\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$
$\Rightarrow x=1,y=1,z=1$