Question

Solve the following system of equations $x - y + z = 4$, $x -2 y + 2z = 9$ and $2x + y + 3z = 1$.

• A. $x = -4$, $y = -3$, $z = 2$
• B. $x = - 1$, $y = -3$, $z - 2$
• C. $x=2$, $y=4$, $z=6$
• D. $x = 3$, $y = 6$ , $z = 9$

Answer 1

Answer: (B)

$x = - 1$, $y = -3$, $z - 2$

Answer 2

Given system of equations is $x - y + z = 4$, $x - 2y + 2z =9$ and $2x + y + 3z = 1$, which is written in the matrix form as $\left[\begin{matrix}1&-1&1\\\ 1&-2&2\\\ 2&1&3\end{matrix}\right]\left[\begin{matrix}x\\\ y\\\ z\end{matrix}\right]=\left[\begin{matrix}4\\\ 9\\\ 1\end{matrix}\right]$ or $AX = B$ where, $A =\left[\begin{matrix}1&-1&1\\\ 1&-2&2\\\ 2&1&3\end{matrix}\right], B=\left[\begin{matrix}4\\\ 9\\\ 1\end{matrix}\right] and X=\left[\begin{matrix}x\\\ y\\\ z\end{matrix}\right]$ Now, $|A| = \left|\begin{matrix}1&-1&1\\\ 1&-2&2\\\ 2&1&3\end{matrix}\right|=1\left(-6-2\right)+1\left(3-4\right)+1\left(1+4\right)=-8-1+5=-4 \ne0$ $\therefore\quad$ A is invertible So, there exists a unique solution $X = A^{-1}B$. Now, cofactor matrix of $A =\left[\begin{matrix}-8&1&5\\\ 4&1&-3\\\ 0&-1&-1\end{matrix}\right]$ $\therefore\quad adj A =\left[\begin{matrix}-8&1&5\\\ 4&1&-3\\\ 0&-1&-1\end{matrix}\right]^{T}=\left[\begin{matrix}-8&4&0\\\ 1&1&-1\\\ 5&-3&-1\end{matrix}\right]$ So, $A^{-1}=\frac{1}{\left|A\right|} \left(adj A\right)=\frac{-1}{4} \left[\begin{matrix}-8&4&0\\\ 1&1&-1\\\ 5&-3&-1\end{matrix}\right]$ Now, $X=A^{-1}B=\frac{-1}{4} \left[\begin{matrix}-8&4&0\\\ 1&1&-1\\\ 5&-3&-1\end{matrix}\right]\left[\begin{matrix}4\\\ 9\\\ 1\end{matrix}\right]$ $\Rightarrow\quad\left[\begin{matrix}x\\\ y\\\ z\end{matrix}\right]=\frac{-1}{4} \left[\begin{matrix}-32+36+0\\\ 4+9-1\\\ 20-27-1\end{matrix}\right]=\frac{-1}{4}\left[\begin{matrix}4\\\ 12\\\ -8\end{matrix}\right]=\left[\begin{matrix}-1\\\ -3\\\ 2\end{matrix}\right]$ On comparing, we get $x = -1$, $y = -3$ and $z = 2$