Question

Solve the system of equations $x + 2y + z = 4$, $- x + y + z = 0$ and $x - 3y + z = 4$.

• A. $x = 2$ , $y = 0$, $z = 2$
• B. $x = 2$,$y = 0$, $z = -2$
• C. $x = -2$, $y = 2$, $z = 0$
• D. $x = -2$, $y - 0$, $z = 2$

Answer 1

Answer: (A)

Given system of equations is $x + 2y + z = 4$, $- x + y + z = 0$ and $x - 3y + z = 4$
In matrix form, it can be written as
$\left[\begin{matrix}1&2&1\\ -1&1&1\\ 1&-3&1\end{matrix}\right]\left[\begin{matrix}x\\ y\\ z\end{matrix}\right]=\left[\begin{matrix}4\\ 0\\ 4\end{matrix}\right]$
or $AX = B$
where, $A =\left[\begin{matrix}1&2&1\\ -1&1&1\\ 1&-3&1\end{matrix}\right], B=\left[\begin{matrix}4\\ 0\\ 4\end{matrix}\right] and X=\left[\begin{matrix}x\\ y\\ z\end{matrix}\right]$
Now, $|A| = 1\left(1 + 3\right) - 2 \left(-1 - 1\right) + 1\left(3 - 1\right)= 4 + 4 + 2 = 10$
$\because\quad\left|A\right|\ne0$, hence unique solution exists.
Cofactor matrix of $A =\left[\begin{matrix}4&2&2\\ -5&0&5\\ 1&-2&3\end{matrix}\right]$
$\therefore \quad adj\left(A\right) =\left[\begin{matrix}4&2&2\\ -5&0&5\\ 1&-2&3\end{matrix}\right]^{T}= \left[\begin{matrix}4&-5&1\\ 2&0&-2\\ 2&5&3\end{matrix}\right]$
$A^{-1}=\frac{1}{\left|A\right|} adj \left(A\right)=\frac{1}{10}\left[\begin{matrix}4&-5&1\\ 2&0&-2\\ 2&5&3\end{matrix}\right]
Now, X=A^{-1}B=\frac{1}{10}\left[\begin{matrix}4&-5&1\\ 2&0&-2\\ 2&5&3\end{matrix}\right]\left[\begin{matrix}4\\ 0\\ 4\end{matrix}\right]$
$\left[\begin{matrix}x\\ y\\ z\end{matrix}\right]=\frac{1}{10} \left[\begin{matrix}16+0+4\\ 8+0-8\\ 8+0+12\end{matrix}\right]=\frac{1}{10} \left[\begin{matrix}20\\ 0\\ 20\end{matrix}\right]=\left[\begin{matrix}2\\ 0\\ 2\end{matrix}\right]$
On comparing the corresponding elements, we get $x = 2$, $y = 0$ and $z = 2$