Question

Use matrix to solve the following system of equations
$x+y+z=3$

$x+2y+3z=4$
$2x+3y+4z=7$

• A. $x=2+k,y=-1-2k,z=-k$ where $k\in R$
• B. $x=2+k,y=1-2k,z=k$ where $k\in R$
• C. $x=-2-k,y=1-2k,z=-k$ where $k\in R$
• D. $x=-2+k,y=-1+2k,z=-k$ where $k\in R$

Answer 1

The correct option is B $x=2+k,y=1-2k,z=k$ where $k\in R$

Given system of equations can be written as
$AX=B$
where $A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 2& 3& 4\end{array}\right]$
$X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ ;$B=\left[\begin{array}{c}3\\ 4\\ 7\end{array}\right]$
Here, $|A|=0$
Now, we will find $\left(adjA\right)B$
$adjA=C^T={\left[\begin{array}{ccc}-1& 2& -1\\ -1& 2& -1\\ 1& -2& 1\end{array}\right]}^T$
$\Rightarrow adjA=\left[\begin{array}{ccc}-1& -1& 1\\ 2& 2& -2\\ -1& -1& 1\end{array}\right]$
Now, $\left(adjA\right)B=\left[\begin{array}{ccc}-1& -1& 1\\ 2& 2& -2\\ -1& -1& 1\end{array}\right]\left[\begin{array}{c}3\\ 4\\ 7\end{array}\right]$
$\Rightarrow \left(adjA\right)B=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
$\Rightarrow \left(adjA\right)B=O$
Hence,the system of equations has infinitely many solutions.
Let $z=k$ where $k\in R$
Then
$x+y=3-k$
$x+2y=4-3k$
Solving these eqns, we get
$y=1-2k;x=2+k$