Question

For what value of a, if any, will the following system of equations in x,y and z have solution?

$2x+3y=4$
$x+y+z=4$
$x+2y-z=a$

• A. Any real number
• B. 0
• C. 1
• D. There is no such value

Answer 1

The correct option is B 0

$[A:B]=\left[\begin{array}{ccc}1& 1& 1:4\\ 2& 3& 0:4\\ 1& 2& -1:a\end{array}\right]$
$R_2\to R_2-2R_1$
$R_3\to R_3-R_1$
$\sim \left[\begin{array}{ccccc}1& 1& 1& :& 4\\ 0& 1& -2& :& -4\\ 0& 1& -2& :& a-4\end{array}\right]$
$R_3\to R_3-R_2$
$\sim \left[\begin{array}{ccccc}1& 1& 1& :& 4\\ 0& 1& -2& :& -4\\ 0& 0& 0& :& a\end{array}\right]$

If a = 0, then

rank(A)=2 and rank([A:B])=2
$\Rightarrow rank\left(A\right)=rank(A:B)$
So consistent.