Question

The system of equaions, given below, has
$x+2y+4z=2$
$4x+3y+z=5$
$3x+2y+3z=1$

• A. a unique solution
• B. two solutions
• C. no solution
• D. more than two solutions

Answer 1

The correct option is A a unique solution

Let $AX=B$
$\left(\begin{array}{ccccc}1& 2& 4& ⋮& 2\\ 4& 3& 1& ⋮& 5\\ 3& 2& 3& ⋮& 1\end{array}\right)$
$R_2\to R_2-4R_3\to R_3-3R_1$
$-\left(\begin{array}{ccccc}1& 2& 4& ⋮& 2\\ 0& 5& -15& ⋮& -3\\ 0& -4& -9& ⋮& -5\end{array}\right)$
$R_3\to 5R3-4R_2$
$\left(\begin{array}{ccccc}1& 2& 4& ⋮& 2\\ 0& 5& -15& ⋮& -3\\ 0& 0& 15& ⋮& -13\end{array}\right)$
=Echelon form
$\because$ Rank A= Rank of [A:B]=3=number of variables.
Hence unique solution exists.