Question

Number of solutions for the system of linear equations, $x+y+z=6,x+2y+3z=14,$ and $x+4y+2z=30$ are

• A. infinite
• B. unique
• C. zero

Answer 1

The correct option is A infinite

Here, $AX=B$
$\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 4& 7\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}6\\ 14\\ 30\end{array}\right]$

Let's calculate $|A|$
$|A|=1(14-12)-1(7-3)+1(4-2)=2-4+2=0$
$\therefore$ The given system of equations either has no solution or infinite number of solution.

Cofactor matricx of A is:
$\left[\begin{array}{ccc}2& -4& 2\\ -3& 6& -3\\ 1& -2& 1\end{array}\right]$

Adjoint of A is:
$\left[\begin{array}{ccc}2& -3& 1\\ -4& 6& -2\\ 2& -3& 1\end{array}\right]$
$\therefore adj\left(AB\right)=\left[\begin{array}{ccc}2& -3& 1\\ -4& 6& -2\\ 2& -3& 1\end{array}\right]\times \left[\begin{array}{c}6\\ 14\\ 30\end{array}\right]$
$=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]=0$

We know that if $|A|=0$ and adj$\left(AB\right)=0$, then there exist infinite solutions.