The system of linear equations gathered x+y+z=6 2x+5y+az=36 x+2y+3z=b gathered has :

Question

The system of linear equations

$\begin{gathered} x+y+z=6 \ 2x+5y+az=36 \ x+2y+3z=b \end{gathered}$

has :

Accepted Answer

For the system to have infinitely many solutions, the determinant of the coefficient matrix ( $\Delta$ ) must be zero.

$\begin{gathered} \Delta=|\begin{array}{ccc}1 \quad  1 \quad  1 \ 2 \quad  5 \quad  a \ 1 \quad  2 \quad  3\end{array}|=0 \ 1(15-2a)-1(6-a)+1(4-5)=0 \ 15-2a-6+a-1=0 \ 8-a=0\Rightarrow a=8 \end{gathered}$

Now, for infinitely many solutions, ${\Delta}_{x}={\Delta}_{y}={\Delta}_{z}=0$.

Check ${\Delta}_{x}$ :

$\begin{gathered} {\Delta}_{x}=|\begin{array}{ccc}6 \quad  1 \quad  1 \ 36 \quad  5 \quad  8 \ b \quad  2 \quad  3\end{array}|=0 \ 6(15-16)-1(108-8b)+1(72-5b)=0 \ 6(-1)-108+8b+72-5b=0 \ -6-108+72+3b=0 \ -42+3b=0\Rightarrow 3b=42\Rightarrow b=14 \ {\Delta}_{y}=|\begin{array}{ccc}1 \quad  6 \quad  1 \ 2 \quad  36 \quad  8 \ 1 \quad  14 \quad  3\end{array}|=1(108-112)-6(6-8)+1(28-36)=-4+12-8=0 \ {\Delta}_{z}=|\begin{array}{ccc}1 \quad  1 \quad  6 \ 2 \quad  5 \quad  36 \ 1 \quad  2 \quad  14\end{array}|=1(70-72)-1(28-36)+6(4-5)=-2+8-6=0 \end{gathered}$

If $a=8$ and $b
e 14$, the system has no solution.

If $a=8$ and $b=14$, the system has infinitely many solutions.

If $a
e 8$, the system has a unique solution regardless of $b$.

Solutions · 1