Question

Use matrices and decide the value of P and Q so that the system has no solution.

x1+x2+3x3=-2
2x1+x2+Px3=3Q
x1+2x2+5x3=-4

Answer 1

reduce it using row echelon

Answer 2

then I get \[\left[\begin{matrix}1 & 0 & 0 & (-(3*y+2)/(x-4) \\ 0 & 1 & 0 & (-2*(3y+2)/(x-4))-2 \\ 0 & 0 & 1 & (3y+2)/(x-4) \end{matrix}\right]\]

so what do I do then?

Answer 3

I wrote it wrong on the last post. x is P and y is Q

Answer 4

then i can set the P=4 since the determinant of the coefficient will then be 0. then I get

\[\left[\begin{matrix}1 & 0 & 0 & -3y+2 \\ 0 & 1 & 0 & -2(3y+2)-2 \\ 0 & 0 & 1 & 3y+2\end{matrix}\right]\]

Answer 5

wrote wrong again y=Q.

but what do I do then?

Answer 6

first find when the determinant is zero and solve for P
that will mean the system has either no solution or infinite solutions
plug in the value you get for P and reduce the system using the technique you tried earlier and find which value(s) of Q make the system inconsistent