Matrix systems of equations.

[WTF?]

Hi and thanks for your time.

I have a problem to which I do not know how to aproach it.

1.For what value n does
2x +4y=n
x +2y=7

Have infinitely many solutions?

The books says 14, but I don't know how they got it.

Another one...
2.For what value of n does
4x - 6y =5
2x +ny=2

Have no solution?
I got 3 for this one. Is it correct?

 

[stapel]

1) To have "infinitely many solutions", the one equation has to be a multiple of the other. So what multiple of "x + 2y" is "2x + 4y"? So what multiple of "7" does "n" need to be?

2) To have "no solutions", the variable parts must be the same (or multiples of each other) but the plain number part (on the other side of the "equals" sign) must be unequal (or at least not the same multiples as the variable parts were). What multiple of "2x" is "4x"? What should "n" be so that "ny" and "-6y" are the same multiple?

Eliz.

 

[soroban]

Hello, WTF!

Here's another approach . . . Try to solve the systems.

1. For what value of $\displaystyle n$ does .$\displaystyle \begin{Bmatrix}2x\,+\,y\:=\:n\\x\,+\,2y\:=\:7\end{Bmatrix}$ .have infinitely many solutions?

A system has infinitely many solutions if we get a true statement, like: $\displaystyle 0\,=\,0$ or $\displaystyle 2\,=\,2$.

Multiply the second equation by -2:

. . . . $\displaystyle 2x\,+\,4y\:=\:\,n$
. . . .-$\displaystyle 2x\,-\,4y\:=\:$-$\displaystyle 14$

Add the equations: .$\displaystyle 0\:=\:n\,-\,14$

We get a true statement "0 = 0" if $\displaystyle n\,=\,14$.

2. For what value of $\displaystyle n$ does .$\displaystyle \begin{Bmatrix}4x\,-\,6y\:=\:5\\2x\,+\,y\:=\:2\end{Bmatrix}$ .have no solution?

A system has no solution if we get a false statement, like $\displaystyle 1 = 2$
. . or we get answers which are undefined.

Multiply the second equation by -2:

. . . . .$\displaystyle 4x\,-\;6y\:=\;5$
. . . .-$\displaystyle 4x\,-\,2ny\:=\,$-$\displaystyle 4$

Add the equations: .-$\displaystyle 6y\,-\,2ny\:=\:1$

. . . Factor: .-\(\displaystyle 2(3\,+\,n)y\:=\:1\:\:\Rightarrow\;\;y\,=\,-\frac{1}{2(n+3)\)

$\displaystyle y$ has a value for every value of $\displaystyle n$ . . . except $\displaystyle n\,=\,$-$\displaystyle 3$


Therefore, when $\displaystyle n\,=\,$-$\displaystyle 3$, the system has no solution.

 

[soroban]

Hello, WTF!

Silly me . . . I just noticed your heading: "Matrix systems of equations".

2) For what value of $\displaystyle n$ does .$\displaystyle \begin{Bmatrix}4x\,-\,6y\:=\:5\\2x\,+\,ny\:=\:2\end{Bmatrix}$ .have no solution?

A system has no solution if the determinant of its coefficients, $\displaystyle D$, equals 0.

So we have: .$\displaystyle D\:=\:\begin{vmatrix}4&-6\\2&n\end{vmatrix}\:=\:0\;\;\Rightarrow\;\;4n\,+\,12\:=\:0\;\;\Rightarrow\;\;n\,=\,-3$