Question

Consider the following system of equations
1. 4x+3y+3z=-8
2. 2x+y+z=-4
3. 3x-2y+(m^2-6)z)=m-4
Determine the value(s) of m for which this system of equations will have:
a. no solution
b. one solution
c. an infinite number of solution

I don't need answers, I just need some explanation..

Answer 1

@SithsAndGiggles can you help me? please

Answer 2

@phi @Hero @Luigi0210

Answer 3

You matrices

Answer 4

*use

Answer 5

i'm not using matrices..

Answer 6

It looks like sith is answering. This is not a trivial idea to explain... it depends on how much you already know.

Answer 7

Oops, nvm

Answer 8

Here is a lecture that gives background
http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-4-square-systems/

Answer 9

\[\begin{cases}4x+3y+3z=-8\\2x+y+z=-4\\3x-2y+(m^2-6)z=m-4\end{cases}\]
\[\begin{pmatrix}4&3&3\\2&1&1\\3&-2&m^2-6\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}-8\\-4\\m-4\end{pmatrix}\]
I know that if the determinant of the coefficient matrix is non-zero, you have exactly one solution. Using a cofactor expansion (along the first column), you have
\[\begin{align*}\begin{vmatrix}4&3&3\\2&1&1\\3&-2&m^2-6\end{vmatrix}&=4\begin{vmatrix}1&1\\-2&m^2-6\end{vmatrix}+2(-1)\begin{vmatrix}3&3\\-2&m^2-6\end{vmatrix}+3\begin{vmatrix}3&3\\1&1\end{vmatrix}\\
&=4\left(m^2-6-(-2)\right)-2\left(3m^2-18-(-6)\right)+3\left(3-3\right)\\
&=8-2m^2\end{align*}\]
The determinant is zero when \(m=\pm2\), so one solution exists for \(m\in(-\infty,-2)\cup(-2,2)\cup(2,\infty)\).

Plug in \(-2\) and \(2\) for \(m\) and see what happens to the system.

Answer 10

@Data_LG2, do you mean you're currently not using matrices, or you're not supposed to?

Answer 11

i'm not currently using matrices,..

Answer 12

But you have learned about determinants and the like? It's best to use this method. Working this out by hand with elimination or substitution looks dreadful.

Answer 13

I just did elementary operations but i'm stuck..
first, I eliminate X with equation 1 and 2, then I stopped when I'm calculating for eqn. 1 and 3..

Answer 14

4x+3y+3z=-8 ...(1)
2x+y+z=-4 ....(2)
\[3x-2y+\left( m ^{2} -6\right)z=m-4 ...(3)\]
2(2)-(1) gives,
4x+2y+2z-4x-3y-3z=-8+8
-y-z=0
y=-z
2(3)-3(2) gives,
\[6x-4y+2\left( m ^{2}-6 \right)z-6x-3y-3z=2m-8+12\]
\[-7y+\left( 2m ^{2}-12-3 \right)z=2m+4\]
\[-7y+\left( 2m ^{2} -15\right)z=2m+4\]
plug y=-z
\[\left( 2m ^{2} -15+7\right)z=2m+4\]
\[z=\frac{ 2\left( m+2 \right) }{ 2\left( m ^{2} -4\right) }\]
z has no solution if
\[m ^{2}-4=0,m ^{2}=4,m=\pm 2\]
\[z=\frac{ m+2 }{ \left( m+2 \right)\left( m-2 \right) }=\frac{ 1 }{m-2 }\]
it has infinite solutions by giving different value to m except -2 and 2