Do simultaneous equations always have a unique solution?

[chwala]

Homework Statement

consider the following system of equations;
$x+2y+mz=0$
$x+3y-z=3$
$mx+8y+5z=6$
where $m$ is a scalar. For what values of m, does the system have a (i) unique solution (ii)many solutions (iii) no solution

Relevant Equations

simultaneous equations

...$2y-2z-2mz=6$
$2y-2z-2mz= mx+8y+5z$
i then let,
$-2z-2mz=5z$
$m=-7/2$
$→2y=mx+8y$
$7x=12y$

 

[Mark44]

Problem Statement: consider the following system of equations;
$x+2y+mz=0$
$x+3y-z=3$
$mx+8y+5z=6$
where $m$ is a scalar. For what values of m, does the system have a (i) unique solution (ii)many solutions (iii) no solution
Relevant Equations: simultaneous equations

...$2y-2z-2mz=6$
$2y-2z-2mz= mx+8y+5z$
i then let,
$-2z-2mz=5z$
$m=-7/2$
$→2y=mx+8y$
$7x=12y$

I think you're heading in the wrong direction.
Look at the system of equations as a matrix equation, $A\vec x = \vec b$, where $A = \begin{bmatrix} 1 & 2 & m \\ 1 & 3 & -1 \\ m & 8 & 5\end{bmatrix}$, $\vec x = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ and $\vec b = \begin{bmatrix} 0 \\ 3 \\ 6 \end{bmatrix}$.

Finding the determinant of A will tell you a lot about whether the system has a unique solution or not.

 

[chwala]

i am getting the equation $3m^2-6m-13=0$ giving $m_1=-1.31$ and $m_2$=$3.31$ since the matrix is non-singular, it follows that the system will have a unique solution. Conversely, if the system is singular then no solution exists. can we use the row reduction echelon form to determine the solutions? am i supposed to think of 'rank' and dimension here? long since i read this...

 

[Mark44]

i am getting the equation $3m^2-6m-13=0$

For the determinant, I get $3m^2 + 6m + 13$. I've checked my work twice, so am pretty sure of it.

Edit: Despite checking my work, I still missed a sign error.

giving $m_1=-1.31$ and $m_2$=$3.31$ since the matrix is non-singular, it follows that the system will have a unique solution. Conversely, if the system is singular then no solution exists. can we use the row reduction echelon form to determine the solutions? am i supposed to think of 'rank' and dimension here? long since i read this...

 

[chwala]

For the determinant, I get $3m^2 + 6m + 13$. I've checked my work twice, so am pretty sure of it.

Your characteristic equation is not correct. Did you consider the + and - place values for the 3 by 3 matrix before coming up with the equation?

 

[LCKurtz]

I let Maple calculate the determinant and I'm agreeing with Chwala.

 

[Mark44]

i am getting the equation $3m^2-6m-13=0$

Apologies, the equation above is correct -- I misread a sign.

giving $m_1=-1.31$ and $m_2$=$3.31$ since the matrix is non-singular, it follows that the system will have a unique solution. Conversely, if the system is singular then no solution exists.

If the determinant is zero, the matrix is singular, so there will be either no solution or an infinite number of them. In this case, because the column of constants is nonzero, there won't be any solutions when the determinant is zero.

If the determinant is nonzero, there will be a unique solution for each value of m for which $3m^2 - 6m - 13 \ne 0$

can we use the row reduction echelon form to determine the solutions? am i supposed to think of 'rank' and dimension here? long since i read this...

The problem doesn't ask for solutions -- just whether there are solutions.

 

[vela]

In this case, because the column of constants is nonzero, there won't be any solutions when the determinant is zero.

A non-zero column of constants doesn't necessarily mean there will be no solutions. For example, the system
\begin{align*}
x+y &= 1 \\
y + z &= 2 \\
x + 2y + z &= 3
\end{align*} has an infinite number of solutions.

Row-reducing the augmented matrix can show you if the system has solutions or not.

 

[Mark44]

A non-zero column of constants doesn't necessarily mean there will be no solutions. For example, the system
\begin{align*}
x+y &= 1 \\
y + z &= 2 \\
x + 2y + z &= 3
\end{align*} has an infinite number of solutions.

By "in this case" I meant when the determinant is zero. If the determinant of the matrix of coefficients is zero, but the column of constants in the augmented matrix is nonzero, there won't be any solutions.

Edit based on @vela's reply in post 10 -- I stand corrected.

 

[vela]

By "in this case" I meant when the determinant is zero. If the determinant of the matrix of coefficients is zero, but the column of constants in the augmented matrix is nonzero, there won't be any solutions.

For the example I gave in my previous post, we have
$$A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 2 & 1 \end{pmatrix}\text{ and }\vec b = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}.$$ The third row of $A$ is the sum of the first two rows, so we have $\det(A) = 0$. According to your statement, this system has no solution because $\vec b \ne 0$. But $(x,y,z)= (1,0,2)$ is a solution.

 

[Mark44]

For the example I gave in my previous post, we have
$$A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 2 & 1 \end{pmatrix}\text{ and }\vec b = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}.$$ The third row of $A$ is the sum of the first two rows, so we have $\det(A) = 0$. According to your statement, this system has no solution because $\vec b \ne 0$. But $(x,y,z)= (1,0,2)$ is a solution.

I take back my earlier remark... Thanks for the correction.