Show consistency of equations

[Gregg]

Homework Statement

Show that the simultanbeous equations

$6x-7y+2z=4$

$6x-y-z=7$

$2x-3y+z=k$

where k is a constant, are consisten only when k=1.

The Attempt at a Solution

Don't know how to start, the determinent of the co-efficient matrix is -9. This means they are independant, which means I can't express multiples of (1) and (2) for (3) right? I tried to get x and y in terms of z, then substitute for (3)... Doesn't work. I need the method.

 

[statdad]

Set up the augmented matrix of your system of equations (since the order in which the equations are given is unimportant, you can set things up this way)

$$\begin{pmatrix} 2 & -3 & \hphantom{-}1 & k \\ 6 & -7 & \hphantom{-}2 & 4\\ 6 & -1 & -1 & 7 \end{pmatrix}$$


Use row operations to reduce this to row reduced echelon form. You'll see why when you reach that form.

 

[Architeuthis Dux]

I would like to clarify the terms used in the problem. Consistency in equations means that there exists a solution that satisfies all the equations simultaneously. In other words, the equations are not contradictory and can be solved together. In this case, we are looking for a value of k that makes the equations consistent.

To start, we can rewrite the equations as a matrix equation:

A * x = b

where A is the coefficient matrix, x is the vector of variables (x, y, z), and b is the constant vector (4, 7, k).

The determinant of the coefficient matrix, as mentioned, is -9. This means that the equations are consistent if and only if the determinant of the augmented matrix [A | b] is also -9. In other words, the equations are consistent if and only if the following equation holds:

6x-7y+2z=4

6x-y-z=7

2x-3y+z=k

has a unique solution.

To find the value of k that satisfies this condition, we can use Gaussian elimination to reduce the augmented matrix to row-echelon form. This can be done by performing elementary row operations on the matrix without changing the solution of the system of equations.

After performing Gaussian elimination, we get the following matrix:

[1 0 -1/3 | (k+2)/3]

[0 1 -1/3 | (k+4)/3]

[0 0 1 | k-1]

From this, we can see that the equations are consistent if and only if k-1=0, which means that k=1.

In conclusion, the equations are consistent only when k=1. Any other value of k would result in an inconsistent system of equations.