Question

Consider the following system of simultaneous equations:
6x+2y-z=1
x+y+z=2
kx+y-z=1.
For what values of k, is there:
i) a unique solution?
ii) no solution?

Answer 1

Are you familiar with determinants?
One way of checking whether or not a given matrix equation has a solution or not is to take the determinant of the square matrix (ignoring the fourth column of the augmented matrix).

Answer 2

\[\left[\begin{matrix}6 & 2 & 1 \\ 1 & 1 & 1 \\k & 1 & -1\end{matrix}\right] \] is the square matrix in question.

Answer 3

how would you do it algebraically then?

Answer 4

The "determinant" of this matrix is 6(-1-1) - 1(-2-1) + k(2-1).
If this matrix equals zero, there's no solution.
If this matrix does not equal zero, there may be a unique solution.

Answer 5

So, what we'll do first is to set this determinant equal to zero and solve for k:
6(-2) -1(-3) + k = 0
-12 + 3 + k + 0
k = ???? is the k value for which the system of linear equations has no solution.

Answer 6

Brainstorm on what to do next. How would you determine a value of k for which the system has just one unique solution?

Answer 7

wouldnt it be all values except for k?

Answer 8

@shamil98

Answer 9

No solution where k = 11/3
Refer to the attached Mathematica presentation.

Answer 10

Cindy: I must apologize; while on the right track towards determining the value of k for which this system of linear equations has no solution, the matrix I presented has one incorrect element.
This matrix should be as follows:

6 2 -1
1 1 1
k 1 -1

As stated before, one way to determine whether or not a given system has NO SOLUTION is to find the determinant of the system matrix (of which the above matrix is an example). If the value of that determinant is ZERO, the system has NO SOLUTION.

Evaluating the determinant, using the method of cofactors, results in
6(-1-1) - 1(-2+1) +k(2+1), or -12 + 1 + 3k.
As our computer expert Robert did, we set this equation equal to zero and solve for k.
The result is k = 11/3, same as Robert obtained.
Conclusion: this system of linear equations has no solution if k happens to be 11/3, and the reason for that is that the system determinant is zero for that k value.

Answer 11

If we focus on solution of systems of linear equations using matrices, specifically on augmented system matrices, there are three possible cases:

Case 1: The system has infinitely many solutions if the augmented matrix can be reduced through row operations to a form in which one of the rows (usually the last row) consists of all zeros:

1 0 0 a
0 1 0 b
0 0 0 0

Case 2: The system has no solutions if the augmented matrix can be reduced through row ops to a form in which one of the rows (usually the last row) consists of all zeros WITH THE EXCEPTION THAT THE VERY LAST VALUE IN THAT ROW IS NOT ZERO:

1 0 0 a
0 1 0 b
0 0 0 K

The reason? 0 0 0 K translates into the equation 0x + 0y + 0z = K, where K is NOT zero.

For example:

0 0 0 5 translates to 0x + 0y + 0z = 5, or 0 = 5, which is obviously not possible.

Case 3: The system has exactly one solution if the augmented matrix can be reduced through row operations to the following form:

1 0 0 a
0 1 0 b
0 0 1 c

or, in other words/symbols, \[\left[ I \right]\left[ X \right]\],

where [I] is the identity matrix and {X] is a 3x1 (column) matrix representing the solution (a, b, c).

Let's hear it for Robert, who obviously knows his stuff where it comes to programming in Mathematica.

Answer 12

@robtobey