Question

Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. X1 + 3x2 = 4 2x1 + kx2 = h

Answer 1

a) The system has a unique solution for
`k\\eq 6` and any value of
`h`, and we say the system is consisted

b) The system has infinite solutions for
`k=6` and
`h=8`

c) The system has no solution for
`k=6` and
`h\\eq 8`

Explanation:

Since we need to base the solutions of the system on one of the independent terms (
`h`), the determinant method is not suitable and therefore we use the Gauss elimination method.

The first step is to write our system in the augmented matrix form:

`\left[\begin{array}cc1&3&4\\2&k&h\end{array}\right]`

The we can use the transformation
`r_0\rightarrow r_0 -2r_1`, obtaining:

`\left[\begin{array}cc1&3&4\\0&k-6&h-8\end{array}\right]`.

Now we can start the analysis:

• If
  `k\\eq 6` then, the system has a unique solution for any value of
  `k`, meaning that the last row will transform back to the equation as:

`(k-6)x_2=h-8\\x_2=h-8/(k-6)`

from where we can see that only in the case of
`k=6` the value of
`x_2` can not be determined.

• if
  `k=6` and
  `h=8` the system has infinite solutions: this is very simple to see by substituting these values in the equation resulting from the last row:

`(k-6)x_2=h-8\\0=0` which means that the second equation is a linear combination of the first one. Therefore, we can solve the first equation to get
`x_1` as a function of
`x_2` o viceversa. Thus,
`x_2` (
`x_1`) is called a parameter since there are no constraints on what values they can take on.

if
`k=6` and
`h\\eq 8` the system has no solution. Again by substituting in the equation resulting from the last row:

`(k-6)x_2=h-8\\0=h-8` which is false for all values of
`h\\eq 8` and since we have something that is not possible
`(0\\eq h-8,\ \forall \ h\\eq 8)` the system has no solution