Choose h and k such that the system has a solution, a unique solution and many solutions.

Question

Im learning linear algebra, and im tasked with choosing $h$ and $k$ such that this system:

$$ \begin{cases} x_1+hx_2=2\\ 4x_1+8x_2=k\\ \end{cases} $$

Has (a) no solution, (b) a unique solution, and (c) many solutions.

First i made an augmented matrix, then performed row reduction:

$$ \left[ \begin{array}{cc|c} 1&h&2\\ 4&8&k \end{array} \right] \sim \left[ \begin{array}{cc|c} 1&h&2\\ 0&8-4h&k-8 \end{array} \right]$$ Continuing row reduction, i get: $$\sim \left[ \begin{array}{cc|c} 1&0&\frac{k-8}{2(h-2)}+\frac{k}{4}\\ 0&1&\frac{k-8}{8-4h} \end{array} \right]$$ Im not sure how to go about solving the problem with the matrix i end up with? Or if im going about the problem in the correct manner?

Answer 1

It may make it easier to see what happens if you leave the ratio in the first row of the "constants column" in your reduced matrix over a common denominator:

$$ \left[\begin{array}{cc|c}1&0&\frac{16 - hk}{4(2-h)}\\0&1&\frac{k-8}{4(2-h)}\end{array}\right]$$

As pointed out by Christiaan Hattingh, there's trouble if $ \ h \ = \ 2 \ $ for any value of $ \ k \ $ . So the system of equations is consistent (one solution) if $ \ h \ \neq \ 2 \ $ . If $ \ h \ = \ 2 \ $ and $ \ k \ = \ 8 \ $ , we would have $ \ \frac{0}{0} \ $ in both entries of the constant column. This would be interpreted to mean that both variables take on "indeterminate" values. The original matrix looks like:

$$ \left[ \begin{array}{cc|c} 1&2&2\\ 4&8&8 \end{array} \right] \ \ , $$

in which the second row is a multiple of the first; thus we have a "dependent" system with the single line equation $ \ x + 2y \ = \ 2 \ $ . The remaining case is the one for which $ \ h \ = \ 2 \ $ and $ \ k \ \neq \ 8 \ , $ giving us a matrix

$$ \left[ \begin{array}{cc|c} 1&2&2\\ 4&8&\neq8 \end{array} \right] \ \ , $$

which is now an "inconsistent" system, one with no solutions. (In the row-reduced augmented matrix at the start of this post, the entries in the constants column become two ratios having a non-zero number divided by zero.)

Answer 2

It is better not to do row reduction when you have unknown parameters such as $h$ and $k$ in the matrix...for example suppose $h=2$, what happens to your last matrix?

Since you have a very simple system you can derive the answers directly, by considering the following:

1. The system will have no solution when the coefficient matrix rows are linearly dependent (one row is a multiple of the other), BUT the augmented matrix rows are linearly independent.
2. The system will have exactly one solution when the coefficient matrix rows are linearly independent.
3. The system will have many solutions when the augmented systems rows are linearly dependent.

Answer 3

Write the system as $A_h x = b_k$.

If $\det A_h \neq 0$ then there is a unique solution $x = A_h^{-1} x_k$.

If $\det A_h = 0$, then there no solution if $b_k \notin {\cal R} A_h$, and many solutions otherwise (since $\ker A_h \neq \{0\}$).

Since $\det A_h = 8-4h$, we see that $\det A_h \neq 0$ iff $h \neq 2$.

If $h=2$, then ${\cal R} A_h = \operatorname{sp}\{ \binom{2}{8} \}$, hence $b_k \notin {\cal R} A_h$ iff $k\neq 8$