Choosing two variables which determine whether a Ax= b has infinite, unique or no solutions.

Question

I've been given a problem regarding the solution set of a matrix:

    Let A = 1  2 -5         1
           -1  3  0 ,   b = 0
            1 -2  k         h

Find the value(s) of k and h such that the matrix equation Ax = b has

(a) a unique solution (b) infinitely many solutions (c) no solutions.

I've made some attempts at this however I find myself stumped. One of my attempts went as follows:

(1) 1 2 -5 |1 R1<= R1+ R2 -1 3 0 |0 1 -2 k |h

(2) 0 5 -5 |1
-1 3 0 |0 1 -2 k |h R3<= R3 + R2

(3) 0 5 -5 |1 R1<=>R2 -1 3 0 |0 0 1 k |h

(4) -1 3 0 |0
0 5 -5 |1 R2<=>R3 0 1 k |h

(5) -1 3 0 |0
0 1 k |h 0 5 -5 |1 R3<= R3 - 5*R2

(6) -1 3 0 |0
0 1 k |h 0 0 -5-(5*k) |1 -5*h

And so thus far I've been trying to find a unique solution using multiples of 5( I'm trying to find the unkown "x1" or rather "z") but It's just headache inducing at the moment.

I can easily pick -1 for k to make x1(or z) 0 and just any arbitary number for h in order to make the matrix have no solution, but determining a value for k and h to find infinitely many solutions and a unique solution escapes me.

Any help would be much appreciated.

Answer 1

Let's start by forming the augmented matrix $[A|b]$ and doing elementary row operations to get the $A$ part into row echelon form. $$\left[\begin{array}{ccc|c}1&2&-5&1\\-1&3&0&0\\1&-2&k&h\end{array}\right] \Rightarrow_{1} \left[\begin{array}{ccc|c}1&2&-5&1\\0&5&-5&1\\0&-4&k+5&h-1\end{array}\right] \Rightarrow_{2} \left[\begin{array}{ccc|c}1&2&-5&1\\0&5&-5&1\\0&0&k+1&h-\frac{1}{5}\end{array}\right]$$

where step 1 is: $$\mathrm{R}_2 \leftarrow \mathrm{R}_2 + \mathrm{R}_1 \\ \mathrm{R}_3 \leftarrow \mathrm{R}_3 - \mathrm{R}_1$$

and step 2 is: $$\mathrm{R}_3 \leftarrow \mathrm{R}_3 + \frac{4}{5}\mathrm{R}_2 $$

The last row on the right is the key the whole problem. Notice that when we do back substitution, the last row represents the equation... $$(k+1)x_3 = h- \frac{1}{5}$$

Now, if $k \ne -1$, then, the the matrix $A$ has full rank and the system $Ax = b$ has a unique solution for any value of $h$. Things get interesting when $(k +1) = 0$. If $k = -1$, then the equation in the last row becomes: $$0\cdot x_3 = h- \frac{1}{5}$$ and in this case, if $h$ is such that we get $0 \ne0$ (ie, $h \ne \frac{1}{5}$), the system is inconsistent - there is no solution. If however, $h = \frac{1}{5}$ then we're left with $0 = 0$ in that last row and the system has infinitely many solutions.

Answer 2

Hints: Here is one approach, but you should try this with Gaussian Elimination on the $3x3$.

Notice that the second row gives $x = 3y$, so we can eliminate this row, rewrite the system as a $2x2$ by substituting that into rows $1$ and $3$ of the original system as:

$$5y - 5z = 1 \\ 1 y + k z = h$$

Some observations:

• The determinant of the $2x2$ is $5(1+k)$. What is the determinant not supposed to be for solutions to exist?
• What happens to this system if you make $k = -1, h = t \ne \dfrac{1}{5}$? In other words, $h = t$ is almost a free variable and can be anything. Just look at the rows of the $2x2$. (No solutions)
• What happens if $k = -1, h = \dfrac{1}{5}$? (Infinite solutions)
• Can you continue with this system and find a unique solution from the two previous statements? If not, try it with Cramer's Rule, Gaussian Elimination or whatever methods you have learned.