This question is not for an assignment, it was on the midterm and I am interested in figuring out how to solve it before the final exam.
cheers,
Determine when the system has a) no solution, b) 1 solution and c) infinitely many solutions
Given the system
\begin{align*}
x + y + 7z & = -7\\
2x + 3y +17z & = -16\\
x +2y +(a^2+1)z & = 3a
\end{align*}
Attempted Solution
using matrix form $$\pmatrix{1&1&7&-7\cr2&3&17&-16\cr1&2&a^2+1&3a\cr}$$ using gaussian elimination to isolate eqn 3 for variable z (sorry about the missing a^2+1 up there html doesn't like me)
line 1 – line 3
$$\pmatrix{1&1&7&-7\cr2&3&17&-16\cr 0&1&7-(a^2+1)&-7-3a\cr}$$
2*line 1 – line 2
$$\pmatrix{1&1&7&-7\cr 0&-1&-3&2 \cr 0&-1&7-(a^2+1)&-7-3a\cr}$$
-1*line2
$$\pmatrix{1 & 1 &7 &-7\\ 0 & 1 & 3 & -2\\ 0 & -1 & 7 – (a^2 + 1) & -7 – 3a}$$
-1*line 2 + line 1
$$\pmatrix{1 & 0 & 4 & -5\\ 0 & 1 & 3 & -2\\ 0 & -1 & 7 – (a^2 + 1) & -7 – 3a}$$
line 2 + line 3
$$\pmatrix{1 & 0 & 4 & -5\\ 0 & 1 & 3 & -2\\ 0 & 0 & 10 – (a^2 + 1) & -9 – 3a}$$
so we get equation 3
$10-(a^2+1)z=-9-3a$
rearranging,
$$(a^2+1)z=-19-3a \rightarrow$$
$$z=19/a^2+1 – 3a/a^2+1 \rightarrow $$
$$z=(19/a^2+1) – (-3/a+1)$$
$$z=19/a^2 + 1 – (-3/a) +1$$
$$z=19/a^2 + 2 + 3/a$$
multiply all terms by $a$
$$za=19/a + 2a + 3$$
$$za=(39/19)a + 3$$
$$za=3a+3$$
$$z=3+3/a$$
wut?
I am unsure how to properly answer the question, please help stack exchange!
My algebra skills are a little lacking, please point out helpfully where my thinking is wrong.
I can solve for z but what does that prove
Best Answer
You have arrived to this (
which is followed by some not very reader friendly part, which I am ignoringEDIT: In the meantime, N. F. Taussig fixed it): $$\left(\begin{array}{ccc|c} 1&1&7&-7\\ 0&-1&-3&2\\ 0&-1&7-(a^2+1)&-7-3a \end{array}\right)= \left(\begin{array}{ccc|c} 1&1&7&-7\\ 0&-1&-3&2\\ 0&-1&6-a^2&-7-3a \end{array}\right)\sim \left(\begin{array}{ccc|c} 1&1&7&-7\\ 0&-1&-3&2\\ 0&0&9-a^2&-9-3a \end{array}\right)= \left(\begin{array}{ccc|c} 1&1&7&-7\\ 0&-1&-3&2\\ 0&0&(3-a)(3+a)&-3(3+a) \end{array}\right) $$For $a=-3$ you have $$\left(\begin{array}{ccc|c} 1&1&7&-7\\ 0&-1&-3&2\\ 0&0&0&0 \end{array}\right)$$ This system has infinitely many solutions.
If $a\ne-3$, you can divide by $a+3$ to get $$ \left(\begin{array}{ccc|c} 1&1&7&-7\\ 0&-1&-3&2\\ 0&0&3-a&-3 \end{array}\right) $$ If $a=3$, then it has no solution. (The last equation is $0x+0y+0z=3$.)
For $a\ne3$ you can see that there is only one solution. (Maybe it would be better if I wrote for $a\ne\pm3$; just to remind that now we only consider the case when $a\ne-3$ and we got an additional condition $a\ne3$ to get unique solution.)
As a sanity check, you can plug $a=3$ in the original system. You can notice that if you subtract the first equation from the second, you get $$x+2y+10z=-9.$$
For $a=3$ the third equation is $$x+2y+10z=9.$$ These two equations are clearly incompatible.
Also for $a=-3$ you can see that the third equation becomes $$x+2y+10z=-9.$$ So in this case the third equation is direct consequence of the first two equations.
I will also point out that if I was doing the Gaussian (or Gauss-Jordan) elimination, I would probably start by subtracting the first row from the second, since then I have two very similar rows, which makes things easier). $$\left(\begin{array}{ccc|c} 1 & 1 & 7 &-7 \\ 2 & 3 &17 &-16\\ 1 & 2 &a^2+1& 3a \end{array}\right)\sim \left(\begin{array}{ccc|c} 1 & 1 & 7 &-7 \\ 1 & 2 &10 &-9\\ 1 & 2 &a^2+1& 3a \end{array}\right)\sim \left(\begin{array}{ccc|c} 1 & 1 & 7 &-7 \\ 1 & 2 &10 &-9\\ 0 & 0 &a^2-9& 3a+9 \end{array}\right)\sim \left(\begin{array}{ccc|c} 1 & 1 & 7 &-7 \\ 1 & 2 &10 &-9\\ 0 & 0 &(a-3)(a+3)& 3(a+3) \end{array}\right)\sim \left(\begin{array}{ccc|c} 1 & 1 & 7 &-7 \\ 0 & 1 & 3 &-2\\ 0 & 0 &(a-3)(a+3)& 3(a+3) \end{array}\right)$$