consider the following linear system
$$x+2y+2z=1\tag{1}$$
$$x+ay+3z=3\tag{2}$$
$$x+11y+az=b\tag{3}$$
in matrix form
$$\pmatrix{1&2&2&1\cr1&a&3&3\cr1&11&a&b\cr}$$
For what values of a will the system have a unique solution, and for which pair of values (a,b) will the system have more than one solution
My attempt at solving it:
-1*line 1 + line 2, and, -1*line 1 + line 3
$$\pmatrix{1&2&2&1\cr0&a-2&1&2\cr0&9
&a-2&b-1\cr}$$
1/(a-2) * line
$$\pmatrix{1&2&2&1\cr0&a-2&1&2\cr0&9/(a-2)&1&(b-1)/(b-2)\cr}$$
-1*line 3 + line 2
$$\pmatrix{1&2&2&1\cr0&((a-5)(a+1))/(a-2)&0&(-b+2a-3)/(a-2)\cr0&9/(a-2)&1&(b-1)/(b-2)\cr}$$
I don't know how to proceed beyond this, help guide me stack exchange!
Also if someone could touch up my matrices formatting that would be soooo cool.
Best Answer
Here is the answer to the updated question.
The original system is equivalent to the following system: $$x+2y+2z=1\tag{1}$$ $$x+ay+3z=3\tag{2}$$ $$x+11y+az=b\tag{3}$$
Solve $y,z$ from (1) and (2) and substitution of it into (3) leads to:
$$\frac{(33+a^2+6b-2a(3+b))}{2(a-3)}+\frac{5+4a-a^2}{2(a-3)}=g(a,b)+f(a)x=0\tag{4}$$
(1)If $f(a)=g(a,b)=0$, then $x$ can take any values.
(3) If $g(a,b)=0$,$f(a)\not=0$, then $x=0$.
(3) If $g(a,b)\not=0$, $f(a)=0$, then $x$ has no solution.
(4) If $g(a,b)\not=0$, $f(a)\not =0$, then $x$ has one solution, $x=-\frac{g(a,b)}{f(a)}$.