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

Question

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.

Answer 1

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, $x=-\frac{g(a,b)}{f(a)}$.

Answer 2

The original system is equivalent to the following under-determined system: $$x+2y+2z+w=0\tag{1}$$ $$x+ay+3z+3w=0\tag{2}$$ $$x+11y+az+bw=0\tag{3}$$

Solve $z,w$ from (1) and (2) and substitution of it into (3) leads to:

$$(3-2a+b)x+(33+a^2+6b-2a(3+b))y=f(a,b)x+g(a,b)y=0\tag{4}$$

(1)If $a=-1,b=-5$ or $a=5,b=7$, then $f(a,b)=g(a,b)=0$. Thus $x,y$ can take any values.

(2) If $a\not=-1,5$ and $b=2a-3$, then $f(a,b)=0$ but $g(a,b)\not=0$, so $y=0$ and $x$ can take any value.

(3) If $a\not=-1,5$ and $b=\frac{33-6a+a^2}{2(a-3)}$, then $g(a,b)=0$ but $f(a,b)\not=0$, so $x=0$ and $y$ can take any value.

(4) If $a\not=-1,5$, $b\not=2a-3,\frac{33-6a+a^2}{2(a-3)}$, then $f(a,b)\not =0,g(a,b)\not=0$, so $x=-\frac{g(a,b)}{f(a,b)}y$ and $y$ can take any value.