Problem with a system of three linear equations with two unknown constants

Question

Given

$$\begin{aligned} x+y+4z &= 1\\ 3x+2y+16z &= 5\\ 4x+2y+(a-1)z &=b-4\end{aligned}$$

find conditions on $a, b \in \mathbb R$ for which:

1. the system has no solutions;
2. the system has only one solution;
3. the system has an infinite number of solutions.

I tried solving to find a point where I found $z = \frac{b-20}{a-25}$. I am not really sure about how I should proceed. Any help would be appreciated.

Answer 1

Your system is$$\left\{\begin{array}{l}x+y+4z=1\\3x+2y+16z=5\\4x+2y+(a-1)z=b-4. \end{array}\right.$$If you subtract from the second line of your system the first line times $3$ and if you subtract from the third line of your system the first line times $4$, you get $$\left\{\begin{array}{l}x+y+4z=1\\-y+4z=2\\-2y+(a-17)z=b-8\end{array}\right.$$Now, if you add to the third line of this system the second line times $-2$, you will get$$\left\{\begin{array}{l}x+y+4z=1\\-y+4z=2\\(a-25)z=b-12.\end{array}\right.$$So:

• if $a\neq25$, the system has one and only one solution;
• if $a=25$ and $b\neq12$, the system has no solutions;
• if $a=25$ and $b=12$, the system has infinitely many solutions.

Answer 2

Well, if the second expression is set to zero, the solutions are

$x=-(7a-111+8b)/(a-25)$, $y=4(2a-42+b)/(a-25)$, and $z=(b+8)/(a-25)$.

With the new setting $=5$, the solutions are

$x=(3a+21-8b)/(a-25)$, $y=2(a-1-2b)/(a-25)$, and $z=(b-12)/(a-25)$.

It's the Maple output of the call

solve({x+y+4*z-1, 3*x+2*y+16*z-5, 4*x+2*y+(a-1)*z-b+4}, {x,y,z});

Answer 3

The determinant of the coefficient matrix is equal to $25-a$, so if $a\ne25$, the system has exactly one solution for any value of $b$. Now, set $a=25$ and solve as usual, but with only one unknown to worry about. I’ll proceed by row-reducing the augmented matrix until it looks like $$\left[\begin{array}{ccc|c}1&1&4&1\\0&-1&4&2\\0&0&0&b-12\end{array}\right].$$ From this we can see that there is an infinite number of solutions when $b=12$ and none (the system is inconsistent) otherwise.