Find a and b such that the system has (a) no solution; (b) many solutions; (c) a single solution

linear algebrasystems of equations

So, I have this system with three equations:

$$\left\{\begin{array}{l}x + 2y + z = 3\\
ay + 5z = 10\\
2x + 7y +az = b\end{array}\right.$$

And I am asked to find $a$ and $b$ such that the system has

  1. no solution;
  2. many solutions;
  3. a single solution.

I have tried to reduce the system matrix but the augmented matrix has as second row $[0\ \ a\ \ 5\ \ 10]$ and I am not sure what to do with that $a$. Since nothing says that $a$ can not be zero, it is illegal to multiply the first line by $-a$ then sum the resulting line to the second line, right?

Any tips on how I can manage to proceed with this problem?

Best Answer

Consider the system matrix$$\begin{bmatrix}1&2&1&3\\0&a&5&10\\2&7&a&b\end{bmatrix}.$$If you multiply the first line by $-2$ and add it to the third one, you'll get$$\begin{bmatrix}1&2&1&3\\0&a&5&10\\0&3&a-2&b-6\end{bmatrix}.$$Now, you can exchange the second and the third lines:$$\begin{bmatrix}1&2&1&3\\0&3&a-2&b-6\\0&a&5&10\end{bmatrix}$$and then divide the second line by $3$:$$\begin{bmatrix}1&2&1&3\\0&1&\frac{a-2}3&\frac{b-6}3\\0&a&5&10\end{bmatrix}.$$And now you can multiply the second line by $-a$ and add it to the third one:$$\begin{bmatrix}1&2&1&3\\0&1&\frac{a-2}3&\frac{b-6}3\\0&0&5-\frac{a^2-2a}3&10-\frac{ab-6a}3\end{bmatrix}.$$Now, consider two cases: when $5-\frac{a^2-2a}3=0$ (that is, when $a=-3$ or $a=5$) and when $5-\frac{a^2-2a}3\ne0$. Can you take it from here?

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