Choose a coefficient so that the system of equations has exactly one solution Choose the value of $a$ from $\{-4, 88\}$ so that the system of equations has exactly one solution:
$$2x +20y+3z=1$$
$$2x +2y+41z=2$$
$$ax -22y-44z=-3$$
I tried solving the system for $a=-4$ and after few steps found out that it was inconsistent. I would like to ask whether there is a faster way to determine the solution (so that I don't have to plug the numbers in). Thanks.
 A: By Cramer's Rule,
$A$ =
$\begin{pmatrix}
2&20&3\\
2&2&41\\
a&-22&-44
\end{pmatrix}$
;
$A_x=
\begin{pmatrix}
1&20&3\\
2&2&41\\
-3&-22&-44
\end{pmatrix}$
;
$A_y=$
$\begin{pmatrix}
2&1&3\\
2&2&41\\
a&-3&-44
\end{pmatrix}$
;
$A_z=$
$\begin{pmatrix}
2&20&1\\
2&2&2\\
a&-22&-3
\end{pmatrix}$
;
When $A$ is non-zero, the equations are linearly independent, so $x=A_x/A$ ....can be uniquely determined.
A: You could look at the determinant of the matrix $$\begin{pmatrix}2&20&3\\2&2&41\\a&-22&-44\end{pmatrix}$$
which is $$2\left|\begin{array}{cc}2&41\\-22&-44\end{array}\right|-20\left|\begin{array}{cc}2&41\\a&-44\end{array}\right|+3\left|\begin{array}{cc}2&2\\a&-22\end{array}\right|$$
$$=2(-88+902)-20(-88-41a)+3(-44-2a)$$
$$=1628+1760+820a-132-6a$$
$$=814a+3256$$
Now just choose one of your values for $a$ which makes this equation non zero.
A: If you simply add your equations, you get $(4+a)x=0$.
Therefore the original system is equivalent to
$$
\begin{align}
2x+20y+3z&=1\\
2x+2y+41z&=2\\
(4+a)x&=0
\end{align}
$$
So you can see that for $a=-4$ you get a system of two equations with three variables. It is not hard to find out that you get infinitely many solutions.
In all other cases, you get $x=0$ from the third equation and the system reduces to
$$
\begin{align}
20y+3z&=1\\
2y+41z&=2
\end{align}
$$
So you have system of two equations with two unknowns. As the left sides are not linearly dependent, there is a solution. And for 2 variables and 2 independent equations you get only one solution.

You wrote: I tried solving the system for $a=-4$ and after few steps found out that it was inconsistent.
This is not correct. Just try $x=\frac{19}{18}$, $y=-\frac1{18}$, $z=0$.
A: If the determinant of the matrix
$$\begin{pmatrix}
2&20&3\\
2&2&41\\
a&-22&-44
\end{pmatrix}$$
is $0$, then the system has no solutions. 
