Choose a coefficient so that the system of equations has exactly one solution

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

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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 $x=\frac{19}{18}$, $y=-\frac1{18}$, $z=0$.