Find the values of $a$, for which this system of linear equations has one solution, no solution, or infinite solutions

Question

Determine the values of a for which the following system of linear equations has no solutions, a unique solution, or infinitely many solutions. You can select 'always', 'never', '$a = $', or '$a \neq$', then specify a value or comma-separated list of values. $$ \begin{align} 2x−6y−4z &= 16\\ ax−y−4z &= 6\\ 2x−3y−4z &= 10 \end{align} $$ This is a problem for my Linear Algebra class, and I can't seem to figure out how to get it into RREF and evaluate.

Answer 1

With $$z=\frac{1}{2}(x-3y-8)$$ (from the first equation) we get $$x(a-2)+5y=-10$$ and $$y=-2$$ in the third equation, so we obtain $$x(a-2)=0$$ Can you finish?

Answer 2

Hint:

Let the equations be $(1), (2), (3)$ in the order you show.

Perform the following three calculations:

$(1)-(2)$

$(1)-(3)$

$(3)-(2)$

The rest should fall out quite nicely.

Answer 3

The criterion to have solutions is is that the matrix of coefficients of the linear system and the augmented matrix have the same rank. This common rank is then the codimension of the affine space of solutions.

So let's perform row reduction on the augmented matrix: \begin{align} &\left[\begin{array}{ccc|c} 2&-6&-4&16\\ a&-1&-4&6\\2&-3&-4&10 \end{array}\right]\rightsquigarrow \left[\begin{array}{ccc|c} 1&-3&-2&8\\ 2&-3&-4&10 \\ a&-1&-4&6 \end{array}\right]\rightsquigarrow\left[\begin{array}{ccc|c} 1&-3&-2&8\\ 0&3&0&-6 \\ 0&3a-1&2a-4&6-8a \end{array}\right]\rightsquigarrow\\ \rightsquigarrow&\left[\begin{array}{ccc|c} 1&-3&-2&8\\ 0&1&0&-2 \\ 0&3a-1&2a-4&6-8a \end{array}\right]\rightsquigarrow \left[\begin{array}{ccc|c} 1&-3&-2&8\\ 0&1&0&-2 \\ 0&0&2a-4&4-2a \end{array}\right]\rightsquigarrow \left[\begin{array}{ccc|c} 1&0&-2&2\\ 0&1&0&-2 \\ 0&0&a-2&2-a \end{array}\right] \end{align}

• if $a\ne2$, we can proceed as follows: $$\left[\begin{array}{ccc|c} 1&0&-2&2\\ 0&1&0&-2 \\ 0&0&1&-1 \end{array}\right]\rightsquigarrow \left[\begin{array}{ccc|c} 1&0&0&0\\ 0&1&0&-2 \\ 0&0&1&-1 \end{array}\right] $$ There is a unique solution (the last column) since the matrix has full rank.
• If $a=2$, we have a row reduced echelon form: $$\left[\begin{array}{ccc|c} 1&0&-2&2\\ 0&1&0&-2 \\ 0&0&0&0 \end{array}\right]$$ which shows, as the matrix and the augmented matrix have rank $2$ has; as a set of solutions, an affine space of dimension $3-2=1$.