Solve the system of linear equations for all values of a

Question

When solving the following matrix for $a$: $$ \left[ \begin{array}{ccc|c} x&2y&az&3\\ (2a-4)x&(2-a)y&(-2a+4)z&0\\ 4x&4y&2az&8 \end{array} \right] $$

When do I put restrictions on which values $a$ can take? Say I divide the second row by $(2-a)$, does this mean that $a$ cannot be 2, or can I only say something like that when the entire matrix is in row echelon form, and I would be left with something like: $$ \left[ \begin{array}{ccc|c} 1&0&0&\frac{2a}{a-2}\\ 0&1&0&\frac{4-5a}{a-2}\\ 0&0&1&\frac{a^2}{a-2} \end{array} \right] $$

Answer 1

I didn't check the detail of calculation but, if it is correct, the solution you have found is valid for $a\neq 2$ (i.e. the system has exactly a solution $\forall a\neq2$).

Now you need also to verify the solution/s for the case $a=2$ in the original system, that is

$$\left[ \begin{array}{ccc|c} 1&2&2&3\\ 4&4&4&8 \end{array} \right]\implies \left[ \begin{array}{ccc|c} 1&2&2&3\\ 1&1&1&2 \end{array} \right]\implies \left[ \begin{array}{ccc|c} 1&2&2&3\\ 0&1&1&1 \end{array} \right]\implies\left[ \begin{array}{ccc|c} 1&0&0&1\\ 0&1&1&1 \end{array} \right]$$ thus for a=2 the system has infinitely many solutions

• $x=1$
• $y+z=1$

Answer 2

from the first and the last equation we get $$x=1$$(after dividing the third equation by $4$ then we get $$2y+az=2$$ and $$(2-a)y+(4-2a)z=4-2a$$ with $$y=\frac{2-az}{2}$$ we get $$\frac{2-a}{2}(2-az)+(4-2a)z=4-2a$$ can you finish? the last equation is equivalent to $$1/2\, \left( a-2 \right) \left( az-4\,z+2 \right)=0$$