Solution of 3 equations in 3 unknowns

Question

Find the value of $c$ which makes it possible to solve:

$$u+v+2w=2,$$ $$2u+3v-w=5,$$ $$3u+4v+w=c$$

Answer 1

HINT: Add the two first equations.

Answer 2

Set up your augmented matrix in the usual way:

$$\left[\begin{array}{rrr|r} 1&1&2&2\\ 2&3&-1&5\\ 3&4&1&c \end{array}\right]\;.$$

Then row-reduce it; reducing the first column, for instance, yields

$$\left[\begin{array}{rrr|c} 1&1&2&2\\ 0&1&-5&1\\ 0&1&-5&c-6 \end{array}\right]\;.\tag{1}$$

Now you can either stop and think about the equations corresponding to the bottom two rows of $(1)$ (what does $c$ have to be in order for them to be consistent?), or finish the row-reduction and then think about what $c$ has to be to avoid having an inconsistent system.

Answer 3

If you add the second and third equation you get 5u+7v=5+c ; if you multiply the second equation by 2 and then add it to the first then you get 4u+u+6v+v=12=5u+7v , so c=7