-2
$\begingroup$

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

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

  • 3
    You have an equation of the form $Ax = b$ with $b = (2,5,c)$. Recall that this equation has a solution if, and only if $rank(A,b) = rank(A)$,2012-10-09

3 Answers 3

3

HINT: Add the two first equations.

3

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.

  • 1
    I don't understand. None of the posters have the (exact) same IP. This is either an exam question, a homework question, or a strange form of malevolence.2012-10-09
2

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