How do you set up and solve the following system using augmented matrices $3 x - 2 y = 8, 6 x - 4 y = 1$ $3x-2y=8, 6x-4y=1$ ?

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Answer 1 · 2017-03-08T23:18:12

No solution

The typesetting here does not support row augmented matrices.

But, if we are going with: $3 x - 2 y = 8, 6 x - 4 y = 1$ $3x-2y=8, 6x-4y=1$ ...then we start with:

$((3, -2),(6, -4))((8),(1))$

However, there is an immediate problem, we have linearly dependent rows, this is a singular matrix :(

$((3, -2),(6, -4)) = ((1/2(6), 1/2(-4)),(6, -4))$

So this either has no solutions or infinitely many:

$((3, -2),(6, -4)) ((x),(y)) = ((8),(1))$

Then go back and look at the equations and you will see that they represent parallel lines:

$3x-2y=color(red)(8)$

$2(3x)-2(2y)=color(blue)(2(1/2))$

How do you set up and solve the following system using augmented matrices 3x-2y=8, 6x-4y=13x−2y=8,6x−4y=13x-2y=8, 6x-4y=1?