How do you solve #x + 4y = 8# and #2x + y = 9# using matrices?

1 Answer
Mar 4, 2017

Answer:

#x = 4, y = 1#

Explanation:

In preamble, we can write these equations in column vector form, ie:

#x ((1),(2)) + y ((4),(1)) = ((8),(9))#

And so they can also be written in matrix form, as we then have a set of rules in matrix algebra which means they are the same thing:

#((1, 4),(2, 1)) ((x),(y)) = ((8),(9))#

That is: #M mathbf x = mathbf b#

From here, we have choices. We can row reduce, but as we are looking at the simplest matrix algebra (I think), we should try to use the inverse matrix #M^(-1)#, the idea being:

#M^(-1)M mathbf x = M^(-1) mathbf b#

#implies I mathbf x = M^(-1) mathbf b#

Generally speaking, for a #2 times 2# matrix, #M = ((a,b),(c,d))#, the inverse #M^(-1) # is:

#M^(-1) = 1/(ad - cb) ((d,-b),(-c,a))#.

So we have:

#((1, 0),(0, 1)) ((x),(y)) =- 1/7 ((1, -4),(-2, 1)) ((8),(9))#

#implies ((x),(y)) =- 1/7 ((-28),(-7))#

#implies ((x),(y)) = ((4),(1))#