How do you solve #3x-y=3# and #-2x+y=2# using matrices?

Question

1619 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-04T09:46:39

The Soln. is #x=5, y=12#.

We can write the given system of eqns., using the Matrix Form, as :

#[(3,-1),(-2,1)][(x), (y)]=[(3),(2)].............(1)#.

Let #A=[(3,-1),(-2,1)] , X=[(x), (y)], and, B=[(3),(2)]#.

Then, #(1)# becomes, #AX=B#.

Now, we know from Algebra that the soln. of #(1)# exists iff #A^-1# exists.

Also, #A^-1# exists iff #detA!=0#, where,

#detA=det|(3,-1),(-2,1)|=3-2=1!=0#. Hence, #A^-1# exists, and, so does the unique soln. of the eqns.

Now, #A^-1=1/detA*adjA=1/1[(1,1),(2,3)]=[(1,1),(2,3)]#

Therefore, the soln. is #X=A^-1B=[(1,1),(2,3)][(3),(2)]#

#:. [(x), (y)]=[(1,1),(2,3)][(3),(2)]=[(5),(12)]#

So, the Soln. is #x=5, y=12#.