How do you solve #x + y = 5 # and #3x – y = –1# using matrices?

1 Answer
Jul 16, 2018

Answer:

#x=1 and y=4#

Explanation:

Here,

#x+y=5to(1)# , #and # #3x-y=-1to(2)#

Let us write in the matrix equation form :

#((1,1),(3,-1))((x),(y))=((5),(-1))#

We take ,

#A=((1,1),(3,-1))# , #X=((x),(y))# ,#and B=((5),(-1))#

#:.AX=B#

Now, #detA=|(1,1),(3,-1)|=-1-3=-4!=0#

#:. "We can say that , " A^-1 " exists"#

Now ,#adjA=((-1,-1),(-3,1))#

#:.A^-1=1/(detA)*adjA#

#:.A^-1=-1/4((-1,-1),(-3,1))#

We have,

#AX=B=>X=A^-1B#

#=>X=-1/4((-1,-1),(-3,1))((5),(-1))#

Using Product of two matrices :

#X=-1/4((-5+1),(-15-1))#

#=>X=-1/4((-4),(-16))#

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

#=>x=1 and y=4#