How do you solve #p-3q=-1# and #5p+16q=5# using matrices?

1 Answer
Bio
Mar 27, 2016

#p = -1/31#

#q = 10/31#

Explanation:

When written in matrix form, the system of linear equations looks like this.

#((1,-3),(5,16)) ((p),(q)) = ((-1),(5))#

We multiply the inverse on both sides.

#((1,-3),(5,16))^{-1} ((1,-3),(5,16)) ((p),(q)) = ((1,-3),(5,16))^{-1} ((-1),(5))#

#((1,0),(0,1)) ((p),(q)) = ((1,-3),(5,16))^{-1} ((-1),(5))#

#((p),(q)) = ((1,-3),(5,16))^{-1} ((-1),(5))#

The inverse of a #2xx2# matrix #A = ((a,b),(c,d))# is given by

#A^{-1} = 1/["det"(A)] ((d,-b),(-c,a))#

#= 1/(ad-bc) ((d,-b),(-c,a))#

In this question, #A = ((1,-3),(5,16))#,

  • #a = 1#
  • #b = -3#
  • #c = 5#
  • #d = 16#

#A^{-1} = 1/((1)(16)-(-3)(5)) ((16,-(-3)),(-5,1))#

#= 1/31 ((16,3),(-5,1))#

So,

#((p),(q)) = ((1,-3),(5,16))^{-1} ((-1),(5))#

#= 1/31 ((16,3),(-5,1)) ((-1),(5))#

#= 1/31 ((-1),(10))#

#= ((-1/31),(10/31))#

This means #p = -1/31# and #q = 10/31#.

You can check your answer by substituting the values of #p# and #q# into the original equations.

#(-1/31)-3(10/31)=-1#

#5(-1/31)+16(10/31)=5#

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