Question

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

Answer 1

`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`