Question

How do you solve `6x+5y = 8` and `3x-y = 7` using matrices?

Answer 1

`x=43/21` & `y=-6/7`

Explanation:

Given equations can be written as follows

`6x+5y-8=0\ ......(1)`

`3x-y-7=0\ ......(2)`

Using matrix method, the solution can be written as

`\frac{x}{(5)(-7)-(-8)(-1)}=\frac{y}{(-8)(3)-(6)(-7)}=\frac{1}{(6)(-1)-(5)(3)}`

`\frac{x}{-43}=\frac{y}{18}=-1/21`

`\implies \frac{x}{-43}=-1/21`

`x=43/21`

`\implies \frac{y}{18}=-1/21`

`y=-6/7`

Answer 2

`color(blue)(x=43/21, y=-6/7)`

Explanation:

Put the coefficients of the equations into a matrix, and put the solutions into 1 column vector and the variables into another vector:

`bbA=[(6,5),(3,-1)]`

`bb(b)=[(8),(7)]`

`bbX=[(x),(y)]`

We now have the matrix equation:

`bb(AX)=bb(b)`

If `bbA` in invertible, then `bb(A^-1)` exists, so:

`bb(A^-1AX)=bb(A^-1)bb(b)`

`bb(IX)=bb(A^-1)bb(b)`

We need to find `bb(A^-1)`

First find the determinant of `bb(A)`

Determinant of `bbA=[(a,b),(c,d)]=ad-bc`

So determinant of:

`bbA=[(6,5),(3,-1)]=6(-1)-(5)(3)=-21`

Swap the elements on the leading diagonal and change the signs on the non-leading diagonal:

`bbA=[(-1,-5),(-3,6)]`

Next multiply by the reciprocal of the determinant:

`bb(A^-1)=[(1/21,5/21),(1/7,-6/21)]`

We now have:

`[(x),(y)]=[(1/21,5/21),(1/7,-6/21)][(8),(7)]=[(8/21+35/21),(8/7-2)]=[(43/21),(-6/7)]`

Hence:

`x=43/21`

`y=-6/7`