Question

How do you solve `((1, -6, 0), (0, 1, -7), (3, 0, 2))X=((1), (4), (11))`?

Answer 1

`X=((4),(0.5),(-0.5))`

Explanation:

Let `X=((a),(b),(c))`
Using Cramer's Rule to solve the augmented matrix
`color(white)("XXXXXXXXXXXxX")acolor(white)("XX")bcolor(white)("Xx")c color(white)("XXX")k`
`color(white)("XXX")((M,"|",k))=((1,-6,0,"|",1),(0,1,-7,"|",4),(3,0,2,"|",11))`

Calculating the Determinants:
`color(white)("XXX")Det(M)= |(1,-6,0),(0,1,-7),(3,0,2)|`

`color(white)("XXXXXXX")=1color(white)("X")[(1xx2)-(0xx-7)]`
`color(white)("XXXXXXXX")-color(white)("X")0[(-6xx2)-(0xx0)]`
`color(white)("XXXXXXXX")+3color(white)("X")[((-6)xx(-7))-(0xx0)]`

`color(white)("XXXXXXX")=128`

Similarly
`color(white)("XXX")Det(M_a)=512`

`color(white)("XXX")Det(M_b)=64`

`color(white)("XXX")Det(M_c)=-64`

By Cramer's Rule:
`color(white)("XXX")a=(Det(M_a))/(Det(M))=512/128=4`

`color(white)("XXX")b=(Det(M_b))/(Det(M))=64/128=0.5`

`color(white)("XXX")c=(Det(M_c))/(Det(M))=(-64)/128=-0.5`

Confession: I used as spreadsheet to do the detailed arithmetic work