Question

If `"A" = [(1,1,1),(1,0,2),(3,1,1)]`, find A^-1. Hence, solve the system of equations x + y + z = 6, x + 2z = 7, 3x + y + z = 12.

Answer 1

Given Matrix `"A" = [(1,1,1),(1,0,2),(3,1,1)]`
To find A^-1, we need cofactors of each element of matrix A.

cofactor of `"a"_11 = (-1)^(1+1) |(0,2),(1,1)| = -2`

cofactor of `"a"_12 = (-1)^(1+2) |(1,2),(3,1)| = -(1 -6) = 5`

cofactor of `"a"_13 = (-1)^(1+3) |(1,0),(3,1)| = 1`

cofactor of `"a"_21 = (-1)^(2+1) |(1,1),(1,1)| = 0`

cofactor of `"a"_22 = (-1)^(2+2) |(1,1),(3,1)| = (1-3) = -2`

cofactor of `"a"_23 = (-1)^(2+3) |(1,1),(3,1)| = - (1-3) = 2`

cofactor of `"a"_31 = (-1)^(3+1) |(1,1),(0,2)| = 2`

cofactor of `"a"_32 = (-1)^(3+2) |(1,1),(1,2)| = -(2 -1) = -1`

cofactor of `"a"_33 = (-1)^(3+3) |(1,1),(1,0)| = -1`

So cofactor of martix of `"A" = [(-2, 5, 1),(0, -2, 2),(2, -1, -1)]`

∵ the transpose of cofactor matrix A is adj (A)

So adj (A) = `[(-2,0,2),(5,-2,-1),(1,2,-1)]`

|A| = 1 (0 -2) -1 (1-6) + 1 (1-0)
= -2 + 5 +1
= 4

And `"A"^-1=1/|"A"|"adj" ("A")`

SO, `"A"^-1  = (1)/(4) [(-2,0,2),(5,-2,-1),(1,2,-1)]`

Now, the given system of eqⁿ is
x + y + z = 6
x + 2z = 7
3x + y + z =12
Writing the above equation in matrix form

`[(1,1,1),(1,0,2),(3,1,1)] [(x),(y),(z)] = [(6),(7),(12)]`

`"A" [(x),(y),(z)] = [(6),(7),(12)]`

`[(x),(y),(z)] = "A"^-1  [(6),(7),(12)]`

`[(x),(y),(z)] = (1)/(4) [(-2,0,2),(5,-2,-1),(1,2,-1)] [(6),(7),(12)]`

` or,  (1)/(4) [(-12+ 0+ 24),(30 - 14 - 12),(6 + 14 - 12)]` 

`or, (1)/(4) [(12),(4),(8)]`

`or, [(x),(y),(z)] = [(3),(1),(2)]`

`i .e  x = 3, y = 1 , z = 2`