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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.
Solution
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`
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