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Question
Find the inverse of the matrices (if it exists).
`[(1,0,0),(0, cos alpha, sin alpha),(0, sin alpha, -cos alpha)]`
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Solution
Let A = `[(1, 0, 0), (0, cos alpha, sin alpha), (0,sin alpha, -cos alpha)]`
|A| = 1(−cos2 α − sin2 α)
= −1 ≠ 0
∴ A−1 exists.
Cij = (−1)i + j Mij
C11 = `(-1)^(1 + 1) |(cos alpha, sin alpha), (sin alpha, -cos alpha)|`
= −cos2 α − sin 2α
= −1
C12 = `(-1)^(1 + 2) |(0, sin alpha), (0, -cos alpha)|` = 0
C13 = `(-1)^(1 + 3) |(0, cos alpha), (0, sin alpha)|` = 0
C21 = `(-1)^(2 + 1) |(0, 0), (sin alpha, -cos alpha)|` = 0
C22 = `(-1)^(2 + 2) |(1, 0), (0, -cos alpha)|` = −cos α
C23 = `(-1)^(2 + 3) |(1, 0), (0, sin alpha)|` = −sin α
C31 = `(-1)^(3 + 1) |(0, 0), (cos alpha, sin alpha)|` = 0
C32 = `(-1)^(3 + 2) |(1, 0), (0, sin alpha)|` = −sin α
C33 = `(-1)^(3 + 3) |(1, 0), (0, cos alpha)|` = cos α
Now, Adj A = `[(-1, 0, 0), (0, -cos alpha, -sin alpha), (0, -sin alpha, cos alpha)]`
A−1 = `1/|A|` (adj A)
= `1/-1 [(-1, 0, 0), (0, -cos alpha, -sin alpha), (0, -sin alpha, cos alpha)]`
= `[(1, 0, 0), (0, cos alpha, sin alpha), (0, sin alpha, -cos alpha)]`
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