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Find the inverse of the matrices (if it exists). [(1,0,0),(0, cos alpha, sin alpha),(0, sin alpha, -cos alpha)] - Mathematics

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प्रश्न

Find the inverse of the matrices (if it exists).

`[(1,0,0),(0, cos alpha, sin alpha),(0, sin alpha, -cos alpha)]`

योग
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उत्तर

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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अध्याय 4: Determinants - Exercise 4.5 [पृष्ठ १३२]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 4 Determinants
Exercise 4.5 | Q 11 | पृष्ठ १३२

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