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Find 12 (A + A') and 12 (A -A') When A=[0ab-a0c-b-c0] - Mathematics

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प्रश्न

Find `1/2` (A + A')  and  `1/2` (A -A') When `A = [(0, a, b),(-a,0,c),(-b,-c,0)]`

योग
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उत्तर

Given `A = [(0, a, b),(-a, 0, c),(-b, -c, 0)]`

So, A' = `[(0, - a, - b),(a, 0, - c),(b, c, 0)] = - [(0, a, b),(-a, 0, c),(-b, -c, 0)]` = - A

Now, `1/2` (A + A') = `1/2  ([(0, a, b),(-a, 0, c),(-b, -c, 0)] - [(0, a, b),(-a, 0, c),(-b, -c, 0)])`

`= [(0,0,0),(0,0,0),(0,0,0)]`

Then,  `1/2` (A - A') = `1/2  ([(0, a, b),(-a, 0,c),(-b, -c, 0)] + [(0, a, b),(-a, 0, c),(-b, -c, 0)])`

`= 1/2 [(0, 2a, 2b),(- 2a, 0, 2c),(- 2b, - 2c, 0)]`

`= [(0, a, b),(-a, 0, c),(-b, -c, 0)]`

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अध्याय 3: Matrices - Exercise 3.3 [पृष्ठ ८९]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 3 Matrices
Exercise 3.3 | Q 9 | पृष्ठ ८९

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