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Find 1/2 (A + A') and 1/2 (A – A'), when A = [(0, a, b),(–a, 0, c),(–b, –c, 0)]

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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 ६७]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 3 Matrices
EXERCISE 3.3 | Q 9. | पृष्ठ ६७

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