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Answer the following question: If A = diag [2 –3 –5], B = diag [4 –6 –3] and C = diag [–3 4 1] then find 2A + B – 5C

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प्रश्न

Answer the following question:

If A = diag [2 –3 –5], B = diag [4 –6 –3] and C = diag [–3 4 1] then find 2A + B – 5C

योग
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उत्तर

A = diag [2 –3 –5]

∴ A = `[(2, 0, 0),(0, -3, 0),(0, 0, -5)]`

B = diag [4 –6 –3]

∴ B = `[(4, 0, 0),(0, -6, 0),(0, 0, -3)]`

C = diag [–3 4 1]

∴ C = `[(-3, 0, 0),(0, 4, 0),(0, 0, 1)]`

2A + B – 5C = 2 diag [2 – 3 – 5] + diag [4 – 6 – 3] – 5 diag [ –3 4 1]

`= 2[(2, 0, 0),(0, -3, 0),(0, 0, -5)] + [(4, 0, 0),(0, -6, 0),(0, 0, -3)] -5[(-3, 0, 0),(0, 4, 0),(0, 0, 1)]`

`= [(4, 0, 0),(0, -6, 0),(0, 0, -10)] + [(4, 0, 0),(0, -6, 0),(0, 0, -3)] - [(-15, 0, 0),(0, 20, 0),(0, 0, 5)]`

`= [(4 + 4 - (-15), 0, 0),(0, -6 - 6 - 20, 0),(0, 0, -10 - 3 - 5)]`

`= [(23, 0, 0),(0, -32, 0),(0, 0, -18)]`

= diag [23 – 32 – 18].

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अध्याय 4: Determinants and Matrices - Miscellaneous Exercise 4(B) [पृष्ठ १००]

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बालभारती Mathematics and Statistics (Arts and Science) Part 1 [English] Standard 11 Maharashtra State Board
अध्याय 4 Determinants and Matrices
Miscellaneous Exercise 4(B) | Q II. (1) (ii) | पृष्ठ १००

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