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तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएसएसएलसी (अंग्रेजी माध्यम) कक्षा १०

Let A = [1213], B = [4915], C = [2012] Show that A(BC) = (AB)C - Mathematics

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प्रश्न

Let A = `[(1, 2),(1, 3)]`, B = `[(4, 0),(1, 5)]`, C = `[(2, 0),(1, 2)]` Show that A(BC) = (AB)C

योग

उत्तर

A = `[(1, 2),(1, 3)]`, B = `[(4, 0),(1, 5)]`, C = `[(2, 0),(1, 2)]`

BC = `[(4, 0),(1, 5)] xx [(2, 0),(1, 2)]`

= `[(8 + 0, 0 + 0),(2 + 5, 0 + 10)]`

= `[(8, 0),(7, 10)]`

A(BC) = `[(1, 2),(1, 3)] xx [(8, 0),(7, 10)]`

= `[(8 + 14, 0 + 20),(8 + 21, 0 + 30)]`

= `[(22, 20),(29, 30)]`   ...(1)

AB = `[(1, 2),(1, 3)] xx [(4, 0),(1, 5)]`

= `[(4 + 2, 0 + 10),(4 + 3, 0 +15)]`

= `[(6, 10),(7, 15)]`

(AB)C = `[(6, 10),(7, 15)] xx [(2, 0),(1, 2)]`

= `[(12 + 10, 0 + 20),(14 + 15, 0 + 30)]`

= `[(22, 20),(29, 30)]`   ...(2)

From (1) and (2) we get

A(BC) = (AB)C

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अध्याय 3: Algebra - Exercise 3.18 [पृष्ठ १५४]

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सामाचीर कलवी Mathematics [English] Class 10 SSLC TN Board
अध्याय 3 Algebra
Exercise 3.18 | Q 7. (i) | पृष्ठ १५४
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