मराठी

If P = [x000y000z] and Q = abc[a000b000c], prove that PQ = abc[xa000yb000zc] = QP

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प्रश्न

If P = `[(x, 0, 0),(0, y, 0),(0, 0, z)]` and Q = `[("a", 0, 0),(0, "b", 0),(0, 0, "c")]`, prove that PQ = `[(x"a", 0, 0),(0, y"b", 0),(0, 0, z"c")]` = QP

बेरीज
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उत्तर

Given that,

P = `[(x, 0, 0),(0, y, 0),(0, 0, z)]` and Q = `[("a", 0, 0),(0, "b", 0),(0, 0, "c")]`

PQ = `[(x, 0, 0),(0, y, 0),(0, 0, z)] [("a", 0, 0),(0, "b", 0),(0, 0, "c")]` 

PQ = `[(x"a" + 0 + 0, 0 + 0 + 0, 0 + 0 + 0),(0 + 0 + 0, 0 + y"b" + 0, 0 + 0 + 0),(0 + 0 + 0, 0 + 0 + 0, 0 + 0 + z"c")]`

PQ = `[(x"a" , 0, 0),(0, y"b", 0),(0, 0, z"c")]`

Now QP = `[("a", 0, 0),(0, "b", 0),(0, 0, "c")] [(x, 0, 0),(0, y, 0),(0, 0, z)]`

QP = `[(x"a" + 0 + 0, 0 + 0 + 0, 0 + 0 + 0),(0 + 0 + 0, 0 + y"b" + 0, 0 + 0 + 0),(0 + 0 + 0, 0 + 0 + 0, 0 + 0 + z"c")]`

QP = `[(x"a", 0, 0),(0, y"b", 0),(0, 0, z"c")]`

Hence, PQ = QP.

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पाठ 3: Matrices - Exercise [पृष्ठ ५५]

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