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Question
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
Solution
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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