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Prove that (bar"a" xx bar"b").(bar"c" xx bar"d") = |bar"a".bar"c" bar"b".bar"c"bar"a".bar"d" bar"b".bar"d"|. - Mathematics and Statistics

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Sum

Prove that `(bar"a" xx bar"b").(bar"c" xx bar"d")` =
`|bar"a".bar"c"    bar"b".bar"c"|`
`|bar"a".bar"d"    bar"b".bar"d"|.`

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Solution

Let `bar"a" xx bar"b" = bar"m"`

LHS = `(bar"a" xx bar"b").(bar"c".bar"d")`

`= bar"m".(bar"c" xx bar"d")`

`= (bar"m" xx bar"c").bar"d"`

`= [(bar"a" xx bar"b") xx bar"c"].bar"d"`

`= [(bar"c".bar"a")bar"b" - (bar"c". bar"b")bar"a"].bar"d"`

`= (bar"c".bar"a")(bar"b".bar"d") - (bar"c".bar"b")(bar"a".bar"d")`

= `|bar"c".bar"a"    bar"c".bar"b"|`
   `|bar"a".bar"d"    bar"b".bar"d"|.`

= `|bar"a".bar"c"    bar"b".bar"c"|`
   `|bar"a".bar"d"    bar"b".bar"d"|.`    .....[Do product is commutative]

= RHS.

Concept: Section Formula
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