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Question
Let A = `[(1, 2),(-1, 3)]`, B = `[(4, 0),(1, 5)]`, C = `[(2, 0),(1, -2)]` and a = 4, b = –2. Show that: (AB)T = BTAT
Solution
We have,
A = `[(1, 2),(-1, 3)]`
B = `[(4, 0),(1, 5)]`
C = `[(2, 0),(1, -2)]`
And a = 4, b = –2
AB = `[(1, 2),(-1, 3)] [(4, 0),(1, 5)]`
= `[(4 + 2, 0 + 10),(-4 + 3, 0 + 15)]`
= `[(6, 10),(-1, 15)]`
∴ (AB)T = `[(6, -1),(10, 15)]`
Now, BTAT = `[(4, 1),(0, 5)] [(1, -1),(2, 3)]`
= `[(4 + 2, -4 + 3),(0 + 10, 0 + 15)]`
= `[(6, -1),(10, 15)]`
= (AB)T
Hence proved.
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