Question

If A = `[(1, 0, 1),(3, 1, 2)], "B" = [(2, 1, -4),(3, 5, -2)] "and"  "C" = [(0, 2, 3),(-1, -1, 0)]`, verify that (A + 2B + 3C)^T = A^T + 2B^T + C^T.

Answer 1

A + 2B + 3C

= `[(1, 0, 1),(3, 1, 2)] + 2[(2, 1, -4),(3, 5, -2)] + 3[(0, 2, 3),(-1, -1, 0)]`

= `[(1, 0, 1),(3, 1, 2)] + [(4, 2, -8),(6, 1, -4)] + [(0, 6, 9),(-3, -3, 0)]`

= `[(1 + 4 + 0, 0 + 2 + 6, 1 - 8 + 9),(3 + 6 - 3, 1 +10 - 3, 2 - 4+ 0)]`

∴ A + 2B + 3C = `[(5, 8, 2),(6, 8, -2)]`

∴ [A + 2B + 3C]^T = `[(5, 6),(8, 8),(2, -2)]`   ...(i)

Now, A^T = `[(1, 3),(0, 1),(1, 2)], "B"^"T" = [(2, 3),(1, 5),(-4, -2)]`

and C^T = `[(0, -1),(2, -1),(3, 0)]`

∴ A^T + 2B^T + 3C^T 

= `[(1, 3),(0, 1),(1, 2)] + 2[(2, 3),(1, 5),(-4, -2)] + 3[(0, -1),(2, -1),(3, 0)]`

= `[(1, 3),(0, 1),(1, 2)] + [(4, 6),(2, 10),(-8, -4)] + [(0, -3),(6, -3),(9, 0)]`

= `[(1 + 4 + 0, 3 + 6 + 3),(0 + 2 + 6, 1 + 10 - 3),(1 - 8 + 9, 2 - 4 + 0)]`

∴ A^T + 2B^T + 3C^T = `[(5, 6),(8, 8),(2, -2)]`     ...(iii)

From (i) and (ii), we get
[A + 2B + 3C]^T = A^T + 2B^T + 3C^T.