Question

Construct an m × n matrix A = [a_ij], where a_ij is given by

a_ij = `|3"i" - 4"j"|/4` with m = 3, n = 4

Answer 1

To construct a 3 × 4 matrices.

A = `["a"_"ij"]_(3 xx 4) = [("a"_11, "a"_12, "a"_13, "a"_14),("a"_21, "a"_22, "a"_24, "a"_24),("a"_31, "a"_32, "a"_33, "a"_34)]`

a₁₁ = `|3 xx 1 - 4 xx 1|/4`

= `|3 - 4|/4`

= `|- 1|/4`

= `1/4`

a₁₂ = `|3 xx 1 - 4 xx 2|/4`

= `|3 - 8|/4`

= `|- 5|/4`

= `5/4`

a₁₃ = `|3 xx 1 - 4 xx 3|/4`

= `|3 - 12|/4`

= `|- 9|/4`

= `9/4`

a₁₄ = `|3 xx 1 - 4 xx 4|/4`

= `|3 - 16|/4`

= `|- 13|/4`

= `13/4`

a₂₁ = `|3 xx 2 - 4 xx 1|/4`

= `|6 - 4|/4`

= `|2|/4`

= `2/4`

= `1/2`

a₂₂ = `|3 xx 2 - 4 xx 2|/4`

= `|6 - 8|/4`

= `|- 2|/4`

= `2/4`

= `1/2`

a₂₃ = `|3 xx 2 - 4 xx 3|/4`

= `|6 - 12|/4`

= `|- 6|/4`

= `6/4`

= `3/2`

a₂₄ = `|3 xx 2 - 4 xx 4|/4`

= `|6 - 16|/4`

= `|- 10|/4`

= `10/4`

= `5/2`

a₃₁ = `|3 xx 3 - 4 xx 1|/4`

= `|9 - 4|/4`

= `|5|/4`

= `5/4`

a₃₂ = `|3 xx 3 - 4 xx 2|/4`

= `|9 - 8|/4`

= `|1|/4`

= `1/4`

a₃₃ = `|3 xx 3 - 4 xx 3|/4`

= `|9 - 12|/4`

= `|- 3|/4`

= `3/4`

a₃₄ = `|3 xx 3 - 4 xx 4|/4`

= `|9 - 16|/4`

= `|- 7|/4`

= `7/4`

∴ The required 3 × 4 matrix is

A = `[(1/4, 5/4, 9/4, 13/4),(1/2, 1/2, 3/2, 5/2),(5/4, 1/4, 3/4, 7/4)]`