`A = [{:(e^(t), e^(-t)"cos"t, e^(-t)"sin"t),(e^(t)-e^(-t), "cos"t-e^(-t)"sin"t, -e^(-t)"sin"t + e^(-t)"cos"t),(e^(t), 2e^(-t)"sin"t, -2e^(-t)"cos"t):}]"then A is"`

Options

A. invertible only when `t =pi`
B. invertible for every `t in R`
C. not invertible for any ` t in R`
D. invertible only when ` t = (pi)/(2)`

Explanation

`|A| = |{:(e^(t), " "e^(-t)"cos"t, " "e^(-t)"sin"t),(e^(t), " "-e^(-t)"cos"t-e^(-t)"sin"t,-e^(-t)"sin"t+e^(-t)"cos"t),(e^(t), " " 2e^(-t)"sin"t, " "-2e^(-t)"cos"t):}|`
`=(e^(t))(e^(-t))(e^(-t)) |{:(1, " cos"t, " sin"t),(1, -"cos"t-"sin"t,-"sin"t+"cos"t),(1, " "2"sin"t, " "-2"cos"t):}| " "("taking common from each column")`
Applying `R_(2) to R_(2) -R_(1) " and " R_(3) to R_(3) -R_(1), " we get " [because e^(t-t) =e^(0) = 1]`
`=e^(-t)|{:(1, " cos"t, " sin"t),(0, -2"cos"t-"sin"t,-2"sin"t+"cos"t),(0, " "2"sin"t-"cos"t, -2"cos"t-"sin"t):}|`
` =e^(-t)((2"cos"t + "sin"t)^(2) + (2"sin"t - "cos"t)^(2)) " " ("expanding along column 1")`
`= e^(-t) (5"cos"^(2)t + 5"sin"^(2)t)`
`= 5e^(-t) " " (because "cos"^(2)t + "sin"^(2)t = 1)`
`rArr |A| = 5e^(-t) ne 0 " for all"t in R`
`therefore "A is invertible for all "t in R " "[because "If"|A| ne 0, "then A is invertible"]`