If a matrix $A$ is such that $3{A^3} + 2{A^2} + 5A + I = 0$, then ${A^{ - 1}}$ is equal to
(A)$ - (3{A^2} + 2A + 5)$ (B) $3{A^2} + 2A + 5$ (C)$3{A^2} - 2A - 5$ (D) None of these

Hint-Here in this question we have to find ${A^{ - 1}}$which is defined as the reciprocal of $A$.To find ${A^{ - 1}}$ we can multiply ${A^{ - 1}}$ in the given equation to introduce ${A^{ - 1}}$ and then transpose the rest of terms.

Complete Step by step solution:
Given equation $3{A^3} + 2{A^2} + 5A + I = 0$
As we have to find ${A^{ - 1}}$, on multiplying$ {A^{ - 1}}$ both side of equation,
$ \Rightarrow {A^{ - 1}}(3{A^3} + 2{A^2} + 5A + I) = {A^{ - 1}}(0)$
$ \Rightarrow 3{A^{ - 1}}({A^3}) + 2{A^{ - 1}}({A^2}) + 5{A^{ - 1}}(A) + {A^{ - 1}}(I) = 0$ [As ${A^{ - 1}}(0) = 0$ and ${A^{ - 1}}k = k{A^{ - 1}}$]
$ \Rightarrow 3{A^2} + 2A + 5{A^0} + {A^{ - 1}} = 0$ [Using${A^{ - 1}}I = {A^{ - 1}}$ and ${A^{ - 1}}{A^n} = {A^{n - 1}}$]
$ \Rightarrow 3{A^2} + 2A + 5 + {A^{ - 1}} = 0$ [Using ${A^0} = I$]
By transposing the terms except \[{A^{ - 1}}\]from L.H.S to R.H.S,
$ \Rightarrow {A^{ - 1}} = - (3{A^2} + 2A + 5)$
Hence the option A is correct.


Note:To find the inverse of the matrix where the matrix is not given, we need to have a polynomial equation of that matrix as in the question.