If I is the unit matrix of order n, where \[k \ne 0\] is a constant then \[adj\left( {KI} \right)\] is equals to
A.\[{K^n}adj\left( I \right)\]
B.\[Kadj\left( I \right)\]
C.\[{K^2}adj\left( I \right)\]
D.\[{K^{n - 1}}adj\left( I \right)\]

Hint: This problem belongs to determinants and matrices. So we will simply use the formula to find the inverse of a matrix. Also we will transpose or shuffle the terms as per our need in the solution. So let’s start!

Complete step-by-step answer:
Given is a unit matrix I.
We will consider matrix A of order \[n \times n\].
Then we know that, inverse of a matrix is given by
\[{\left( A \right)^{ - 1}} = \dfrac{{adj\left( A \right)}}{{\det \left( A \right)}}\]
Now a new term K comes that acts as constant but \[k \ne 0\]
So now we can write,
\[{\left( {KA} \right)^{ - 1}} = \dfrac{{adj\left( {KA} \right)}}{{\det \left( {KA} \right)}}\]
We are interested in finding \[adj\left( {KA} \right)\]
So we can write,
\[adj\left( {KA} \right) = {\left( {KA} \right)^{ - 1}}.\det \left( {KA} \right)\]
We also know that, \[\det \left( {\lambda A} \right) = {\lambda ^n}\det A\]
So applying this in the case above,
\[adj\left( {KA} \right) = {\left( {KA} \right)^{ - 1}}.{K^n}\det \left( A \right)\]
\[adj\left( {KA} \right) = {K^{ - 1}}{\left( A \right)^{ - 1}}.{K^n}\det \left( A \right)\]
Taking the common base separately we can write,
\[adj\left( {KA} \right) = {\left( A \right)^{ - 1}}.{K^{n - 1}}\det \left( A \right)\]
Rearranging the terms,
\[adj\left( {KA} \right) = {K^{n - 1}}\det \left( A \right){\left( A \right)^{ - 1}}\]
Now as we had written initially we will use the formula of inverse of a matrix,
\[adj\left( {KA} \right) = {K^{n - 1}}\det \left( A \right)\dfrac{{adj\left( A \right)}}{{\det \left( A \right)}}\]
Now cancelling the common determinant term,
\[adj\left( {KA} \right) = {K^{n - 1}}adj\left( A \right)\]
Now just replace the matrix A with unit matrix I.
\[adj\left( {KI} \right) = {K^{n - 1}}adj\left( I \right)\]
This is the correct answer.
 So option D is the correct option.
So, the correct answer is “Option D”.

Note: Here note that we are given with a matrix of order n. but when the order of the matrix is mentioned then do replace the n is the formula by the order of the matrix. Or we can say that the formula we found above is for general matrix of order \[n \times n\]
Adjoint matrix , determinant of matrix and inverse of matrix are the different terms related to matrix.