To determine when the inverse of a non-singular diagonalizable matrix A, denoted as A^-1, is diagonalizable, we first need to understand the properties of non-singular matrices and their relationship with eigenvalues.

Non-Singular Matrix Eigenvalues

A matrix A is defined as non-singular if and only if its determinant is non-zero. For a diagonalizable matrix, the determinant is the product of its eigenvalues. Therefore, if a matrix A is non-singular, none of its eigenvalues ($\lambda_1, \lambda_2, \lambda_3$) can be zero. If any eigenvalue is zero, the determinant would be zero, making the matrix singular, and its inverse would not exist.

The problem statement clearly mentions that A is a non-singular diagonalisable matrix. This is a crucial piece of information. It means that A's eigenvalues must all be non-zero for A to be non-singular and for A^-1 to exist.

Inverse Matrix Diagonalizability

If a matrix A is diagonalizable and non-singular, then its inverse A^-1 always exists and is also diagonalizable. The eigenvalues of A^-1 are the reciprocals of the eigenvalues of A. That is, if $\lambda_1, \lambda_2, \lambda_3$ are the eigenvalues of A, then $1/\lambda_1, 1/\lambda_2, 1/\lambda_3$ are the eigenvalues of A^-1. For these reciprocal eigenvalues to exist, none of the original eigenvalues ($\lambda_i$) can be zero.

Eigenvalues Option Analysis

Let's examine the given options for the eigenvalues $\lambda_1, \lambda_2, \lambda_3$ to see which one satisfies the condition that A is a non-singular matrix (i.e., all eigenvalues are non-zero):

• Option 1: $\lambda_1 = 2, \lambda_2 = 0, \lambda_3 = -1$.
  This set of eigenvalues includes $\lambda_2 = 0$. If an eigenvalue is zero, the matrix A would be singular, and its inverse A^-1 would not exist. This contradicts the premise that A is a non-singular matrix. Therefore, A^-1 cannot be diagonalizable under these conditions, as it doesn't even exist.
• Option 2: $\lambda_1 = 0, \lambda_2 = 3, \lambda_3 = -2$.
  This set includes $\lambda_1 = 0$. Similar to Option 1, if an eigenvalue is zero, A would be singular, and A^-1 would not exist. Thus, this option is incorrect.
• Option 3: $\lambda_1 = -1, \lambda_2 = 2, \lambda_3 = -3$.
  In this set, all eigenvalues ($\lambda_1 = -1, \lambda_2 = 2, \lambda_3 = -3$) are non-zero. This means that the determinant of A would be $(-1)(2)(-3) = 6 \neq 0$. Therefore, A is a non-singular matrix. Since A is given as diagonalizable and we've confirmed it's non-singular with these eigenvalues, its inverse A^-1 will also be diagonalizable. The eigenvalues of A^-1 would be $1/(-1), 1/2, 1/(-3)$, which are $-1, 1/2, -1/3$.
• Option 4: $\lambda_1 = -3, \lambda_2 = 1, \lambda_3 = 0$.
  This set includes $\lambda_3 = 0$. As explained for Options 1 and 2, the presence of a zero eigenvalue means A is singular, and A^-1 does not exist. Hence, this option is also incorrect.

Based on the analysis, the only option where all eigenvalues are non-zero, allowing A to be a non-singular matrix and ensuring the existence and diagonalizability of A^-1, is Option 3.

Conclusion on Diagonalizable Matrix

For a non-singular diagonalisable matrix A, its inverse A^-1 is always diagonalizable, provided A itself is non-singular. The condition for A to be non-singular is that all its eigenvalues must be non-zero. Option 3 is the only one where all eigenvalues are non-zero.