Final answer:
The matrix will be diagonalizable if its characteristic polynomial has three linearly independent eigenvectors, which is determined using the values a, b, and c in the quadratic formula. The example constants given (a = 1.00, b = 10.0, c = -200) illustrate how to calculate the polynomial's roots, which are the eigenvalues of the matrix.
Step-by-step explanation:
For a matrix to be diagonalizable, it must have enough eigenvectors to form a basis of the vector space. The matrix M given as:
M = [\[0 \quad 0 \quad a\\ 1 \quad 0 \quad 3\\ 0 \quad 1 \quad 0\]]
will be diagonalizable if the characteristic polynomial, which is derived from det(M - \lambda\cdot I), has three linearly independent eigenvectors. The characteristic polynomial of M can be determined by calculating det(M - \lambda\cdot I), resulting in a polynomial of degree 3, typically of the form at^2 + bt + c = 0.
For the example constants given, a = 1.00, b = 10.0, and c = -200, the solutions to the characteristic polynomial are found by applying the quadratic formula:
\[ \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Plugging in the values into the quadratic formula, we find the roots of the polynomial, which correspond to the eigenvalues of M. If the resulting eigenvalues are distinct, this indicates that M has a set of linearly independent eigenvectors and is thus diagonalizable. For instance:
\[ \frac{-10.0 \pm \sqrt{10.0^2 - 4 \cdot 1.00 \cdot (-200)}}{2 \cdot 1.00} \]
It is crucial to note that if the eigenvalues are not distinct, additional checks for linear independence of eigenvectors must be performed to confirm diagonalizability.