Question

Let \( M = \begin{bmatrix} 3 & -1 & -2 \\ 0 & 2 & 4 \\ 0 & 0 & 1 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \). If \(6M^{-1} = M^2 - 6M + \alpha I\) for some \(\alpha \in \mathbb{R}\), then the value of \(\alpha\) is equal to ................

Hint:

When you see a matrix polynomial equation, especially one involving the inverse like \(M^{-1}\), think of the Cayley-Hamilton Theorem. It's often the fastest way to solve the problem without having to compute matrix powers or inverses explicitly.

Answer 1

Step 1: Understanding the Concept:
This problem can be solved efficiently using the Cayley-Hamilton Theorem, which states that every square matrix satisfies its own characteristic equation.
Step 2: Key Formula or Approach:
1. Find the characteristic polynomial, \(p(\lambda) = \det(M - \lambda I)\), of the matrix M.
2. By the Cayley-Hamilton Theorem, \(p(M) = 0\).
3. Manipulate the resulting matrix equation to match the form given in the problem and identify the value of \(\alpha\).
Step 3: Detailed Calculation:
The matrix M is an upper triangular matrix. The eigenvalues of a triangular matrix are its diagonal entries.
The eigenvalues of M are \(\lambda_1 = 3\), \(\lambda_2 = 2\), and \(\lambda_3 = 1\).
The characteristic polynomial is: \[ p(\lambda) = (\lambda - 3)(\lambda - 2)(\lambda - 1) \] Expanding the polynomial: \[ p(\lambda) = (\lambda^2 - 5\lambda + 6)(\lambda - 1) = \lambda^3 - \lambda^2 - 5\lambda^2 + 5\lambda + 6\lambda - 6 \] \[ p(\lambda) = \lambda^3 - 6\lambda^2 + 11\lambda - 6 \] By the Cayley-Hamilton Theorem, the matrix M must satisfy \(p(M) = 0\): \[ M^3 - 6M^2 + 11M - 6I = 0 \] We are given an equation involving \(M^{-1}\). Since none of the eigenvalues are zero, M is invertible, and \(M^{-1}\) exists. We can multiply the equation above by \(M^{-1}\): \[ M^{-1}(M^3 - 6M^2 + 11M - 6I) = M^{-1}(0) \] \[ M^2 - 6M + 11I - 6M^{-1} = 0 \] Now, we rearrange this equation to isolate the \(6M^{-1}\) term: \[ M^2 - 6M + 11I = 6M^{-1} \] The problem states that \(6M^{-1} = M^2 - 6M + \alpha I\). Comparing this with our derived equation: \[ M^2 - 6M + \alpha I = M^2 - 6M + 11I \] By direct comparison, we find: \[ \alpha = 11 \] Step 4: Final Answer:
The value of \(\alpha\) is 11.
Step 5: Why This is Correct:
The solution correctly applies the Cayley-Hamilton theorem. The characteristic equation derived from the matrix's eigenvalues is manipulated into the form given in the question, allowing for a direct comparison to find the unknown scalar \(\alpha\).