Question

If \( A = \begin{pmatrix} 2 & 4 & 1 \\ 0 & 2 & -1 \\ 0 & 0 & 1 \end{pmatrix} \) satisfies \( A^3 + \mu A^2 + \lambda A - 4I_3 = 0 \), then the respective values of \( \lambda \) and \( \mu \) are:

• A. -5, 8
• B. 8, -5
• C. 5, -8
• D. -8, 5

Hint:

For triangular matrices, the eigenvalues are simply the entries on the main diagonal. This makes finding the characteristic polynomial much faster. For a 3x3 matrix, the characteristic equation is also \( x^3 - (\text{tr}(A))x^2 + (\text{sum of cofactors of diagonal elements})x - (\det(A)) = 0 \). Here, tr(A)=2+2+1=5, det(A)=2*2*1=4, giving \(x^3 - 5x^2 + ... - 4 = 0\), quickly confirming \(\mu=-5\).

Answer 1

The Correct Option is B

Step 1: Introduction:
The Cayley-Hamilton Theorem states that a square matrix satisfies its characteristic equation. This problem involves finding the values of \( \lambda \) and \( \mu \) by comparing the characteristic equation of matrix A with a given polynomial equation.

Step 2: Method:
The characteristic equation is found using \( \det(A - xI) = 0 \). Since A is an upper triangular matrix, its eigenvalues are the diagonal entries: \(x_1=2, x_2=2, x_3=1\). The characteristic polynomial is calculated as \( (x-x_1)(x-x_2)(x-x_3) = 0 \): \[ (x-2)(x-2)(x-1) = 0 \]\[ (x^2-4x+4)(x-1) = 0 \]\[ x(x^2-4x+4) - 1(x^2-4x+4) = 0 \]\[ x^3 - 4x^2 + 4x - x^2 + 4x - 4 = 0 \]\[ x^3 - 5x^2 + 8x - 4 = 0 \]
Step 3: Solution:
Applying the Cayley-Hamilton Theorem, matrix A satisfies the equation: \[ A^3 - 5A^2 + 8A - 4I = 0 \]Given the equation: \[ A^3 + \mu A^2 + \lambda A - 4I_3 = 0 \]Comparing coefficients yields: \[ \mu = -5 \]\[ \lambda = 8 \]
Step 4: Answer:
The values are \( \lambda=8 \) and \( \mu=-5 \).