Step 1: Concept Overview:
This problem links the trace and determinant of a 2x2 matrix to the trace of its square. The Cayley-Hamilton theorem, which states a matrix satisfies its own characteristic equation, is key.
Step 2: Key Formula:
For a \( 2 \times 2 \) matrix A, the characteristic equation is:
\[ \lambda^2 - \text{trace}(A)\lambda + \det(A) = 0 \]By Cayley-Hamilton:
\[ A^2 - \text{trace}(A)A + \det(A)I = 0 \]Rearrange to solve for \( A^2 \) and then take the trace.
Step 3: Detailed Solution:
Substitute given values into the Cayley-Hamilton equation:
\[ A^2 - 5A + 4I = 0 \]Solve for \( A^2 \):
\[ A^2 = 5A - 4I \]Take the trace of both sides, using \( \text{trace}(X+Y) = \text{trace}(X) + \text{trace}(Y) \) and \( \text{trace}(c X) = c \cdot \text{trace}(X) \):
\[ \text{trace}(A^2) = \text{trace}(5A - 4I) \]\[ \text{trace}(A^2) = \text{trace}(5A) - \text{trace}(4I) \]\[ \text{trace}(A^2) = 5 \cdot \text{trace}(A) - 4 \cdot \text{trace}(I) \]Given \( \text{trace}(A) = 5 \). The trace of the \( 2 \times 2 \) identity matrix \( I = \begin{pmatrix} 1 & 0
0 & 1 \end{pmatrix} \) is \( 1+1=2 \).
\[ \text{trace}(A^2) = 5(5) - 4(2) = 25 - 8 = 17 \]Alternative Eigenvalue Method:Let eigenvalues of A be \( \lambda_1 \) and \( \lambda_2 \). Then:
- \( \text{trace}(A) = \lambda_1 + \lambda_2 = 5 \)- \( \det(A) = \lambda_1 \lambda_2 = 4 \)Eigenvalues of \( A^2 \) are \( \lambda_1^2 \) and \( \lambda_2^2 \).
- \( \text{trace}(A^2) = \lambda_1^2 + \lambda_2^2 \)Find \( \lambda_1^2 + \lambda_2^2 \):
\[ \lambda_1^2 + \lambda_2^2 = (\lambda_1 + \lambda_2)^2 - 2\lambda_1\lambda_2 = (5)^2 - 2(4) = 25 - 8 = 17 \]
Step 4: Final Answer:
The value of \( \text{trace}(A^2) \) is 17.