Question

Let A be a \( 2 \times 2 \) matrix with \( \det(A) = 4 \) and \( \text{trace}(A) = 5 \). Then the value of \( \text{trace}(A^2) \) is:

• A. 10
• B. 13
• C. 17
• D. 18

Hint:

For any \( n \times n \) matrix A, there is a useful identity: \( \text{trace}(A^2) = (\text{trace}(A))^2 - 2\det(A) \). This is derived from the eigenvalue method shown above and works specifically for the \(2 \times 2\) case. Memorizing this can lead to a very fast solution.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question relates the trace and determinant of a 2x2 matrix to the trace of its square. We can use the Cayley-Hamilton theorem, which states that a matrix satisfies its own characteristic equation.

Step 2: Key Formula or Approach:
For a \( 2 \times 2 \) matrix A, the characteristic equation is given by: \[ \lambda^2 - \text{trace}(A)\lambda + \det(A) = 0 \] By the Cayley-Hamilton theorem, the matrix A satisfies this equation: \[ A^2 - \text{trace}(A)A + \det(A)I = 0 \] We can rearrange this equation to find an expression for \( A^2 \) and then take the trace.

Step 3: Detailed Explanation:
Substitute the given values into the Cayley-Hamilton equation: \[ A^2 - 5A + 4I = 0 \] Rearrange to solve for \( A^2 \): \[ A^2 = 5A - 4I \] Now, take the trace of both sides of the equation. We use the properties that \( \text{trace}(X+Y) = \text{trace}(X) + \text{trace}(Y) \) and \( \text{trace}(c X) = c \cdot \text{trace}(X) \). \[ \text{trace}(A^2) = \text{trace}(5A - 4I) \] \[ \text{trace}(A^2) = \text{trace}(5A) - \text{trace}(4I) \] \[ \text{trace}(A^2) = 5 \cdot \text{trace}(A) - 4 \cdot \text{trace}(I) \] We are 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 Method using Eigenvalues: Let the eigenvalues of A be \( \lambda_1 \) and \( \lambda_2 \). Then: - \( \text{trace}(A) = \lambda_1 + \lambda_2 = 5 \) - \( \det(A) = \lambda_1 \lambda_2 = 4 \) The eigenvalues of \( A^2 \) are \( \lambda_1^2 \) and \( \lambda_2^2 \). - \( \text{trace}(A^2) = \lambda_1^2 + \lambda_2^2 \) We can find \( \lambda_1^2 + \lambda_2^2 \) from the known sums and products: \[ \lambda_1^2 + \lambda_2^2 = (\lambda_1 + \lambda_2)^2 - 2\lambda_1\lambda_2 = (5)^2 - 2(4) = 25 - 8 = 17 \]
Step 4: Final Answer:
The value of \( \text{trace}(A^2) \) is 17.