Let A = a Show that det (A-\I₂) = λ²−tr (4)\+det(A), where tr (A) =a+d. c d (In general, the trace of a square matrix is the sum of the entries on the main diagonal.)
Let A = a Show that det (A-\I₂) = λ²−tr (4)\+det(A), where tr (A) =a+d. c d (In general, the trace of a square matrix is the sum of the entries on the main diagonal.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:**Matrix Determinant and Trace**
Let \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \). Show that \(\det(A - \lambda I_2) = \lambda^2 - \text{tr}(A)\lambda + \det(A)\), where \(\text{tr}(A) = a + d\).
(In general, the trace of a square matrix is the sum of the entries on the main diagonal.)
This exercise involves finding the determinant of the matrix \( A - \lambda I_2 \), where \( I_2 \) is the 2x2 identity matrix and \( \lambda \) is a scalar. The goal is to express this determinant in terms of \(\lambda\), the trace of \( A \), and the determinant of \( A \).
**Explanation:**
1. **Determinant**: The determinant, denoted as \(\det\), is a scalar value that can be computed from the elements of a square matrix and provides insights into the properties of the matrix such as invertibility.
2. **Trace**: The trace of a square matrix, \(\text{tr}(A)\), is the sum of the elements on the main diagonal, which in this case is \( a + d \).
The expression to show involves understanding how the matrix \( A - \lambda I_2 \) adjusts each element and consequently affects the determinant calculation. The formula expressed shows a polynomial representation in terms of \(\lambda\), linking the matrix's trace and determinant to its eigenvalues. This is fundamental in linear algebra, reflecting properties such as eigenvalues and characteristic polynomials.
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