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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
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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**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.
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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