Chapter 12: Problem 54
In Exercises \(54-57\), verify the Cayley-Hamilton theorem for the given matrix. Note: You may also want to investigate the polyvalm command. $$ \left[\begin{array}{rr} 5 & 2 \\ -6 & -2 \end{array}\right] $$
Short Answer
Expert verified
The matrix satisfies its characteristic polynomial, verifying the Cayley-Hamilton theorem.
Step by step solution
01
Write the characteristic polynomial
For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the characteristic polynomial \( p(\lambda) \) is given by \( \lambda^2 - \text{trace}(A) \cdot \lambda + \det(A) \). Calculate the trace and determinant of \( A = \begin{bmatrix} 5 & 2 \ -6 & -2 \end{bmatrix} \). The trace is \( \text{trace}(A) = 5 + (-2) = 3 \), and the determinant is \( \det(A) = 5(-2) - 2(-6) = -10 + 12 = 2 \). Hence the characteristic polynomial is \( \lambda^2 - 3\lambda + 2 \).
02
Verify the polynomial with the matrix
The Cayley-Hamilton theorem states that a matrix satisfies its own characteristic polynomial. Substituting the matrix \( A \) into the polynomial \( p(A) = A^2 - 3A + 2I \), where \( I \) is the identity matrix \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \).
03
Calculate \( A^2 \)
To calculate \( A^2 \), multiply the matrix \( A \) by itself: \[ A^2 = \begin{bmatrix} 5 & 2 \ -6 & -2 \end{bmatrix} \cdot \begin{bmatrix} 5 & 2 \ -6 & -2 \end{bmatrix} = \begin{bmatrix} (5 \cdot 5 + 2 \cdot (-6)) & (5 \cdot 2 + 2 \cdot (-2)) \ (-6 \cdot 5 + (-2) \cdot (-6)) & (-6 \cdot 2 + (-2) \cdot (-2)) \end{bmatrix} = \begin{bmatrix} 13 & 6 \ -18 & -8 \end{bmatrix} \].
04
Substitute back into the polynomial
Substitute \( A^2 \) and \( 3A \) into the polynomial: \[ p(A) = A^2 - 3A + 2I = \begin{bmatrix} 13 & 6 \ -18 & -8 \end{bmatrix} - 3 \begin{bmatrix} 5 & 2 \ -6 & -2 \end{bmatrix} + 2 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \].
05
Calculate \( 3A \)
Compute \( 3A \): \[ 3A = 3 \begin{bmatrix} 5 & 2 \ -6 & -2 \end{bmatrix} = \begin{bmatrix} 15 & 6 \ -18 & -6 \end{bmatrix} \].
06
Calculate \( 2I \)
Compute \( 2I \): \[ 2I = 2 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} \].
07
Final Calculation
Perform the final calculation: \[ p(A) = \begin{bmatrix} 13 & 6 \ -18 & -8 \end{bmatrix} - \begin{bmatrix} 15 & 6 \ -18 & -6 \end{bmatrix} + \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} \].The result is the zero matrix, confirming that the Cayley-Hamilton theorem holds for this matrix.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Polynomial
The characteristic polynomial is a special polynomial equation associated with a matrix, which plays a significant role in linear algebra, especially in the study of matrix properties. To find it, start with a square matrix, say \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \). The characteristic polynomial, \( p(\lambda) \), for a 2x2 matrix is given by:
In this problem, the matrix is \( \begin{bmatrix} 5 & 2 \ -6 & -2 \end{bmatrix} \). To find the characteristic polynomial, we need to compute:
- \( \lambda^2 - \text{trace}(A) \cdot \lambda + \det(A) \)
In this problem, the matrix is \( \begin{bmatrix} 5 & 2 \ -6 & -2 \end{bmatrix} \). To find the characteristic polynomial, we need to compute:
- Trace of \( A \), which is the sum of the diagonal elements, \( 5 + (-2) = 3 \).
- Determinant of \( A \), calculated as \( 5 \cdot (-2) - 2 \cdot (-6) = -10 + 12 = 2 \).
Matrix Operations
Matrix operations are fundamental techniques used to manipulate matrices in order to solve problems, including verifying the Cayley-Hamilton theorem.
In our example, the steps involved in matrix operations include multiplication, scalar multiplication, and addition/subtraction.
In our example, the steps involved in matrix operations include multiplication, scalar multiplication, and addition/subtraction.
Matrix Multiplication
To find \( A^2 \), the matrix \( A \) is multiplied by itself:- Find each element by multiplying corresponding rows and columns.
- For the first element, calculate \((5 \cdot 5 + 2 \cdot (-6)),\) resulting in \(13\).
- Repeat for all elements to get \( A^2 = \begin{bmatrix} 13 & 6 \ -18 & -8 \end{bmatrix} \).
Scalar Multiplication
Scalar multiplication means multiplying every element in a matrix by a scalar value:- Calculate \( 3A \) where each element of \( A \) is multiplied by 3.
- \( 3A = \begin{bmatrix} 15 & 6 \ -18 & -6 \end{bmatrix} \).
Identity Matrix
The identity matrix is a special type of matrix that acts like the number 1 in matrix multiplication. It doesn't change the other matrix when involved in a multiplication.
For a 2x2 matrix, the identity matrix \( I \) looks like this:
In this exercise, the identity matrix is used in the expression \( 2I \), where:
For a 2x2 matrix, the identity matrix \( I \) looks like this:
- \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \)
In this exercise, the identity matrix is used in the expression \( 2I \), where:
- Each element of \( I \) is multiplied by 2.
- This results in \( 2I = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} \).
Determinant Computation
Determinants are numbers that can be calculated from square matrices and are fundamental in determining properties of matrices like invertibility and solving systems of linear equations.
For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant \( \det(A) \) is calculated by:
For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant \( \det(A) \) is calculated by:
- \( ad - bc \)
- If the determinant equals 0, the matrix is not invertible (or singular).
- If non-zero, the matrix is invertible.