Question

Using Cayley-Hamilton theorem, the value of \(A^{4}-4 A^{3}-5 A^{2}-A+2 I\) when \(A=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right]\) is. (Anna, 2009)

Short answer

The result is \( \begin{bmatrix} -173 & -110 \\ -220 & -228 \end{bmatrix} \).

Step-by-step solution

Find the Characteristic Polynomial

Given matrix is \( A = \begin{bmatrix} 1 & 2 \ 4 & 2 \end{bmatrix} \). To find the characteristic polynomial, we calculate \( \det(A - \lambda I) \), where \( I \) is the identity matrix and \( \lambda \) is a scalar. Thus, \( A - \lambda I = \begin{bmatrix} 1-\lambda & 2 \ 4 & 2-\lambda \end{bmatrix} \). The determinant is \((1-\lambda)(2-\lambda) - 8 = \lambda^2 - 3\lambda - 6 \). So the characteristic polynomial is \( \lambda^2 - 3\lambda - 6 = 0 \).

Apply Cayley-Hamilton Theorem

By the Cayley-Hamilton theorem, a square matrix satisfies its own characteristic equation. So \( A^2 - 3A - 6I = 0 \). Therefore, \( A^2 = 3A + 6I \).

Calculate Powers of A

Use the relation \( A^2 = 3A + 6I \) to calculate higher powers. For \( A^3 \), multiply \( A^2 \, A = (3A + 6I)A = 3A^2 + 6A = 9A + 6A^2 = 9A + 18I \). Since \( A^2 = 3A + 6I \), substitute to find \( 6A^2 = 18A + 36I \). Thus, \( A^3 = 9A + 18I \). For \( A^4 \), use \( A^3 \), so \( A^4 = A^3 \, A = (9A + 18I)A = 9A^2 + 18A \). Substitute \( A^2 = 3A + 6I \) again to get \( A^4 = 27A + 54I + 18A = 45A + 54I \).

Simplify Expression

Substitute the known expressions from the previous steps into the original expression: \( A^4 - 4A^3 - 5A^2 - A + 2I \). Plug in \( A^4 = 45A + 54I \), \( A^3 = 21A + 36I \) and \( A^2 = 3A + 6I \): \[ (45A + 54I) - 4(21A + 36I) - 5(3A + 6I) - A + 2I \]. Simplify to: \( 45A + 54I - 84A - 144I - 15A - 30I - A + 2I \). This results in \( (45 - 84 - 15 - 1)A + (54 - 144 - 30 + 2)I = -55A - 118I \).

Calculate the Final Result

Now that the expression is simplified to \(-55A - 118I\), calculate \(-55A\) and \(-118I\). For \( A = \begin{bmatrix} 1 & 2 \ 4 & 2 \end{bmatrix} \), \(-55A = \begin{bmatrix} -55 & -110 \ -220 & -110 \end{bmatrix} \) and \(-118I = \begin{bmatrix} -118 & 0 \ 0 & -118 \end{bmatrix} \). Hence the result is \( \begin{bmatrix} -173 & -110 \ -220 & -228 \end{bmatrix} \).

Key concepts

Characteristic Polynomial

The characteristic polynomial is a fundamental tool in linear algebra used to study matrices. It is derived from the determinant of a specific matrix called the 'characteristic matrix.' For a given square matrix \( A \), the characteristic polynomial is obtained by subtracting \( \lambda I \) (where \( \lambda \) is a scalar and \( I \) is the identity matrix) from \( A \) and then calculating the determinant of the resultant matrix. In mathematical terms, the characteristic polynomial is expressed as \( \det(A - \lambda I) \). This polynomial is crucial for determining important properties of the matrix, such as its eigenvalues.

To find the characteristic polynomial of matrix \( A = \begin{bmatrix} 1 & 2 \ 4 & 2 \end{bmatrix} \), we start by calculating the matrix \( A - \lambda I \):
\[ A - \lambda I = \begin{bmatrix} 1 - \lambda & 2 \ 4 & 2 - \lambda \end{bmatrix} \]

Next, we find the determinant of this matrix to obtain the characteristic polynomial:
\[ \det(A - \lambda I) = (1 - \lambda)(2 - \lambda) - 8 = \lambda^2 - 3\lambda - 6 \]

This equation, \( \lambda^2 - 3\lambda - 6 = 0 \), reveals that \( A \) has eigenvalues that can be found by solving it. These eigenvalues are pivotal in solving many linear algebra problems, including those involving the Cayley-Hamilton theorem.

Matrix Calculations

Matrix calculations involve performing mathematical operations on matrices, such as addition, multiplication, and finding powers. These operations help in transforming and simplifying matrix expressions, making them an integral part of linear algebra.

In our exercise, we were tasked with finding higher powers of the matrix \( A \), specifically \( A^4 \). To accomplish this, we first established relationships between various powers of \( A \) using Cayley-Hamilton theorem principles.

• First, using the characteristic polynomial, it was determined that \( A^2 = 3A + 6I \).
• Then, to compute \( A^3 \), we multiplied \( A^2 \) by \( A \), which led to \( A^3 = 9A + 18I \) after substituting back \( A^2 = 3A + 6I \).
• Finally, multiplying \( A^3 \) by \( A \) gave us \( A^4 = 45A + 54I \).

Each calculation was strategically conducted to simplify the expression, demonstrating the essential role matrix calculations play in advancing our understanding of matrix properties.

Linear Algebra Concepts

Linear algebra is a branch of mathematics concerning linear equations, vectors, matrices, and their transformations. Several key concepts in linear algebra are crucial for tackling problems like the one discussed in this exercise.

One of the central concepts is the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic equation. This theorem allows us to reduce complex matrix powers into linear combinations of simpler matrices, which is invaluable for simplifying matrix expressions. Here, recognizing that \( A^2 = 3A + 6I \) allows us to computationally evaluate higher powers of matrix \( A \) more easily.

Other important linear algebraic concepts involve understanding and manipulating matrices, determinants, and identity matrices. These tools enable us to solve systems of linear equations, explore vector spaces, and analyze linear transformations effectively. By applying these concepts, students can develop critical problem-solving skills in mathematics and related fields.