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Chapter 11: Problem 1

Find the eigenvalues of the following matrices: a \(A=\left(\begin{array}{rr}7 & 3 \\ 0 & -4\end{array}\right)\) b \(A=\left(\begin{array}{ll}5 & -2 \\ 4 & -1\end{array}\right)\) c \(A=\left(\begin{array}{rr}-1 & 4 \\ 2 & 1\end{array}\right)\)

Short Answer

Expert verified
Matrix a: \(7, -4\), Matrix b: \(2+\sqrt{17}, 2-\sqrt{17}\), Matrix c: \(i, -i\).

Step by step solution

01

Title - Set Up the Characteristic Equation

For each matrix, set up the characteristic equation given by \(\text{det}(A - \lambda I) = 0\).
02

Step 2a - Matrix a: Calculate Characteristic Polynomial

The characteristic polynomial for matrix a is: \(\text{det}\left(\begin{array}{rr}7 - \lambda & 3 \ 0 & -4 - \lambda\end{array}\right) = (7 - \lambda)(-4 - \lambda)\). Solve the quadratic equation \(\lambda^2 - 3\lambda - 28 = 0\).
03

Step 3a - Matrix a: Solve for Eigenvalues

Solve \((\lambda - 7)(\lambda + 4) = 0\). The solutions are \lambda = 7\ and \lambda = -4\.
04

Step 2b - Matrix b: Calculate Characteristic Polynomial

The characteristic polynomial for matrix b is: \(\text{det}\left(\begin{array}{ll}5 - \lambda & -2 \ 4 & -1 - \lambda\end{array}\right) = (5 - \lambda)(-1 - \lambda) - (-2 \cdot 4)\). Simplify to \(\lambda^2 - 4\lambda - 13=0\).
05

Step 3b - Matrix b: Solve for Eigenvalues

Solve \((\lambda - 2 + \sqrt{17})(\lambda - 2 - \sqrt{17}) = 0\). The solutions are \lambda = 2 + \sqrt{17}\ and \lambda = 2 - \sqrt{17}\.
06

Step 2c - Matrix c: Calculate Characteristic Polynomial

The characteristic polynomial for matrix c is: \(\text{det}\left(\begin{array}{rr}-1 - \lambda & 4 \ 2 & 1 - \lambda\end{array}\right) = (-1 - \lambda)(1 - \lambda) - (4 \cdot 2)\). Simplify to \(\lambda^2 + 1 = 0\).
07

Step 3c - Matrix c: Solve for Eigenvalues

Solve \(\lambda^2 = -1\). The solutions are \lambda = i\ and \lambda = -i\.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Characteristic Equation
The characteristic equation is essential for finding eigenvalues. For any given matrix **A**, the characteristic equation is obtained by subtracting \( \lambda I \) from \(A\) and setting the determinant to zero. Here, \(I\) is the identity matrix of the same dimension as \(A\), and \(\lambda\) is a scalar, which represents the eigenvalues we need to find. Mathematically, this is represented as \[ \text{det}(A - \lambda I) = 0 \].

Breaking it down:
  • **Step 1**: Subtract \(\lambda I\) from the matrix \(A \). This means subtracting \(\lambda\) from each diagonal element of **A**.
  • **Step 2**: Compute the determinant of the resultant matrix. This involves multiplying the difference terms and accounting for the signs of the permutations.
  • **Step 3**: Set the determinant to zero, resulting in a characteristic equation which is typically a polynomial in \(\lambda\).
This equation forms the basis for determining eigenvalues. Eigenvalues are solutions to this polynomial equation.
Characteristic Polynomial
The characteristic polynomial is obtained from the characteristic equation, specifically, it is the polynomial formed when calculating the determinant \( \text{det}(A - \lambda I)\).

Consider matrix **A =** \( \left( \begin{array}{ll} 5 & -2 \ \ 4 & -1 \end{array} \right) \). Subtracting \(\lambda I\) gives us **A - \lambda I =** \( \left( \begin{array}{ll} 5 - \lambda & -2 \ \ 4 & -1 - \lambda \end{array} \right) \). The determinant of this matrix is:
  • \[ \text{det}(5 - \lambda & -2 \ 4 & -1 - \lambda) = (5 - \lambda)(-1 - \lambda) - (4 \cdot -2) = \lambda^2 - 4\lambda - 13 = 0 \]
The polynomial, \( \lambda^2 - 4\lambda - 13 = 0 \), is the characteristic polynomial. The solutions to this polynomial, the values of \(\lambda\) that satisfy the equation, are the eigenvalues.
Quadratic Equation
The characteristic polynomial often reduces to a quadratic equation, especially in the case of 2x2 matrices. Recognizing and solving quadratic equations is pivotal in finding eigenvalues.

A standard quadratic equation takes the form \[ ax^2 + bx + c = 0 \]. The solutions to this equation can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

For example, in matrix **A** from the step-by-step solution: **a** = 1, **b** = -4, and **c** = -13, giving us the quadratic equation \( \lambda^2 - 4\lambda - 13 = 0 \). Plug these values into the quadratic formula:
  • **Step 1**: Compute the discriminant: \( b^2 - 4ac = (-4)^2 - 4(1)(-13) = 16 + 52 = 68 \)
  • **Step 2**: Plug into the quadratic formula: \( \lambda = \frac{4 \pm \sqrt{68}}{2} = 2 \pm \sqrt{17} \)
So, the eigenvalues are \( \lambda = 2 + \sqrt{17} \) and \( \lambda = 2 - \sqrt{17} \).

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Most popular questions from this chapter

Find the values of \(\lambda\) that give a non-trivial solution of $$ (\mathrm{A}-\lambda \mathbf{I}) \mathbf{u}=\mathbf{0} $$ for \(\mathbf{a} \mathbf{A}=\left(\begin{array}{rr}-1 & 1 \\ 9 & -1\end{array}\right)\) b \(\mathrm{A}=\left(\begin{array}{rr}1 & 3 \\ -6 & 5\end{array}\right)\) \(\left[\mathbf{u}=\left(\begin{array}{l}x \\ y\end{array}\right)\right.\) and \(\mathbf{I}\) is the identity matrix \(]\)

By applying Kirchhoff's law and Ohm's law to a circuit, we obtain the following equations: $$ \begin{aligned} 3 i_{1}-5 i_{2}+3 i_{3} &=7.5 \\ 2 i_{1}+i_{2}-7 i_{3} &=-17.5 \\ -10 i_{1}+4 i_{2}+5 i_{3} &=16 \end{aligned} $$ where \(i_{1}, i_{2}\) and \(i_{3}\) represent currents. By using the inverse matrix method, find the values of \(i_{1}, i_{2}\) and \(i_{3}\).

Find the inverse matrix of the following: a \(\mathbf{A}=\left(\begin{array}{ll}5 & 4 \\ 3 & 1\end{array}\right)\) b \(\mathbf{A}=\left(\begin{array}{ll}3 & 6 \\ 7 & 8\end{array}\right)\) c \(\mathbf{A}=\left(\begin{array}{cc}7 & 1 \\ 14 & 2\end{array}\right)\)

Find \(\mathrm{A}^{\mathrm{T}}\) (A transposed) for the following: \(\mathbf{a} \mathbf{A}=\left(\begin{array}{rr}-2 & 3 \\ 1 & 5\end{array}\right) \quad\) b \(\mathbf{A}=\left(\begin{array}{rc}1 / 3 & -2 / 5 \\\ -3 / 7 & \pi\end{array}\right)\) \(\mathbf{c} \mathrm{A}=\left(\begin{array}{rrr}7 & 3 & 4 \\ 2 & 6 & 1 \\ -3 & -3 & 1\end{array}\right)\) \(\mathbf{d} \mathrm{A}=\left(\begin{array}{rr}1.17 & 1.36 \\ 9.39 & -1.45 \\\ 2.11 & 5.20\end{array}\right)\) e \(A=\left(\begin{array}{rrrr}-7 & 2 & 1 & 5 \\ 3 & 6 & -4 & 7 \\ 8 & 3 & -3 & 5 \\ -4 & 6 & 7 & 0\end{array}\right)\) f \(A=\left(\begin{array}{rrr}1 & -3 & 7 \\ -9 & 4 & 6 \\ -4 & 2 & 8 \\ 9 & 19 & 11\end{array}\right)\)

For \(\mathbf{A}=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\) and \(\mathbf{B}=\left(\begin{array}{ll}a & c \\ b & d\end{array}\right)\), show that \(\operatorname{det} \mathbf{A}=\operatorname{det} \mathbf{B}\).
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