Math Problem Statement

Find the characteristic polynomial of A. Use x for the variable in your polynomial. You do not need to factor your polynomial.

Solution

To find the characteristic polynomial of the given matrix AA, we compute the determinant of AxIA - xI, where II is the identity matrix of the same size as AA and xx is a variable. Let's work through this step-by-step.

Matrix AA:

-5 & -2 & 0 \\ 12 & 5 & 0 \\ 12 & 4 & -3 \end{bmatrix}$$ ### $$A - xI$$: Subtract $$x$$ times the identity matrix from $$A$$: $$A - xI = \begin{bmatrix} -5 - x & -2 & 0 \\ 12 & 5 - x & 0 \\ 12 & 4 & -3 - x \end{bmatrix}$$ ### Characteristic Polynomial: The characteristic polynomial is given by: $$\text{det}(A - xI)$$ #### Determinant Calculation: Expanding along the first row: $$\text{det}(A - xI) = \begin{vmatrix} -5-x & -2 & 0 \\ 12 & 5-x & 0 \\ 12 & 4 & -3-x \end{vmatrix}$$ The determinant simplifies because the third column contains a $$0$$, meaning the expansion only involves the first two rows: $$\text{det}(A - xI) = (-5-x) \cdot \begin{vmatrix} 5-x & 0 \\ 4 & -3-x \end{vmatrix} - (-2) \cdot \begin{vmatrix} 12 & 0 \\ 12 & -3-x \end{vmatrix}$$ #### Sub-matrix Determinants: 1. For the first term: $$\begin{vmatrix} 5-x & 0 \\ 4 & -3-x \end{vmatrix} = (5-x)(-3-x) - (0)(4) = (-3-x)(5-x)$$ $$= -15 + 3x - 5x - x^2 = -x^2 - 2x - 15$$ 2. For the second term: $$\begin{vmatrix} 12 & 0 \\ 12 & -3-x \end{vmatrix} = (12)(-3-x) - (0)(12) = -36 - 12x$$ #### Substitute Back: $$\text{det}(A - xI) = (-5-x)(-x^2 - 2x - 15) - (-2)(-36 - 12x)$$ Expand both terms: 1. Expanding the first: $$(-5-x)(-x^2 - 2x - 15) = (-5)(-x^2) + (-5)(-2x) + (-5)(-15) + (-x)(-x^2) + (-x)(-2x) + (-x)(-15)$$ $$= 5x^2 + 10x + 75 + x^3 + 2x^2 + 15x$$ $$= x^3 + 7x^2 + 25x + 75$$ 2. Expanding the second: $$(-2)(-36 - 12x) = 72 + 24x$$ #### Combine: $$\text{det}(A - xI) = (x^3 + 7x^2 + 25x + 75) - (72 + 24x)$$ $$= x^3 + 7x^2 + (25x - 24x) + (75 - 72)$$ $$= x^3 + 7x^2 + x + 3$$ ### Final Answer: The characteristic polynomial is: $$x^3 + 7x^2 + x + 3$$ Would you like further details or assistance with understanding this calculation? Here are some questions that may interest you: 1. How do we interpret the roots of the characteristic polynomial? 2. Why is the characteristic polynomial important in linear algebra? 3. Can we solve this cubic polynomial for its eigenvalues? 4. What is the role of the identity matrix in the calculation? 5. How does the determinant expand based on the cofactor method? **Tip:** Always check if simplifying assumptions, such as zeros in rows or columns, can make determinant calculations

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Algebra
Determinants
Eigenvalues
Characteristic Polynomial

Formulas

Characteristic Polynomial: det(A - xI)

Theorems

Cayley-Hamilton Theorem
Determinant Expansion by Cofactors

Suitable Grade Level

College Level (Linear Algebra Course)