Solving Quadratic Equations And Linear Systems Matrix Method

2. Finding a Quadratic Equation with Roots a² + b and a + b²

Let's explore how to determine a quadratic equation given the roots of another quadratic equation. In this case, we are given that a and b are the roots of the quadratic equation x² - 5x + 2 = 0. Our goal is to find a new quadratic equation whose roots are a² + b and a + b². To achieve this, we will utilize Vieta's formulas, which provide a powerful connection between the coefficients of a polynomial and its roots. This approach allows us to construct the desired quadratic equation systematically.

Firstly, we need to find the sum and product of the roots of the given equation x² - 5x + 2 = 0. According to Vieta's formulas, the sum of the roots (a + b) is equal to the negation of the coefficient of the x term divided by the coefficient of the x² term. In this case, a + b = -(-5)/1 = 5. Similarly, the product of the roots (ab) is equal to the constant term divided by the coefficient of the x² term, so ab = 2/1 = 2. These values, a + b = 5 and ab = 2, will be crucial in our subsequent calculations.

Now, let's denote the roots of the new quadratic equation as α and β, where α = a² + b and β = a + b². To construct this quadratic equation, we need to find the sum (α + β) and the product (αβ) of these new roots. The sum α + β can be expressed as a² + b + a + b², which can be rearranged as a² + b² + a + b. We know that (a + b)² = a² + 2ab + b², so a² + b² = (a + b)² - 2ab. Substituting the values we found earlier, a + b = 5 and ab = 2, we get a² + b² = 5² - 2(2) = 25 - 4 = 21. Therefore, α + β = 21 + 5 = 26.

Next, we need to find the product of the roots, αβ = (a² + b)(a + b²). Expanding this expression, we get a³ + a²b² + ab + b³. This can be rewritten as a³ + b³ + a²b² + ab. To simplify further, we can use the identity a³ + b³ = (a + b)(a² - ab + b²). We already know that a + b = 5 and a² + b² = 21, so a² - ab + b² = 21 - 2 = 19. Thus, a³ + b³ = 5 * 19 = 95. Now, we can substitute the values back into the expression for αβ: αβ = 95 + (2)² + 2 = 95 + 4 + 2 = 101.

Finally, we can construct the quadratic equation using the sum and product of the roots. A quadratic equation with roots α and β can be written in the form x² - (α + β)x + αβ = 0. Substituting the values we found, α + β = 26 and αβ = 101, the required quadratic equation is x² - 26x + 101 = 0. This equation has the roots a² + b and a + b², as desired. The systematic application of Vieta's formulas and algebraic manipulation allowed us to solve this problem efficiently.

3. Solving a System of Linear Equations Using the Matrix Method

The matrix method provides a powerful and systematic way to solve systems of linear equations. This method is particularly useful for systems with multiple variables and equations, where traditional methods like substitution or elimination can become cumbersome. The core idea behind the matrix method is to represent the system of equations in matrix form and then use matrix operations to find the solution. In this section, we will delve into the steps involved in solving a system of linear equations using the matrix method, highlighting the key concepts and techniques.

To begin, let's consider a general system of linear equations. A system of n linear equations with n variables can be represented in the form:

a₁₁x₁ + a₁₂x₂ + ... + a₁ₙxₙ = b₁
a₂₁x₁ + a₂₂x₂ + ... + a₂ₙxₙ = b₂
...
aₙ₁x₁ + aₙ₂x₂ + ... + aₙₙxₙ = bₙ

where x₁, x₂, ..., xₙ are the variables, aᵢⱼ are the coefficients, and bᵢ are the constants. This system can be represented in matrix form as AX = B, where:

• A is the coefficient matrix:

  | a₁₁ a₁₂ ... a₁ₙ |
  | a₂₁ a₂₂ ... a₂ₙ |
  | ... ... ... ... |
  | aₙ₁ aₙ₂ ... aₙₙ |
  

• X is the variable matrix:

  | x₁ |
  | x₂ |
  | ... |
  | xₙ |
  

• B is the constant matrix:

  | b₁ |
  | b₂ |
  | ... |
  | bₙ |
  

To solve for X, we need to find the inverse of matrix A, denoted as A⁻¹, provided that A is invertible (i.e., its determinant is non-zero). If A⁻¹ exists, we can multiply both sides of the equation AX = B by A⁻¹ on the left:

A⁻¹AX = A⁻¹B

Since A⁻¹A is the identity matrix I, we have:

IX = A⁻¹B

Which simplifies to:

X = A⁻¹B

Thus, the solution to the system of linear equations is given by the product of the inverse of the coefficient matrix and the constant matrix. The process of finding the solution involves the following steps:

1. Write the system of equations in matrix form AX = B. This involves identifying the coefficient matrix A, the variable matrix X, and the constant matrix B.
2. Calculate the determinant of matrix A. The determinant, denoted as |A|, determines whether the matrix is invertible. If |A| = 0, the matrix is singular and the system may have no solution or infinitely many solutions. If |A| ≠ 0, the matrix is invertible and a unique solution exists.
3. Find the inverse of matrix A (A⁻¹). The inverse can be found using various methods, such as the adjoint method or Gaussian elimination. The adjoint method involves finding the adjugate of A (the transpose of the cofactor matrix) and dividing it by the determinant of A. Gaussian elimination involves performing row operations on the augmented matrix [A | I] to transform A into the identity matrix, and the resulting matrix on the right will be A⁻¹.
4. Multiply A⁻¹ by B to find the solution matrix X (X = A⁻¹B). This matrix multiplication yields the values of the variables x₁, x₂, ..., xₙ.

The matrix method offers several advantages. It provides a systematic and organized approach to solving linear systems, making it easier to handle large systems with many variables. The use of matrix operations allows for efficient computation, especially with the aid of computers and software. Additionally, the determinant of the coefficient matrix provides valuable information about the nature of the solution, indicating whether a unique solution exists or if the system is inconsistent or has infinitely many solutions. Overall, the matrix method is a powerful tool for solving systems of linear equations in various fields of mathematics, science, and engineering.

By understanding and applying the steps of the matrix method, one can effectively solve systems of linear equations and gain insights into their solutions. The method's systematic approach and computational efficiency make it a valuable tool for solving complex problems in various domains.