Problem 30 - Algebra 2

Question

Solve each equation for each variable. $$ \left[\begin{array}{cc}{x^{2}} & {4} \\ {-2} & {y^{2}}\end{array}\right]=\left[\begin{array}{rr}{9} & {4} \\ {-2} & {5 y}\end{array}\right] $$

Step-by-Step Solution

Answer

x = -3 or x = 3, y = 0 or y = 5.

Step 1: Identify corresponding elements

Compare each element of the matrices and set them equal to each other. The first element gives the equation: x^2 = 9. The off-diagonal elements are already equal and do not give an equation: 4 = 4 and -2 = -2. The last element gives the equation: y^2 = 5y.

Step 2: Solve for x

Solve the equation x^2 = 9. This can be factored as (x + 3)(x - 3) = 0, which gives the solutions x = -3 or x = 3.

Step 3: Solve for y

To solve y^2 = 5y, you can factor out y to get y(y - 5) = 0, which gives the solutions y = 0 or y = 5.

Key Concepts

Matrix Algebra

Matrix algebra is a fundamental branch of mathematics that deals with the study of matrices and the ways in which they can be manipulated. It's similar to how we perform operations with numbers, but instead, we do them with arrays of numbers or functions.

In the context of the given exercise, we are dealing with a matrix equation where two matrices are set to be equal, which implies that their corresponding elements must also be equal. To find the solution, we compare the corresponding elements and arrive at individual equations for each unknown variable.

Importance of Matrix Comparisons

Comparing matrices element by element is a basic operation in matrix algebra that helps us break down complex problems into simpler ones. In our case, by equating each element of the two given matrices, we can isolate the variables and solve for them individually.

Factoring Quadratics

Factoring quadratics is a crucial technique in algebra for simplifying expressions and solving quadratic equations. A quadratic equation is typically of the form \( ax^2 + bx + c = 0 \), and factoring involves expressing it as \( (mx + n)(px + q) = 0 \), where \( m, n, p, \) and \( q \) are numbers that when multiplied out, return to the original quadratic equation.

In this exercise, we see two instances of factoring quadratics. In the first instance, \( x^2 = 9 \), we factor the equation as \( (x + 3)(x - 3) = 0 \). Similarly, in the second instance, the equation \( y^2 = 5y \) is factored by taking a common factor of y out, leading to \( y(y - 5) = 0 \).

Zero Product Property

After factoring, we use the zero product property to solve for the variables. This property states if a product of factors is zero, at least one of the factors must be zero. So, by setting each factor equal to zero, we solve for the unknown variables. Factoring is advantageous because it transforms a quadratic equation into a form that makes finding solutions straightforward.

Systems of Equations

A system of equations is a set of two or more equations that share common variables. The goal is to find a solution that satisfies all equations simultaneously. Systems of equations can be solved using various methods, including substitution, elimination, and graphing.

However, when dealing with matrix equations, we treat each element's equality as a separate equation that forms a system. By doing so, we can solve for each variable independently when possible.

Matrix Equations as Systems

In the given exercise, the system is derived from the equality of the matrices' corresponding elements. This is akin to having two separate equations \( x^2 = 9 \) and \( y^2 = 5y \) bound together by the matrix structure, and we must solve each within the context of the system. This provides an organized approach to solving multiple equations and is a clear example of how matrix algebra conveniently represents and solves systems of equations.