Problem 40 - Precalculus: Mathematical for Calculus

Question

Solve for \(x\) and \(y\) $$3\left[\begin{array}{ll} x & y \\ y & x \end{array}\right]=\left[\begin{array}{rr} 6 & -9 \\ -9 & 6 \end{array}\right]$$

Step-by-Step Solution

Answer

\( x = 2 \) and \( y = -3 \).

Step 1: Matching Matrices

We start by equating the matrix on the left with the matrix on the right. The equation given is \( 3\begin{bmatrix} x & y \ y & x \end{bmatrix} = \begin{bmatrix} 6 & -9 \ -9 & 6 \end{bmatrix} \). This means that the corresponding elements must be equal. Thus, we form the following equations: 1. \( 3x = 6 \) 2. \( 3y = -9 \)

Step 2: Solve for x

Using the first equation from Step 1, \( 3x = 6 \), we solve for \( x \). \[ x = \frac{6}{3} = 2 \]

Step 3: Solve for y

Using the second equation from Step 1, \( 3y = -9 \), we solve for \( y \). \[ y = \frac{-9}{3} = -3 \]

Step 4: Verification

We should check our solution by verifying that the computed values of \( x \) and \( y \) satisfy the original matrix equation. Thus, we substitute \( x = 2 \) and \( y = -3 \) back into the matrix: \[ 3\begin{bmatrix} 2 & -3 \ -3 & 2 \end{bmatrix} = \begin{bmatrix} 6 & -9 \ -9 & 6 \end{bmatrix} \] The calculations yield, \[ \begin{bmatrix} 6 & -9 \ -9 & 6 \end{bmatrix} \], confirming that our values are correct.

Key Concepts

Solving Systems of Equations

A system of equations consists of two or more equations with the same set of variables. In this exercise, the system involves the variables \( x \) and \( y \). Each element in the matrices corresponds to an equation in the system. For example, the element \( 3x \) in the original matrix corresponds to \( 6 \) in the solution matrix, thus forming the equation \( 3x = 6 \).

Here's a simple step-by-step to solve it:

• Identify corresponding elements in both matrices to form individual equations.
• Use elementary algebra methods like division and multiplication to solve the equations.
• Verify solutions by substituting values back into the original equations.

This is crucial to ensure your calculated values for \( x \) and \( y \) actually solve the given matrix equation.

Matrix Multiplication

Matrix multiplication involves multiplying a number (a scalar) by every element in the matrix or multiplying two matrices together. In this case, we are multiplying the scalar \( 3 \) with each element in the matrix \( \begin{bmatrix} x & y \ y & x \end{bmatrix} \) to equate it to another matrix.

The steps involved are:

• Multiply \( x \) and \( y \) by 3 to get the left side of the equation.
• Each element in the resulting matrix must match the corresponding element in the given matrix \( \begin{bmatrix} 6 & -9 \ -9 & 6 \end{bmatrix} \).

This process helps us establish the set of equations that will allow us to find the values of \( x \) and \( y \). Remember, matrix multiplication, particularly with a scalar, is straightforward as it only involves multiplying each term by the given number.

Elementary Algebra

Elementary algebra provides the foundation for solving equations like \( 3x = 6 \) and \( 3y = -9 \). It revolves around simple operations such as addition, subtraction, multiplication, and division.

To solve for \( x \) and \( y \), follow these basic rules:

• Isolate the variable on one side of the equation. For example, with \( 3x = 6 \), divide both sides by 3 to find \( x = 2 \).
• Apply the same process to find \( y \) from \( 3y = -9 \) to get \( y = -3 \).

These simple steps allow us to manipulate equations and solve for unknowns. By understanding these principles, you can approach even the most complex algebra problems with confidence.