Problem 36 - Algebra and Trigonometry

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

The solution is \( x = 2 \) and \( y = -3 \).

Step 1: Understand the Matrix Equation

The problem provides a matrix equation where you have matrix multiplications on the left side and a constant matrix on the right side. The equation is \( 3\begin{bmatrix} x & y \ y & x \end{bmatrix} = \begin{bmatrix} 6 & -9 \ -9 & 6 \end{bmatrix} \). Our goal is to find the values of \( x \) and \( y \) that satisfy this equation.

Step 2: Set Up Matrix Multiplication Equations

Multiply the matrix on the left by 3, by distributing the multiplication into each element of the matrix: \[ \begin{bmatrix} 3x & 3y \ 3y & 3x \end{bmatrix} = \begin{bmatrix} 6 & -9 \ -9 & 6 \end{bmatrix} \]. Now, you have two pairs of equations to solve based on matching elements.

Step 3: Solve the First Equation

From the matrix equation, equate the first row, first column elements: the equation is \( 3x = 6 \). Solve for \( x \) by dividing both sides of the equation by 3, which gives \( x = 2 \).

Step 4: Solve the Second Equation

Next, equate the first row, second column elements: the equation is \( 3y = -9 \). Solve for \( y \) by dividing both sides of the equation by 3, which gives \( y = -3 \).

Step 5: Verify the Solution

Ensure the values of \( x = 2 \) and \( y = -3 \) satisfy all parts of the original matrix equation by checking the second row equation: \( 3y = -9 \) (which confirms \( y = -3 \)), and \( 3x = 6 \) (which confirms \( x = 2 \)). Both equations are consistent.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. The main goal when dealing with systems of equations is to find a solution that satisfies all given equations simultaneously. In our exercise, we have a quadratic matrix equation where variables \(x\) and \(y\) appear in a system.
This means we're dealing with two equations obtained by comparing corresponding elements from two matrices. Here's how it breaks down:

• From the first row and first column: \(3x = 6\)
• From the first row and second column: \(3y = -9\)

So, solving this system means finding \(x\) and \(y\) such that both conditions are met.
Understanding the structure of a system of equations is crucial in linear algebra, as it provides a systematic way of organizing and solving various equations together.

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra and involves a systematic way of multiplying two matrices together. It's different from scalar multiplication and requires following specific rules. In our problem, we initially see a matrix scaled by a scalar 3.
The multiplication here is straightforward:

• Each entry in the original matrix is multiplied by the scalar (in this case, the number 3).
• This process results in a new matrix where each element \(a_{ij}\) has become \(3 \times a_{ij}\).

After this step, we have a new matrix that can be compared for equality with another matrix.
Understanding this process is vital for solving matrix equations because it simplifies the problem, allowing us to transform complex matrix relationships into more manageable, element-wise equations.

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. It's a powerful tool that allows for solving various mathematical problems efficiently.
In our exercise, we use concepts from linear algebra - such as representing a system of equations in matrix form. Here is why linear algebra is essential:

• It allows for compact representation of equations using matrices, saving space and effort.
• Operations like multiplication and solving systems are systematic and can be scaled up to handle bigger systems.

The practical application of these ideas simplifies the process of finding solutions to problems that might otherwise seem complex. Linear algebra provides the framework necessary to approach and solve real-world problems in engineering, physics, computer science, and more.