Chapter 11: Problem 36
Solve for \(x\) and \(y\) $$ 3\left[\begin{array}{ll}{x} & {y} \\ {y} & {x}\end{array}\right]=\left[\begin{array}{rr}{6} & {-9} \\ {-9} & {6}\end{array}\right] $$
Short Answer
Expert verified
The solution is \( x = 2 \) and \( y = -3 \).
Step by step solution
01
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.
02
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.
03
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 \( x = 2 \).
04
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 \( y = -3 \).
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
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: \(3y = -9\)
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:
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.
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 \(3 \times a_{ij}\).
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:
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.