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Solve for \(x, y, z\), and \(w\). $$ \left[\begin{array}{cc} x+y & x+z \\ y+z & w \end{array}\right]=\left[\begin{array}{ll} 3 & 4 \\ 5 & 4 \end{array}\right] $$

Short Answer

Expert verified
The short version of the answer is: \(x=1, \; y=2, \; z=3, \; w=4\).

Step by step solution

01

Equate Corresponding Elements of Both Matrices

Equate the elements of both matrices. This will give us a system of four linear equations. $$ \begin{cases} x+y = 3 \\ x+z = 4 \\ y+z = 5 \\ w = 4 \end{cases} $$
02

Extract the Value of \(w\)

From our system of equations, we can already see the value of \(w\) directly. Therefore, we extract that value and keep it aside: $$ w = 4 $$
03

Solve the rest of the system to find \(x, y,\) and \(z\)

Now that we have found the value of \(w\), let's solve the system of equations for the remaining variables: $$ \begin{cases} x+y = 3 \\ x+z = 4 \\ y+z = 5 \\ \end{cases} $$ First, solve for \(x\) in terms of \(y\) or \(z\). We can do that by subtracting the first equation from the second equation: $$ (x+z)-(x+y) = 4-3 \Rightarrow z-y = 1 \Rightarrow z = y+1 \hspace{1cm}(1) $$ Substitute equation (1) into the third equation to solve for \(y\): $$ y+(y+1) = 5 \Rightarrow 2y+1=5 \Rightarrow 2y=4 \Rightarrow y=2 \hspace{1cm}(2) $$ Now substitute equation (2) into equation (1) to solve for \(z\): $$ z = 2+1 \Rightarrow z = 3 \hspace{1cm}(3) $$ Lastly, substitute equation (2) into the first equation to solve for \(x\): $$ x+2=3 \Rightarrow x=1 \hspace{1cm}(4) $$
04

Write the solution for \(x, y, z,\) and \(w\)

Now that we have found the values for all our variables, let's write it down: $$ x=1, \; y=2, \; z=3, \; w=4 $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

System of Linear Equations
Understanding the system of linear equations is crucial when solving mathematical problems involving multiple unknown variables. A system is simply a collection of two or more equations with a set of variables. In our exercise, we're dealing with a system that consists of equations related to the variables x, y, z, and w. The main goal in such systems is to find a solution that satisfies all equations simultaneously, which often requires the use of specific methods to manipulate and reduce the complexity of the equations.

When working with matrices, these systems can be visually represented and solutions can be methodically approached, improving clarity and reducing the likelihood of errors. A key aspect of solving these systems is keeping the equations balanced; whatever operation you perform on one side of the equation must be applied to the other side as well to maintain equality. This concept forms the foundation for various algebraic methods used in solving linear equations.
Matrix Equality
Matrix equality is a concept that allows us to equate two matrices of the same dimensions, setting corresponding elements equal to each other. In the presented exercise, we are comparing a matrix comprised of variable expressions with a constant 2x2 matrix. The principle of matrix equality tells us that for the two matrices to be equal, each corresponding element must be identical. This leads us to form a system of linear equations. Each equation originates from the equality of a specific position in the matrices, providing a structured manner to approach the problem.

Matrix equality is a simple yet powerful tool that simplifies complex problems. It is essentially the 'translation' step in taking a matrix and turning it into a system of more manageable linear equations. By efficiently utilizing matrix equality, we avoid confusion that can arise from less organized methods when dealing with systems involving multiple unknowns.
Algebraic Solutions to Matrices
Solving matrix equations algebraically involves several steps, often including substitution or elimination processes. Algebraic solutions to matrix equations can be derived by converting the equations into an equivalent system where standard algebraic techniques are used. The exercise showcases this by transforming matrix equality into a system of equations.

Once we have the system, techniques such as addition, subtraction, substitution, and even multiplication by scalars are used to isolate variables and find solutions. These methods may seem familiar as they are the same approaches used in solving traditional algebraic equations. They are applied here in a systematic manner to deduce the values of the variables within the context of the original matrix problem. Algebra provides a robust set of tools that equip us to handle these tasks efficiently, ensuring consistency and accuracy in the results.
Substitution Method
The substitution method is one of the fundamental techniques for solving systems of linear equations. It entails solving one equation for a single variable and then substituting that solution into the other equations. This effectively reduces the number of equations and variables, simplifying the system.

The exercise illustrates this beautifully. After deducing that w = 4 directly, we can use substitution to find the values of x, y, and z. By expressing z in terms of y, and then substituting this expression into another equation, we start a chain reaction that allows us to solve for each variable in turn. The essence of this method is to break down complex, multi-variable problems into simpler, single-variable equations which can be solved with basic algebra. Substitution is not just a mechanical process, but rather a strategic move in the problem-solving challenge.

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Most popular questions from this chapter

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