Problem 71 - Intermediate Algebra

Question

Solve each system. $$ \left\\{\begin{array}{l} w+x+y+z=3 \\ w-x+y+z=1 \\ w+x-y+z=1 \\ w+x+y-z=3 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(w = 1\), \(x = 1\), \(y = 1\), \(z = 0\).

Step 1: Set Up the Matrix

Write the system of equations in matrix form. The system is:\(\begin{bmatrix} 1 & 1 & 1 & 1 \1 & -1 & 1 & 1 \1 & 1 & -1 & 1 \1 & 1 & 1 & -1 \end{bmatrix} \times \begin{bmatrix} w \ x \ y \ z \end{bmatrix} = \begin{bmatrix} 3 \ 1 \ 1 \ 3 \end{bmatrix}\).

Step 2: Add and Subtract Equations

Subtract the second equation from the first: \((w + x + y + z) - (w - x + y + z) = 3 - 1\),resulting in: \(2x = 2\), so \(x = 1\).Subtract the third equation from the first:\((w + x + y + z) - (w + x - y + z) = 3 - 1\),resulting in: \(2y = 2\), so \(y = 1\).Subtract the fourth equation from the first:\((w + x + y + z) - (w + x + y - z) = 3 - 3\),resulting in: \(2z = 0\), so \(z = 0\).

Step 3: Substitute Back to Find w

Now substitute the found values of \(x\), \(y\), and \(z\) into any equation to solve for \(w\). Using the first equation:\(w + 1 + 1 + 0 = 3\),\(w + 2 = 3\), \(w = 1\).

Step 4: Verify the Solution

Insert \(w = 1\), \(x = 1\), \(y = 1\), and \(z = 0\) back into all equations to ensure they hold true:1. \(1 + 1 + 1 + 0 = 3\) ✔️2. \(1 - 1 + 1 + 0 = 1\) ✔️3. \(1 + 1 - 1 + 0 = 1\) ✔️4. \(1 + 1 + 1 - 0 = 3\) ✔️All equations are satisfied.

Key Concepts

Matrix Form

To solve a system of equations, placing them into matrix form is highly beneficial. This approach allows for a more organized method of dealing with multiple equations. Here's a brief explanation:

- A system of equations consists of several linear equations that we aim to solve simultaneously.
- The matrix form involves representing these equations as a matrix equation, where a single matrix contains the coefficients, another matrix contains the variables, and a third contains the constants from the right-hand side of the equations.

In our problem, the input matrix, representing coefficients of variables, is fashioned like so:\[ \begin{bmatrix} 1 & 1 & 1 & 1 \ 1 & -1 & 1 & 1 \ 1 & 1 & -1 & 1 \ 1 & 1 & 1 & -1 \end{bmatrix} \]This matrix will multiply with the variable matrix \[ \begin{bmatrix} w \ x \ y \ z \end{bmatrix} \]and produce a result matrix \[ \begin{bmatrix} 3 \ 1 \ 1 \ 3 \end{bmatrix} \].

With this, we cleanly encapsulate the system into a format that's ready for various matrix-solving procedures, such as Gaussian elimination or the use of matrix inverses. This neat structural setup lays the groundwork for efficiently finding solutions.

Substitution Method

Once you've leveraged the matrix form, the next step frequently involves using methods like substitution. The substitution method is a trustworthy strategy for solving systems of equations, primarily when they come in a manageable size. Here's how it works:

- The substitution method entails resolving one equation for one variable and then substituting that solution into the other equations.
- The goal is to gradually reduce the system until each variable is isolated.

In the given exercise, we isolated each variable through subtraction simplicity:

• By subtracting the second equation from the first and finding \(x = 1\).
• Repeating a similar subtraction for the third, isolating \(y = 1\).
• Finally, subtracting the fourth finds \(z = 0\).

We peel away the layers until only the first variable remains. By plugging \(x\), \(y\), and \(z\) into any equation (using the first here), finding \(w = 1\) became immediate. Substitution thus streamlines the solution process by strategically isolating variables.

Solution Verification

Once potential solutions are unearthed, the final task is to ensure they fit the system perfectly. This is where solution verification steps in. Verifying the solution is the bumper at the end of problem-solving to ensure accuracy.

What does verification involve?

• It requires substituting the determined values of all variables back into the original equations.
• Each equation in the system should be checked to ensure that both sides of every equation are equal upon substituting the found variables.

For our exercise, upon substituting \(w = 1\), \(x = 1\), \(y = 1\), and \(z = 0\), each equation simplified to hold true:- First: \(1 + 1 + 1 + 0 = 3\)- Second: \(1 - 1 + 1 + 0 = 1\)- Third: \(1 + 1 - 1 + 0 = 1\)- Fourth: \(1 + 1 + 1 - 0 = 3\)

All equations checked out, confirming the solution's validity. Verification is a vital step, ensuring no computational errors were made during solving.