Solve Equations With Matrices: Gaussian Elimination

Hey guys! Today, we're diving into how to solve a system of equations using matrices. Specifically, we'll be using the Gaussian elimination method combined with back-substitution. This is a super useful technique in linear algebra, and it's gonna help us tackle systems of equations like the one below:

$$\left\{\begin{aligned} 3 x+4 y-3 z & =21 \\ 2 x-4 y-3 z & =40 \\ x-3 y & =14 \end{aligned}\right.$$

So, buckle up, and let's get started!

What is Gaussian Elimination?

Gaussian elimination, at its heart, is a systematic way to transform a system of linear equations into an equivalent system that's easier to solve. The goal is to get the matrix into what's called row-echelon form or, even better, reduced row-echelon form. This makes it super straightforward to read off the solutions. Essentially, you're using elementary row operations to simplify the matrix. These operations include:

1. Swapping two rows.
2. Multiplying a row by a non-zero constant.
3. Adding a multiple of one row to another row.

The beauty of Gaussian elimination lies in its ability to methodically eliminate variables, making the system of equations more manageable. By strategically applying these row operations, we can transform the original matrix into an upper triangular form, where all the elements below the main diagonal are zero. This simplified form allows us to easily solve for the variables using back-substitution.

The process begins by constructing an augmented matrix, which combines the coefficients of the variables and the constants on the right-hand side of the equations. Each row of the augmented matrix represents one equation in the system. The Gaussian elimination procedure then involves performing row operations to systematically eliminate the variables from the equations. The goal is to create a leading 1 (also known as a pivot) in each row and column, and then use row operations to eliminate all other entries in that column. This process continues until the matrix is in row-echelon form, where the leading entries form a staircase pattern.

Once the matrix is in row-echelon form, we can proceed with back-substitution to solve for the variables. Starting from the last row, we solve for the last variable in terms of the others. Then, we substitute this value into the previous row to solve for the next variable, and so on, until we have solved for all the variables in the system. This method ensures that we obtain the correct solutions to the system of equations.

Step-by-Step Solution

1. Write the Augmented Matrix

First, let's represent our system of equations as an augmented matrix:

$$\begin{bmatrix} 3 & 4 & -3 & | & 21 \\ 2 & -4 & -3 & | & 40 \\ 1 & -3 & 0 & | & 14 \end{bmatrix}$$

2. Apply Gaussian Elimination

Our goal is to get this matrix into row-echelon form. Let's start by swapping row 1 and row 3 to get a 1 in the top-left corner:

$$\begin{bmatrix} 1 & -3 & 0 & | & 14 \\ 2 & -4 & -3 & | & 40 \\ 3 & 4 & -3 & | & 21 \end{bmatrix}$$

Next, we'll eliminate the 2 in the second row and the 3 in the third row. To do this, we perform the following row operations:

• R2 = R2 - 2 * R1
• R3 = R3 - 3 * R1

This gives us:

$$\begin{bmatrix} 1 & -3 & 0 & | & 14 \\ 0 & 2 & -3 & | & 12 \\ 0 & 13 & -3 & | & -21 \end{bmatrix}$$

Now, let's divide the second row by 2 to get a 1 in the second row, second column:

$$\begin{bmatrix} 1 & -3 & 0 & | & 14 \\ 0 & 1 & -3/2 & | & 6 \\ 0 & 13 & -3 & | & -21 \end{bmatrix}$$

Next, eliminate the 13 in the third row by performing the row operation:

• R3 = R3 - 13 * R2

This results in:

$$\begin{bmatrix} 1 & -3 & 0 & | & 14 \\ 0 & 1 & -3/2 & | & 6 \\ 0 & 0 & 39/2 - 3 & | & -21 - 78 \end{bmatrix}$$

Simplify the third row:

$$\begin{bmatrix} 1 & -3 & 0 & | & 14 \\ 0 & 1 & -3/2 & | & 6 \\ 0 & 0 & 33/2 & | & -99 \end{bmatrix}$$

Now, multiply the third row by 2/33 to get a 1 in the third row, third column:

$$\begin{bmatrix} 1 & -3 & 0 & | & 14 \\ 0 & 1 & -3/2 & | & 6 \\ 0 & 0 & 1 & | & -6 \end{bmatrix}$$

3. Back-Substitution

Now that we have our matrix in row-echelon form, we can use back-substitution to solve for x, y, and z. Starting from the bottom row:

• z = -6

Next, substitute z into the second row:

• y - (3/2) * z = 6
• y - (3/2) * (-6) = 6
• y + 9 = 6
• y = -3

Finally, substitute y into the first row:

• x - 3 * y = 14
• x - 3 * (-3) = 14
• x + 9 = 14
• x = 5

4. The Solution

So, the solution to the system of equations is:

• x = 5
• y = -3
• z = -6

In other words, (x, y, z) = (5, -3, -6).

Verification

Let's quickly verify our solution by plugging these values back into the original equations:

1. 3x + 4y - 3z = 3(5) + 4(-3) - 3(-6) = 15 - 12 + 18 = 21 (Correct!)
2. 2x - 4y - 3z = 2(5) - 4(-3) - 3(-6) = 10 + 12 + 18 = 40 (Correct!)
3. x - 3y = 5 - 3(-3) = 5 + 9 = 14 (Correct!)

Our solution checks out!

Why Gaussian Elimination is Awesome

Gaussian elimination is a powerful technique for several reasons. First off, it's systematic. This means that if you follow the steps correctly, you're guaranteed to find the solution (or determine that no solution exists). It's like a recipe for solving systems of equations! Plus, it's versatile. It works for any system of linear equations, regardless of the number of equations or variables. Whether you're dealing with two equations and two unknowns or a hundred equations and a hundred unknowns, Gaussian elimination can handle it.

Moreover, Gaussian elimination is fundamental. It's a cornerstone of linear algebra, and it's used as a building block for more advanced techniques. Understanding Gaussian elimination will give you a solid foundation for tackling more complex problems in math, science, and engineering. It's also practical. Many real-world problems can be modeled as systems of linear equations, and Gaussian elimination provides a way to solve these problems efficiently. From circuit analysis to economic modeling, Gaussian elimination is a valuable tool for solving a wide range of practical problems.

Common Mistakes to Avoid

When performing Gaussian elimination, it's easy to make mistakes, especially when dealing with fractions or negative numbers. One common mistake is arithmetic errors. Be careful when performing row operations, and double-check your calculations to avoid errors. Another common mistake is incorrect row operations. Make sure you're performing the row operations correctly, and that you're not accidentally swapping rows or multiplying by the wrong constant. Additionally, some people make sign errors. Pay close attention to the signs of the numbers when performing row operations, and be careful not to make any sign errors. To avoid these mistakes, it's helpful to work neatly and keep track of your row operations. This will make it easier to spot errors and correct them. It's also a good idea to practice Gaussian elimination on a variety of problems to build your skills and confidence.

Conclusion

And there you have it! We've successfully solved a system of equations using matrices and the Gaussian elimination method with back-substitution. I hope this step-by-step guide was helpful. Keep practicing, and you'll become a pro at solving systems of equations in no time! Remember, the key is to be systematic and careful with your calculations. Happy solving!