Row Operation: Transforming A Matrix With (1/5)R1

Hey guys! Today, we're diving into the world of matrices and row operations. Specifically, we're going to tackle a problem where we need to transform a matrix by replacing one of its rows using scalar multiplication. It might sound a bit technical, but trust me, it's a fundamental concept in linear algebra and super useful for solving systems of equations and other cool stuff.

Understanding Row Operations

Before we jump into the problem, let's quickly recap what row operations are. In essence, they're a set of rules we can apply to a matrix that allows us to manipulate it without changing its underlying solution. Think of it like rearranging equations in a system – you're changing the way it looks, but the answers remain the same.

There are three main types of row operations:

1. Swapping rows: You can interchange any two rows in the matrix.
2. Multiplying a row by a scalar: You can multiply all the elements in a row by a non-zero constant.
3. Adding a multiple of one row to another: You can add a scalar multiple of one row to another row.

These operations are the bread and butter of solving linear systems using methods like Gaussian elimination and Gauss-Jordan elimination. They allow us to systematically simplify the matrix until we can easily read off the solution. Now, let's focus on the specific operation we're dealing with today: multiplying a row by a scalar.

When we talk about multiplying a row by a scalar, what we mean is that we're taking every single element in that row and multiplying it by a constant number. This might seem simple, but it's a powerful tool for scaling the entries in our matrix to make them easier to work with. For example, if we have a row with large numbers, multiplying by a fraction can reduce them to smaller, more manageable values. Or, if we want to create a leading 1 in a row (which is often a key step in solving systems of equations), we can multiply the entire row by the reciprocal of the current leading entry. This is precisely what we're doing in the given problem, where we're replacing R1 (row 1) by (1/5)R1. This means we'll be taking each element in the first row and multiplying it by 1/5, effectively dividing each of those elements by 5. This operation is crucial for simplifying the matrix and moving towards a solution. Remember, the goal here is to manipulate the matrix in a way that makes it easier to solve the system of equations it represents. By strategically using scalar multiplication, we can transform the matrix into a form where the solutions are readily apparent. So, let's get our hands dirty and apply this operation to the given matrix!

The Problem: Scaling Row 1

Okay, let's get to the nitty-gritty. We're given the following matrix:

[ 5 -40 15 20
 -7 0 -5 1
 2 7 4 0 ]

And our mission, should we choose to accept it (we do!), is to replace the first row ($R_1$) with rac{1}{5} R_1. This means we're going to multiply each element in the first row by rac{1}{5}.

Let's break it down step by step:

• First element: 5 * rac{1}{5} = 1
• Second element: -40 * rac{1}{5} = -8
• Third element: 15 * rac{1}{5} = 3
• Fourth element: 20 * rac{1}{5} = 4

So, the new first row will be [1 -8 3 4]. The other rows remain unchanged. Now, let's put it all together to get the transformed matrix.

The Transformed Matrix

After applying the row operation, the transformed matrix looks like this:

[ 1 -8 3 4
 -7 0 -5 1
 2 7 4 0 ]

See? It wasn't so bad! We successfully scaled the first row by a factor of rac{1}{5}. This is a crucial step in many matrix operations, and now you've got a handle on it.

Applying the row operation of multiplying a row by a scalar is a fundamental technique in linear algebra, and this example demonstrates how it works in practice. By multiplying the first row by 1/5, we've simplified the entries in that row, which can make subsequent operations easier. This is a common strategy when solving systems of equations or performing other matrix manipulations. For instance, this step might be part of a larger process like Gaussian elimination, where the goal is to transform the matrix into row-echelon form. In row-echelon form, the leading coefficient (the first non-zero number from the left) in each row is 1, and each leading coefficient is in a column to the right of the leading coefficient of the row above it. This form makes it straightforward to solve for the variables in the corresponding system of equations. So, while this specific operation might seem isolated, it's actually a building block for more complex procedures. Understanding how to perform scalar multiplication on rows is essential for mastering these techniques and becoming proficient in linear algebra. It's also a great way to improve your numerical skills and attention to detail, as even a small mistake in the calculation can throw off the entire result. So, practice makes perfect! Keep working with these operations, and you'll soon become a matrix transformation master.

Why This Matters

You might be wondering,