Find Matrix B: A Step-by-Step Solution

Aug 4, 2025 by Sebastian Müller 39 views

Hey everyone! Let's dive into a fun matrix problem today. We're given matrix A and the result of A multiplied by another matrix B (AB). Our mission, should we choose to accept it, is to find out what matrix B is. Don't worry, it's not as daunting as it sounds! We'll break it down step by step, making it super easy to follow. So, grab your thinking caps, and let's get started!

The Problem

Okay, first things first, let's lay out the problem. We've got this matrix A:

A = 
\begin{bmatrix}
2 & 4 & -2 \\
4 & -5 & 7 \\
2 & 7 & 5
\end{bmatrix}

And we know that when we multiply A by some other matrix B, we get:

AB = 
\begin{bmatrix}
24 \\
-46 \\
-2
\end{bmatrix}

Our goal is to figure out what matrix B is. Sounds like a puzzle, right? Well, it is, but a very solvable one! We're going to use our knowledge of matrix multiplication and a little bit of algebra to crack this code. So, let's jump into the nitty-gritty and see how we can find this mystery matrix B.

Understanding Matrix Multiplication

Before we jump into solving for matrix B, let's quickly recap matrix multiplication. This is super important because it's the foundation of how we're going to solve this problem. Remember, you can't just multiply matrices willy-nilly; there are rules! The main rule is that for two matrices to be multiplied, the number of columns in the first matrix must be equal to the number of rows in the second matrix. This might sound like a mouthful, but it's pretty straightforward once you see it in action.

In our case, matrix A is a 3x3 matrix (3 rows and 3 columns). The result of AB is a 3x1 matrix (3 rows and 1 column). This tells us something crucial about matrix B: it must be a 3x1 matrix. Why? Because the 3 columns of A have to match the number of rows in B, and the resulting matrix has 1 column, which means B itself can only have one column. This is our first big clue! So, we now know that B looks something like this:

B = 
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}

Where x, y, and z are the unknown values we're trying to find. Now that we've got this figured out, we can set up some equations based on the matrix multiplication process. Understanding this fundamental aspect of matrix multiplication is key, guys! It's what allows us to move forward and actually solve for those unknowns. So, let's keep this in mind as we move on to the next step, where we'll start building our equations.

Setting up the Equations

Alright, now for the fun part: setting up the equations! This is where we take our understanding of matrix multiplication and turn it into a system of equations that we can actually solve. Remember how matrix multiplication works? We multiply the rows of the first matrix by the columns of the second matrix. In our case, we're multiplying matrix A by matrix B, and we know the result. This gives us a set of equations.

Let's break it down. When we multiply the first row of A by matrix B, we get our first equation:

(2 * x) + (4 * y) + (-2 * z) = 24

See how that works? We're taking each element in the first row of A, multiplying it by the corresponding element in B, and then adding them all up. This gives us the first element in the resulting matrix AB, which is 24. We can do the same thing for the second and third rows of A. This gives us two more equations:

(4 * x) + (-5 * y) + (7 * z) = -46
(2 * x) + (7 * y) + (5 * z) = -2

Now we have a system of three equations with three unknowns (x, y, and z). This is awesome because we have the tools to solve this! We can use methods like substitution, elimination, or even matrix inversion to find the values of x, y, and z. This is where our algebra skills come into play. Setting up these equations correctly is super crucial, guys. It's the foundation upon which we'll build our solution. So, let's take a deep breath, make sure we've got everything right, and then move on to solving this system of equations!

Solving the System of Equations

Okay, we've arrived at the heart of the problem: solving the system of equations. We've got three equations and three unknowns, which means we can use a variety of methods to find our solution. Let's rewrite those equations here, just to keep them fresh in our minds:

2x + 4y - 2z = 24
4x - 5y + 7z = -46
2x + 7y + 5z = -2

There are several ways to tackle this. We could use substitution, where we solve for one variable in terms of the others and then substitute that into the other equations. We could also use elimination, where we add or subtract multiples of the equations to eliminate variables. For this example, let's use elimination, as it can be quite efficient for this type of system. Our goal is to strategically combine these equations to knock out variables one by one until we can isolate the value of each unknown. This might sound a bit like a puzzle, but trust me, it's a very methodical process.

First, notice that the 'x' terms in the first and third equations have the same coefficient (2). This makes it super easy to eliminate 'x'! We can subtract the first equation from the third equation. Doing so, we get:

(2x + 7y + 5z) - (2x + 4y - 2z) = -2 - 24

Simplifying this, we get:

3y + 7z = -26

This is our new equation number 4. We've successfully eliminated 'x' and now have an equation with just 'y' and 'z'. Awesome! Now, let's eliminate 'x' again, but this time using the first and second equations. To do this, we can multiply the first equation by 2 and then subtract it from the second equation:

(4x - 5y + 7z) - 2(2x + 4y - 2z) = -46 - 2(24)

Simplifying, we get:

-13y + 11z = -94

This is our new equation number 5. Now we have two equations (4 and 5) with two unknowns (y and z). We're getting closer! We can now solve this smaller system for 'y' and 'z'. To do this, we can multiply equation 4 by 13 and equation 5 by 3. This will allow us to eliminate 'y'.

Multiplying equation 4 by 13, we get:

39y + 91z = -338

Multiplying equation 5 by 3, we get:

-39y + 33z = -282

Now, we can add these two equations together:

(39y + 91z) + (-39y + 33z) = -338 + (-282)

Simplifying, we get:

124z = -620

Dividing both sides by 124, we finally find:

z = -5

Yes! We've found the value of 'z'! Now that we know 'z', we can substitute it back into equation 4 to find 'y':

3y + 7(-5) = -26
3y - 35 = -26
3y = 9
y = 3

Fantastic! We've found the value of 'y' as well. Now, with 'y' and 'z' in hand, we can substitute them back into any of our original equations to find 'x'. Let's use the first equation:

2x + 4(3) - 2(-5) = 24
2x + 12 + 10 = 24
2x + 22 = 24
2x = 2
x = 1

Boom! We've done it! We've found the values of x, y, and z. Now we know what matrix B is.

The Solution: Matrix B

Alright, after all that equation-solving fun, we've finally arrived at the solution! We've found the values of x, y, and z, which means we can now construct matrix B. Remember, matrix B looked like this:

B = 
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}

And we found that x = 1, y = 3, and z = -5. So, we can plug these values into our matrix:

B = 
\begin{bmatrix}
1 \\
3 \\
-5
\end{bmatrix}

There you have it! This is the matrix B that, when multiplied by matrix A, gives us the result we were given in the problem. We cracked the code! It's always super satisfying to solve a problem like this, especially when it involves a bit of algebra and matrix multiplication. So, give yourself a pat on the back for sticking with it and following along. We've successfully navigated this matrix maze, guys! This is how math can be engaging and interesting, and even useful in many applications.

Conclusion

So, guys, we've successfully navigated the world of matrices and solved for matrix B! We started with the basics of matrix multiplication, set up a system of equations, and then used elimination to find our unknowns. It was a journey, but we made it! Remember, the key to solving these kinds of problems is to break them down into smaller, manageable steps. Don't get overwhelmed by the big picture; focus on one step at a time, and you'll get there.

Whether you're a student tackling linear algebra or just someone who enjoys a good mathematical puzzle, I hope this walkthrough has been helpful. Matrix problems can seem intimidating at first, but with a little bit of practice and the right approach, they become much more approachable. And who knows, maybe this will even spark your interest in exploring more advanced mathematical concepts! Keep practicing, keep exploring, and most importantly, keep enjoying the process of learning. Until next time, happy problem-solving!