Finding The Y Value For A System Of Equations Matrix Solution

Hey guys! Today, we're diving into a fun math problem where we need to find the $y$-value for the solution of a system of equations. But here's the twist: we have the system presented as a matrix. Don't worry, it's not as scary as it looks! We'll break it down step by step. So, let's jump right in and see how to tackle this. Our main goal here is to translate this matrix back into equations, and then use those equations to solve for the value of $y$. Remember, matrices are just a neat way of organizing equations, so we're really just doing some good old algebra with a fancy presentation.

Understanding the Matrix Representation

First things first, let's understand what this matrix is telling us. The matrix we're given is:

$$\left[\begin{array}{cccc} 2 & 3 & 1 & 3 \\ 0 & 1 & 3 & -11 \\ 3 & 0 & 2 & -2 \end{array}\right]$$

This matrix represents a system of three linear equations with three variables. Usually, we call these variables $x$, $y$, and $z$. The first three columns of the matrix correspond to the coefficients of $x$, $y$, and $z$, respectively, and the last column represents the constants on the right side of the equations. So, let's translate this matrix into its equation form. The first row represents the first equation:

$\qquad 2x + 3y + 1z = 3$

The second row gives us the second equation:

$\qquad 0x + 1y + 3z = -11$

Which simplifies to:

$\qquad y + 3z = -11$

And the third row corresponds to the third equation:

$\qquad 3x + 0y + 2z = -2$

Which simplifies to:

$\qquad 3x + 2z = -2$

Now we have a system of three equations:

1. $2x + 3y + z = 3$
2. $y + 3z = -11$
3. $3x + 2z = -2$

Solving the System of Equations

Okay, so now we have our system of equations. Our mission is to find the value of $y$, so we need to use some algebraic techniques to solve this system. There are a few ways we can do this, such as substitution, elimination, or using matrices (again!). Let's use a combination of substitution and elimination, as it’s pretty straightforward for this setup. We already have equation (2) nicely set up to help us find $y$ if we know $z$. So, let’s focus on eliminating variables to get there.

Step 1: Isolate $y$ in Equation 2

Equation 2 is already quite simple:

$\qquad y + 3z = -11$

We can express $y$ in terms of $z$:

$\qquad y = -11 - 3z$

This is super helpful because now we have an expression for $y$ that we can plug into other equations if needed. But remember, we still need to find the value of $z$ first!

Step 2: Eliminate $x$ from Equations 1 and 3

Let's look at equations 1 and 3. They both have $x$ and $z$, which is great because if we can eliminate $x$, we'll have an equation with just $z$, which we can solve. Here are the equations again:

1. $2x + 3y + z = 3$
2. $3x + 2z = -2$

To eliminate $x$, we can multiply the first equation by 3 and the third equation by -2. This will give us $6x$ and $-6x$, which will cancel each other out when we add the equations. So, let's do it:

Multiply equation 1 by 3:

$\qquad 3(2x + 3y + z) = 3(3)$ $\qquad 6x + 9y + 3z = 9$

Multiply equation 3 by -2:

$\qquad -2(3x + 2z) = -2(-2)$ $\qquad -6x - 4z = 4$

Now, let’s add these two new equations together:

$\qquad (6x + 9y + 3z) + (-6x - 4z) = 9 + 4$ $\qquad 9y - z = 13$

Step 3: Substitute $y$ in the New Equation

We have a new equation: $9y - z = 13$. We also have an expression for $y$ in terms of $z$ from Step 1: $y = -11 - 3z$. Let’s substitute this into our new equation:

$\qquad 9(-11 - 3z) - z = 13$ $\qquad -99 - 27z - z = 13$ $\qquad -99 - 28z = 13$

Step 4: Solve for $z$

Now we have a simple equation with just $z$. Let's solve for it:

$\qquad -28z = 13 + 99$ $\qquad -28z = 112$ $\qquad z = \frac{112}{-28}$ $\qquad z = -4$

Alright! We found the value of $z$! Now we can use this to find the value of $y$.

Step 5: Solve for $y$

Remember our expression for $y$ from Step 1? We have $y = -11 - 3z$. Now that we know $z = -4$, we can plug it in:

$\qquad y = -11 - 3(-4)$ $\qquad y = -11 + 12$ $\qquad y = 1$

Final Answer

We did it! We found the $y$-value for the solution to the system of equations. The value of $y$ is 1. This whole process might seem long, but each step is pretty straightforward. It's all about breaking down the problem and tackling it piece by piece. Remember, the key is to translate the matrix into equations, and then use your algebra skills to solve for the variable you need. You guys rock! Keep practicing, and you'll become matrix-solving masters in no time!

In summary, to find the $y$-value for the solution to the system of equations represented by the matrix, we first translated the matrix into a system of linear equations. Then, we used substitution and elimination methods to solve for $z$, and finally, we substituted the value of $z$ back into the equation to find the value of $y$. The final answer is that the $y$-value is 1.

This problem highlights the importance of understanding how matrices represent systems of equations and how to use algebraic techniques to solve them. It's a fundamental concept in linear algebra and has wide applications in various fields, including engineering, computer science, and economics.

So, next time you see a matrix, don't be intimidated! Remember that it's just a way of organizing equations, and you have the tools to solve them. Keep up the great work, guys, and happy solving! We've really nailed how to extract the $y$-value from a system presented in matrix form. It's all about translating, simplifying, and solving. You've got this!