Finding Coordinates D(m, N) By Solving A Matrix Equation

In the realm of linear algebra, matrices and vectors play a crucial role in representing and solving systems of equations. This article delves into a fascinating problem involving matrix multiplication and vector equality. We are given a matrix equation: $\left(\begin{array}{cc}m & 3 \\ 1 & -2\end{array}\right)\binom{2}{-n}=\binom{m-2}{n}$, and our mission is to determine the coordinates of the point $D(m, n)$. This exploration will not only reinforce your understanding of matrix operations but also showcase how these concepts can be applied to pinpoint specific points in a coordinate plane. Let's embark on this mathematical journey together!

Unveiling the Matrix Equation

At the heart of our problem lies the matrix equation:

$\left(\begin{array}{cc}m & 3 \\ 1 & -2\end{array}\right)\binom{2}{-n}=\binom{m-2}{n}$

To decipher this equation, we need to perform matrix multiplication on the left-hand side. Recall that matrix multiplication involves multiplying corresponding elements of rows and columns and then summing the results. In this case, we have a 2x2 matrix multiplying a 2x1 vector, which will result in a 2x1 vector. Let's perform the multiplication step by step.

Matrix multiplication is a fundamental operation in linear algebra, and understanding it is crucial for solving various problems involving systems of equations, transformations, and more. When multiplying a matrix by a vector, we essentially transform the vector into a new vector based on the rules defined by the matrix. In our equation, the matrix $\left(\begin{array}{cc}m & 3 \\ 1 & -2\end{array}\right)$ acts as a transformation on the vector $\binom{2}{-n}$. To fully grasp the mechanics of this transformation, let's break down the process.

First, consider the first row of the matrix, which is $\begin{bmatrix} m & 3 \end{bmatrix}$. We multiply each element of this row by the corresponding element in the vector $\binom{2}{-n}$ and then sum the results. This gives us $m * 2 + 3 * (-n)$, which simplifies to $2m - 3n$. This value becomes the first element of the resulting vector. Next, we move on to the second row of the matrix, which is $\begin{bmatrix} 1 & -2 \end{bmatrix}$. We repeat the same process, multiplying each element of this row by the corresponding element in the vector $\binom{2}{-n}$ and summing the results. This yields $1 * 2 + (-2) * (-n)$, which simplifies to $2 + 2n$. This value becomes the second element of the resulting vector. Therefore, the result of the matrix multiplication is the vector $\binom{2m-3n}{2+2n}$. Now, we can rewrite the original matrix equation with this result, setting the stage for solving for the unknowns m and n. By equating the corresponding elements of the vectors, we create a system of linear equations that we can solve using various techniques, such as substitution or elimination.

After performing the matrix multiplication, we get:

$\binom{2m - 3n}{2 + 2n} = \binom{m - 2}{n}$

Now, for two vectors to be equal, their corresponding components must be equal. This gives us two equations:

1. $2m - 3n = m - 2$
2. $2 + 2n = n$

Solving the System of Equations

We now have a system of two linear equations with two unknowns, m and n. Let's solve this system to find the values of m and n. We can start by simplifying the equations.

The first equation, $2m - 3n = m - 2$, can be simplified by subtracting m from both sides, resulting in $m - 3n = -2$. This rearranged equation provides a clearer relationship between m and n. The second equation, $2 + 2n = n$, can be simplified by subtracting 2n from both sides, which yields $2 = -n$. This immediately gives us the value of n as -2. Having found the value of n, we can now substitute it back into the first equation to solve for m. This substitution allows us to transform the first equation into an equation with only one unknown, making it straightforward to solve. By substituting n = -2 into $m - 3n = -2$, we get $m - 3(-2) = -2$, which simplifies to $m + 6 = -2$. Subtracting 6 from both sides, we find that m = -8. Therefore, we have successfully determined the values of m and n by solving the system of linear equations. The values of m and n are crucial for finding the coordinates of point D, which is the ultimate goal of the problem.

Let's start with the second equation, $2 + 2n = n$. Subtracting 2n from both sides, we get:

$2 = -n$

Multiplying both sides by -1, we find:

$n = -2$

Now that we have the value of n, we can substitute it into the first equation, $2m - 3n = m - 2$:

$2m - 3(-2) = m - 2$

Simplifying, we get:

$2m + 6 = m - 2$

Subtracting m from both sides:

$m + 6 = -2$

Subtracting 6 from both sides:

$m = -8$

Thus, we have found that $m = -8$ and $n = -2$.

Locating Point D(m, n)

We have successfully determined the values of m and n to be -8 and -2, respectively. Now, we can find the coordinates of point $D(m, n)$. Since $m = -8$ and $n = -2$, the coordinates of point D are $(-8, -2)$.

Point D's location in the coordinate plane is now precisely determined. The coordinates (-8, -2) signify that the point is situated 8 units to the left of the origin along the x-axis and 2 units below the origin along the y-axis. Visualizing this point on a graph helps solidify our understanding of its position relative to the axes and the origin. In the broader context of the problem, the coordinates of point D are the solution we sought after decoding the matrix equation and solving the resulting system of linear equations. This process highlights the power of linear algebra in representing and solving geometric problems. By translating the matrix equation into a system of algebraic equations, we were able to leverage algebraic techniques to find the numerical values of m and n, which in turn gave us the coordinates of point D. The connection between linear algebra and geometry is a fundamental concept in mathematics, and this problem serves as a compelling illustration of this connection.

Therefore, the point $D(m, n)$ is located at $(-8, -2)$.

Conclusion

In this article, we successfully navigated a problem involving matrix multiplication and vector equality. By performing matrix multiplication, setting up a system of linear equations, and solving for the unknowns, we determined the coordinates of the point $D(m, n)$ to be $(-8, -2)$. This exercise demonstrates the power of linear algebra in solving geometric problems and reinforces our understanding of matrix operations and vector equality. Understanding these concepts is crucial for tackling more complex problems in mathematics, physics, and computer science.

This problem exemplifies how seemingly abstract mathematical concepts like matrices and vectors can have concrete applications in geometry. The process of translating a matrix equation into a geometric representation, specifically the coordinates of a point, showcases the interconnectedness of different branches of mathematics. Moreover, the techniques we employed to solve this problem, such as matrix multiplication and solving systems of linear equations, are widely applicable in various fields beyond mathematics. For instance, in computer graphics, matrices are used extensively for transformations like rotations, scaling, and translations, while in physics, vectors are used to represent forces and velocities. Therefore, mastering these fundamental concepts not only enhances one's mathematical abilities but also opens doors to a wide range of opportunities in STEM fields.

Furthermore, the problem-solving approach we adopted can be generalized to a variety of similar problems involving matrix equations and vector equality. The key steps include performing the matrix operations, setting up a system of equations by equating corresponding components, and then solving the system using algebraic techniques. By following this methodical approach, one can effectively tackle a wide range of problems in linear algebra and related fields. In conclusion, this article has provided a comprehensive solution to the given matrix equation problem, highlighting the importance of linear algebra concepts and their applications in diverse fields. The journey from the initial matrix equation to the final coordinates of point D underscores the beauty and power of mathematics in unraveling complex problems.