Solving For Matrix A A Step By Step Guide

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When it comes to solving for matrices in linear algebra, understanding the fundamental operations is key. In this article, we will break down the steps required to solve for a matrix, using a specific example to illustrate the process. Our focus will be on how to isolate the unknown matrix by applying basic matrix arithmetic. We'll cover the initial problem setup, the methodology for isolating the matrix, and the step-by-step calculations involved. Whether you're a student tackling linear algebra or a professional refreshing your skills, this guide will provide a clear and concise approach to solving for matrices. Let's dive in and explore the world of matrix algebra!

Understanding the Problem

In this problem, we are given an equation involving matrix A and asked to solve for A. The equation is:

A+[3−4−1124]=[16−50−9]A+\begin{bmatrix} 3 & -4 & -11 & 24 \end{bmatrix} = \begin{bmatrix} 16 & -5 & 0 & -9 \end{bmatrix}

The goal here is to isolate matrix A on one side of the equation. This involves using the principles of matrix addition and subtraction, which are fundamental operations in linear algebra. Before we dive into the solution, it's crucial to understand what matrices are and how they behave under these operations.

A matrix, in its simplest form, is a rectangular array of numbers arranged in rows and columns. The dimensions of a matrix are defined by the number of rows and the number of columns it contains. For example, a matrix with mm rows and nn columns is referred to as an mimesnm imes n matrix. The matrices in our given equation are both 1imes41 imes 4 matrices, meaning they have one row and four columns. This is important because matrix addition and subtraction are only defined for matrices of the same dimensions.

Matrix addition and subtraction are performed element-wise. This means that you add or subtract the corresponding elements in the matrices. For example, if you have two matrices, B and C, both of size mimesnm imes n, their sum D = B + C is obtained by adding the elements in the same positions in B and C. Mathematically, this can be expressed as:

Dij=Bij+CijD_{ij} = B_{ij} + C_{ij}

Where DijD_{ij} represents the element in the ii-th row and jj-th column of matrix D, and similarly for BijB_{ij} and CijC_{ij}. The same principle applies to subtraction, where you subtract the corresponding elements.

Now that we understand the basics of matrix addition and subtraction, we can approach the problem of isolating matrix A. The key idea is to use the additive inverse property, which states that for any matrix B, there exists a matrix -B such that B + (-B) = 0, where 0 represents the zero matrix (a matrix with all elements equal to zero). In our case, we will subtract the matrix [3−4−1124]\begin{bmatrix} 3 & -4 & -11 & 24 \end{bmatrix} from both sides of the equation to isolate A. This process will become clearer as we walk through the steps in the next section.

Isolating Matrix A

The primary objective in solving for matrix A is to isolate it on one side of the equation. This is akin to solving for a variable in a regular algebraic equation. The approach we'll use here is to subtract the matrix [3−4−1124]\begin{bmatrix} 3 & -4 & -11 & 24 \end{bmatrix} from both sides of the given equation. This maintains the equality while moving the known matrix away from A. Let's break down the steps involved:

Starting with the original equation:

A+[3−4−1124]=[16−50−9]A + \begin{bmatrix} 3 & -4 & -11 & 24 \end{bmatrix} = \begin{bmatrix} 16 & -5 & 0 & -9 \end{bmatrix}

To isolate A, we subtract the matrix [3−4−1124]\begin{bmatrix} 3 & -4 & -11 & 24 \end{bmatrix} from both sides of the equation. This is a valid operation because subtracting the same matrix from both sides does not change the equality. The equation now looks like this:

A+[3−4−1124]−[3−4−1124]=[16−50−9]−[3−4−1124]A + \begin{bmatrix} 3 & -4 & -11 & 24 \end{bmatrix} - \begin{bmatrix} 3 & -4 & -11 & 24 \end{bmatrix} = \begin{bmatrix} 16 & -5 & 0 & -9 \end{bmatrix} - \begin{bmatrix} 3 & -4 & -11 & 24 \end{bmatrix}

On the left side of the equation, the matrix [3−4−1124]\begin{bmatrix} 3 & -4 & -11 & 24 \end{bmatrix} and its negative cancel each other out, leaving us with just A. This is because any matrix added to its negative results in the zero matrix, which does not affect the value of A. So, the equation simplifies to:

A=[16−50−9]−[3−4−1124]A = \begin{bmatrix} 16 & -5 & 0 & -9 \end{bmatrix} - \begin{bmatrix} 3 & -4 & -11 & 24 \end{bmatrix}

Now that A is isolated, we need to perform the subtraction on the right side of the equation to find the values of the elements in matrix A. This involves subtracting the corresponding elements in the two matrices. The next section will detail the calculations involved in this subtraction, providing a clear understanding of how each element in matrix A is derived. This step-by-step approach ensures that the solution is both accurate and easy to follow.

Performing the Subtraction

With matrix A now isolated on one side of the equation, the next crucial step in solving for A is to perform the subtraction of the matrices on the right side. This involves subtracting the corresponding elements of the two matrices. Let's walk through this process methodically to ensure accuracy.

We have the equation:

A=[16−50−9]−[3−4−1124]A = \begin{bmatrix} 16 & -5 & 0 & -9 \end{bmatrix} - \begin{bmatrix} 3 & -4 & -11 & 24 \end{bmatrix}

To perform the subtraction, we subtract the corresponding elements in each matrix. This means we subtract the first element of the second matrix from the first element of the first matrix, the second element from the second, and so on. The calculations are as follows:

  • For the first element: 16−3=1316 - 3 = 13
  • For the second element: −5−(−4)=−5+4=−1-5 - (-4) = -5 + 4 = -1
  • For the third element: 0−(−11)=0+11=110 - (-11) = 0 + 11 = 11
  • For the fourth element: −9−24=−33-9 - 24 = -33

Now, we assemble these results into a new matrix, which will be our solution for matrix A. The elements we calculated become the elements of matrix A:

A=[13−111−33]A = \begin{bmatrix} 13 & -1 & 11 & -33 \end{bmatrix}

This resulting matrix is the solution to the original equation. Each element in A is the result of the subtraction performed on the corresponding elements of the matrices on the right side of the equation. This step-by-step calculation ensures that we arrive at the correct solution. In the next section, we will summarize the final solution and briefly discuss how you can verify your answer to ensure its correctness. Understanding each step of this process is crucial for mastering matrix algebra and solving more complex problems in the future.

Final Solution and Verification

After performing the necessary steps, we have successfully solved for matrix A. The final solution we obtained is:

A=[13−111−33]A = \begin{bmatrix} 13 & -1 & 11 & -33 \end{bmatrix}

This matrix represents the solution to the original equation:

A+[3−4−1124]=[16−50−9]A + \begin{bmatrix} 3 & -4 & -11 & 24 \end{bmatrix} = \begin{bmatrix} 16 & -5 & 0 & -9 \end{bmatrix}

To ensure the correctness of our solution, it is always a good practice to verify the answer. This can be done by substituting the calculated matrix A back into the original equation and checking if the equation holds true. Let's perform this verification step.

Substitute A into the original equation:

[13−111−33]+[3−4−1124]=[16−50−9]\begin{bmatrix} 13 & -1 & 11 & -33 \end{bmatrix} + \begin{bmatrix} 3 & -4 & -11 & 24 \end{bmatrix} = \begin{bmatrix} 16 & -5 & 0 & -9 \end{bmatrix}

Now, we add the matrices on the left side:

  • First element: 13+3=1613 + 3 = 16
  • Second element: −1+(−4)=−5-1 + (-4) = -5
  • Third element: 11+(−11)=011 + (-11) = 0
  • Fourth element: −33+24=−9-33 + 24 = -9

The resulting matrix is:

[16−50−9]\begin{bmatrix} 16 & -5 & 0 & -9 \end{bmatrix}

Comparing this to the right side of the original equation, we see that they are the same:

[16−50−9 ]=[16−50−9]\begin{bmatrix} 16 & -5 & 0 & -9 \ \end{bmatrix} = \begin{bmatrix} 16 & -5 & 0 & -9 \end{bmatrix}

Since the left side equals the right side, our solution for matrix A is correct. This verification step provides confidence in the accuracy of our calculations and the final result. In conclusion, solving for a matrix involves isolating the unknown matrix using basic algebraic principles, performing matrix operations correctly, and verifying the solution to ensure accuracy. This process is fundamental in linear algebra and has wide applications in various fields, including engineering, computer science, and economics.