Problem 66 - College Algebra

Question

Write each system in the form \(A X=B\). Then solve the system by entering \(A\) and \(B\) into your graphing utility and computing \(A^{-1} B\). $$ \begin{aligned} x-y &=1 \\ 6 x+y+20 z &=14 \\ y+3 z &=1 \end{aligned} $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x=1\), \(y=-1\), and \(z=1\).

Step 1: Express the system in the form AX=B

First, gather the coefficients of the equations into matrix \(A\). Based on the system of equations,\[ \begin{aligned}x-y &=1 \6x+y+20z &=14 \y+3z &=1\end{aligned} \]we get\[A = \begin{bmatrix}1 & -1 & 0 \6 & 1 & 20 \0 & 1 & 3\end{bmatrix} \]and \[B = \begin{bmatrix}1 \14 \1\end{bmatrix}\]So we can say, \[AX= B\]where\[X = \begin{bmatrix}x \y \z\end{bmatrix}\]

Step 2: Find Inverse of Matrix A

Now, calculate the inverse of matrix A using a graphing utility as directed. You should find \[A^{-1}= \begin{bmatrix}0.2 & 0.2 & -0.4 \0 & -0.2 & 0.8 \0 & 0 & 0.333\end{bmatrix} \]

Step 3: Compute A^{-1}B

Finally, multiply matrix A's inverse by matrix B to solve the system. This multiplication operation can also be performed using the graphing utility. \[A^{-1}B = \begin{bmatrix}0.2 & 0.2 & -0.4 \0 & -0.2 & 0.8 \0 & 0 & 0.333\end{bmatrix} \times \begin{bmatrix}1 \14 \1\end{bmatrix} = \begin{bmatrix}x \y \z\end{bmatrix} = \begin{bmatrix}1 \-1 \1\end{bmatrix}\]

Key Concepts

Matrix Inversion

Matrix inversion is a crucial concept when solving systems of linear equations. In essence, finding the inverse of a matrix is analogous to finding the reciprocal of a number; but instead of dividing, we apply a more complex mathematical process. The inverse of a matrix A, denoted as \( A^{-1} \), exists only if A is square and non-singular, meaning it has a non-zero determinant.

To compute \( A^{-1} \), several methods exist, including elementary row operations, adjugate and determinant method, or computational algorithms typically found in graphing utilities. When we have \( A^{-1} \), the system \( AX = B \) can be solved by multiplying both sides by \( A^{-1} \) to obtain \( X = A^{-1}B \), thus isolating X, which represents the vector of variables. This process is straightforward in a computational environment, saving a significant amount of time and minimizing computational errors.

Augmented Matrices

Augmented matrices represent an alternative to handling systems of linear equations efficiently. An augmented matrix is a combination of two matrices that include all coefficients of the system along with the constants from the equations. It is often written as a single matrix separated by a line where the right side of the line represents the constants.

In the context of the exercise, although the solution does not explicitly mention the augmented matrix, it's a key concept in understanding similar methods like Gaussian elimination. Augmented matrices facilitate operations that lead to finding solutions without changing the original system's properties. Being proficient in interpreting and manipulating augmented matrices is fundamental when applying linear algebra techniques to solve systems of equations.

Graphing Utility

A graphing utility is an invaluable tool for students and professionals alike. It is a piece of software or a feature in a calculator that enables the graphing of complex equations and performing matrix operations, such as inversion and multiplication. When the system of equations is provided, a graphing utility can swiftly generate matrix A and vector B, compute the inverse of A, if it exists, and multiply it with B to find the solution vector X.

The advantage of using a graphing utility becomes evident as the size and complexity of the system increase. It ensures accurate calculation, saves time, and allows the user to focus on the interpretation of the solution. Therefore, becoming adept at utilizing such tools not only enhances understanding of linear algebra concepts but also expedites the problem-solving process as seen in the exercise.

Gaussian Elimination

Gaussian elimination is a systematic method for solving systems of linear equations. It is based on performing row operations on the augmented matrix to reduce it to its Row Echelon Form (REF) or even further to the Reduced Row Echelon Form (RREF). The goal is to create a series of pivot positions, ultimately simplifying the system to a point where the solutions can be read off directly or back-substituted to find each variable.

In terms of utility and connection to the original exercise, Gaussian elimination gives an alternate avenue to approach solving systems without matrix inversion, which might be preferred when the inverse is difficult to calculate or when the system is not conducive to inversion. Through this method, the student gains a deeper understanding of linear systems' structural properties and develops algebraic skills that are essential for solving problems in a non-computational manner.