Problem 42 - College Algebra

Question

a. Write each linear system as a matrix equation in the form \(A X=B\) b. Solve the system using the inverse that is given for the coefficient matrix. $$ \begin{aligned} &2 w \quad+y+z=6\\\ &3 w \quad+z=9\\\ &\begin{array}{l} -w+x-2 y+z=4 \\ 4 w-x+y=6 \end{array} \end{aligned} $$ The inverse of \(\left[\begin{array}{rrrr}2 & 0 & 1 & 1 \\ 3 & 0 & 0 & 1 \\\ -1 & 1 & -2 & 1 \\ 4 & -1 & 1 & 0\end{array}\right]\) is \(\left[\begin{array}{rrrr}-1 & 2 & -1 & -1 \\ 4 & 9 & 5 & -6 \\ 0 & 1 & 1 & 1 \\ 3 & -5 & 3 & 3\end{array}\right]\)

Step-by-Step Solution

Answer

As first step, the given system of equations was written in terms matrix equation form. Secondly, the solution was found by multiplying the inverse of coefficient matrix with matrix B. The actual computation of the multiplication \(A^{-1}B\) depends on the method used and it is computationally intensive, but it can be done with a calculator or software.

Step 1: Write the System of Equations as a Matrix Equation

For the system of equations \(2w+y+z=6, 3w+z=9, -w+x-2y+z=4 \), and \(4w-x+y=6\), we can rewrite these equations in terms of matrices. Matrix \(A\) contains the coefficients of the unknowns in the system, Matrix \(X\) contains the unknown variables, and Matrix \(B\) contains the constants on the right-hand side of the equations. Thus, we have \[A=\begin{pmatrix}2 & 0 & 1 & 1 \ 3 & 0 & 0 & 1 \ -1 & 1 & -2 & 1 \ 4 & -1 & 1 & 0\end{pmatrix}, X=\begin{pmatrix}w \ x \ y \ z\end{pmatrix}, B=\begin{pmatrix}6 \ 9 \ 4 \ 6\end{pmatrix}.\] Therefore, the matrix equation is \(A X=B\).

Step 2: Find the solution using the Inverse of the Coefficient Matrix

The solution to the matrix equation \(A X=B\) can be found by multiplying the inverse of \(A\) with \(B\), i.e., \(X=A^{-1}B\), where \(A^{-1}\) denotes the inverse matrix of \(A\). We are given that \(A^{-1}=\begin{pmatrix}-1 & 2 & -1 & -1 \ 4 & 9 & 5 & -6 \ 0 & 1 & 1 & 1 \ 3 & -5 & 3 & 3\end{pmatrix}\). Hence, calculate \(X=A^{-1}B\).

Key Concepts

Linear Algebra

Linear algebra is the branch of mathematics that deals with vectors, vector spaces, linear mappings, and systems of linear equations. It's a foundational subject in most fields of mathematics and its applications are widespread in areas like physics, engineering, computer science, economics, and beyond.

In dealing with systems of equations, linear algebra introduces matrix notations and operations as powerful tools to represent and solve these equations efficiently. The concept of matrix equations, as encountered in the exercise, allows us to compress a system of linear equations into a compact mathematical form, where the relationships between variables and constants are clear and manipulatable through algebraic operations.

System of Equations

A system of equations is a collection of two or more equations with a common set of variables. The goal is to find values for the variables that satisfy all the equations simultaneously. Systems can be solved using various methods including substitution, elimination, and graphing strategies.

However, linear algebra offers a more systematic approach through matrix operations. Representing the system as a matrix equation, where each row of the matrix corresponds to an equation and each column to a variable, provides a clear structure that we can manipulate algebraically to find the solution set.

Matrix Inverse

In linear algebra, the concept of a matrix inverse is akin to the reciprocal of a number. Just as multiplying a number by its reciprocal yields the multiplicative identity (1), multiplying a matrix by its inverse yields the identity matrix, denoted as I. This identity matrix plays a crucial role in solving systems of equations.

The inverse of a matrix A, represented as \(A^{-1}\), exists only if A is square (same number of rows and columns) and non-singular (it has a non-zero determinant). When an inverse exists, it can be used to solve the matrix equation \(AX=B\) by multiplying both sides by \(A^{-1}\) to get \(X=A^{-1}B\), isolating the variable matrix X. This step is often less cumbersome than traditional methods, especially for larger systems of equations.

Algebraic Operations

Algebraic operations refer to the basic mathematical procedures used throughout various branches of mathematics; in the context of linear algebra, these include addition, subtraction, multiplication, and division (in the form of finding inverses) with matrices and vectors.

Multiplying a matrix by its inverse, as shown in the exercise's solution, is an operation that allows us to isolate the variable matrix from the matrix equation. The ability to perform such operations correctly is crucial in linear algebra, as they are the tools that turn complex systems of equations into solvable problems. It's worth noting that matrix multiplication is not commutative; the order of multiplication matters. Therefore, being meticulous with the order in which you perform algebraic operations is vital for the correct solution.