Problem 37 - 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. $$ \begin{array}{l} 2 x+6 y+6 z=8 \\ 2 x+7 y+6 z=10 \\ 2 x+7 y+7 z=9 \end{array} $$ The inverse of $$\left[\begin{array}{lll}2 & 6 & 6 \\ 2 & 7 & 6 \\ 2 & 7 & 7\end{array}\right]\( is \)\left[\begin{array}{rrr}7 & 0 & 3 \\ -1 & 1 & 0 \\\ 0 & 1 & 1\end{array}\right]$$

Step-by-Step Solution

Answer

The values for \(x\), \(y\), and \(z\) are 56, 0, and 12 respectively.

Step 1: Write the linear system as a matrix equation

The linear system can be presented as a matrix equation \(A X=B\) as follows: \(A = \left[\begin{array}{lll}2 & 6 & 6 \ 2 & 7 & 6 \ 2 & 7 & 7\end{array}\right]\), \(X = \left[\begin{array}{l}x \ y \ z\end{array}\right]\), and \(B = \left[ \begin{array}{l}8 \ 10 \ 9\end{array}\right]\). So the matrix equation is \[ \left[\begin{array}{lll}2 & 6 & 6 \ 2 & 7 & 6 \ 2 & 7 & 7\end{array}\right] \left[\begin{array}{l}x \ y \ z\end{array}\right] = \left[ \begin{array}{l}8 \ 10 \ 9\end{array}\right]\]

Step 2: Use the given inverse Matrix to solve the system

Given the inverse of matrix \(A\) as \[\left[\begin{array}{rrr}7 & 0 & 3 \ -1 & 1 & 0 \ 0 & 1 & 1\end{array}\right]\]. Multiply both sides of the system by the inverse to solve for \(X\), that is \(A^{-1}A=I\) and \(A^{-1}B=C\) So we get \(C = \left[\begin{array}{c} 56 \ 0 \ 12 \end{array}\right] = X\)

Step 3: Solve for X

We can then write \(X=[x, y, z] = C = [56, 0, 12]\), solving for \(x\), \(y\), and \(z\)

Key Concepts

Understanding Linear Systems

In algebra, linear systems consist of two or more linear equations involving the same set of variables. These systems can help us find common solutions based on the equations provided. For instance, each equation will have the variables placed in order with their coefficients.

They're represented as:

• Equation 1: \[a_1x + b_1y + c_1z = d_1\]
• Equation 2: \[a_2x + b_2y + c_2z = d_2\]
• Equation 3: \[a_3x + b_3y + c_3z = d_3\]

Using a matrix equation makes it easier to manage and manipulate these equations all at once. By converting these systems of linear equations into a matrix form, students can take advantage of different mathematical tools like matrix operations.

Exploring Matrix Inverse

A matrix inverse is useful in solving matrix equations because it plays the same role as division does with numbers in arithmetic. If \(A\) is a square matrix, its inverse, denoted as \(A^{-1}\), is defined as a matrix that when multiplied with \(A\) yields the identity matrix \(I\).

This is expressed as:

• \(AA^{-1} = A^{-1}A = I\)

For the matrix inverse to exist, the matrix must be non-singular, meaning it has a non-zero determinant. If we have an inverse matrix, we can solve nonsingular linear systems by multiplying both sides of the matrix equation by the inverse.

Keep in mind that not all matrices have inverses. The matrix's determinant can help determine whether an inverse exists.

The Process of Matrix Multiplication

Matrix multiplication involves taking the rows of the first matrix and multiplying them with the columns of the second matrix. Each element of the resulting matrix is a product of row-column multiplications. Let's break down the steps:

• Align the first matrix's rows with the second matrix's columns.
• Multiply corresponding elements and add them up to get each element of the result.

This method is crucial when solving systems of equations through matrix equations. In context, after finding a matrix inverse, multiplying it by the vector or another matrix assists in isolating the solution vector.

Efficient computation of matrix multiplication requires an understanding of suitable conditions, such as ensuring the number of columns in the first matrix equals the number of rows in the second matrix.

Techniques for Solving Systems of Equations Using Matrices

Systems of equations can become much easier to solve using matrices and their inverses. Here's a step-by-step approach:

• First, write the system in matrix form, identifying the coefficient matrix \(A\), variable matrix \(X\), and the constants matrix \(B\).
• Next, if given the inverse, multiply both sides by the inverse matrix \(A^{-1}\) to get the identity and solve for \(X\).
• Finish by calculating the matrix multiplication \(A^{-1}B\), providing the values for the variables.

This method, when applicable, vastly simplifies complex systems of equations. The primary goal is to clearly isolate the variable matrix through effective manipulation of the given equations, ensuring accurate solutions.