Problem 39 - 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. $$ \begin{array}{rr} x-y+z= & 8 \\ 2 y-z= & -7 \\ 2 x+3 y & =1 \end{array} $$ The inverse of \(\left[\begin{array}{rrr}1 & -1 & 1 \\ 0 & 2 & -1 \\ 2 & 3 & 0\end{array}\right]\) is \(\left[\begin{array}{rrr}3 & 3 & 1 \\ -2 & -2 & 1 \\\ 4 & 5 & 2\end{array}\right]\)

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 10\), \(y = -12\), \(z = 8\)

Step 1: Formulate the Matrix Equation

The system of equations can be written as a matrix equation of the form \(AX = B\) where \(A\) is the matrix of coefficients, \(X\) is the matrix of variables and \(B\) is the matrix of constant terms. In this case, \(A = \left[1, -1, 1; 0, 2, -1; 2, 3, 0\right]\), \(X = [x, y, z]^T\) and \(B = [8, -7, 1]^T\)

Step 2: Pre-Multiply by the Inverse

Given the inverse of \(A\), to solve for \(X\), pre-multiply both sides of the equation by \(A^{-1}\). This will isolate \(X\) on one side of the equation. The inverse of \(A\) is given as \(A^{-1} = \left[3, 3, 1; -2, -2, 1; 4, 5, 2\right]\). So, if we multiply both sides of the matrix equation by \(A^{-1}\), we get \(A^{-1}AX = A^{-1}B\). Since \(A^{-1}A\) gives the identity matrix, we have \(X = A^{-1}B\).

Step 3: Calculate the Solution

Now, multiply the matrix \(A^{-1}\) by the matrix \(B\). Performing the multiplication, we get \(X = \left[3, 3, 1; -2, -2, 1; 4, 5, 2\right] * [8, -7, 1]^T = [10, -12, 8]^T\). So, the solutions to the system of equations are \(x = 10\), \(y = -12\), \(z = 8\)

Key Concepts

Understanding Linear Systems

A linear system is a collection of one or more linear equations involving the same set of variables. For example, the given system consists of the equations:

• \( x - y + z = 8 \)
• \( 2y - z = -7 \)
• \( 2x + 3y = 1 \)

Our goal is to find values for \(x\), \(y\), and \(z\) that satisfy all equations simultaneously. These systems are important in mathematical modeling of real-world situations like physics and engineering projects.
By converting these equations into matrix form, we utilize powerful mathematical tools to solve them more systematically.

The Magic of Matrix Inversion

Matrix inversion is a crucial concept when dealing with systems of linear equations. A matrix inverse is akin to the reciprocal of a number. It’s the matrix which, when multiplied with the original matrix, yields the identity matrix (similar to how multiplying a number by its reciprocal gets you one).
For a square matrix \(A\), the inverse is denoted by \(A^{-1}\). In our problem, we are provided with matrix \(A\) as:

• \(\begin{bmatrix} 1 & -1 & 1 \ 0 & 2 & -1 \ 2 & 3 & 0 \end{bmatrix}\)

And its inverse:

• \(\begin{bmatrix} 3 & 3 & 1 \ -2 & -2 & 1 \ 4 & 5 & 2 \end{bmatrix}\)

Using the inverse matrix, we can easily solve the linear system by simplifying the matrix equation \(AX = B\).

Role of the Coefficient Matrix

In matrix equations like \(AX = B\), the matrix \(A\) is known as the coefficient matrix. It contains the coefficients of the variables in the linear equations. Here, our coefficient matrix is:

• \(\begin{bmatrix} 1 & -1 & 1 \ 0 & 2 & -1 \ 2 & 3 & 0 \end{bmatrix}\).

This matrix essentially encodes how each variable affects the equations. By examining this matrix, we can identify the relationships between the variables in the system.
The coefficient matrix helps to transform the linear system into matrix form, paving the way for matrix tools to solve the linear systems more efficiently. This transformation can dramatically reduce the complexity involved in solving multiple equations.

Solving Equations Using Matrix Equations

Solving a system of equations through matrix equations involves several steps. First, rewrite the system into the form \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the column of variables, and \(B\) is the constant matrix. For our given example, these matrices become:

• \(A = \begin{bmatrix} 1 & -1 & 1 \ 0 & 2 & -1 \ 2 & 3 & 0 \end{bmatrix}\)
• \(X = \begin{bmatrix} x \ y \ z \end{bmatrix}\)
• \(B = \begin{bmatrix} 8 \ -7 \ 1 \end{bmatrix}\)

With the inverse of \(A\) given, solve for \(X\) by calculating \(A^{-1}B\). This multiplication gives the solution to the system:
\[X = \begin{bmatrix} 3 & 3 & 1 \ -2 & -2 & 1 \ 4 & 5 & 2 \end{bmatrix} \begin{bmatrix} 8 \ -7 \ 1 \end{bmatrix} = \begin{bmatrix} 10 \ -12 \ 8 \end{bmatrix}.\]This result means that \( x = 10 \), \( y = -12 \), and \( z = 8 \), effectively solving the equation system.