Problem 63 - 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 \(A^{-1} B\). $$ \begin{aligned} x-y+z &=-6 \\ 4 x+2 y+z &=9 \\ 4 x-2 y+z &=-3 \end{aligned} $$

Step-by-Step Solution

Answer

The matrix form \(AX=B\) of the system of equations is: \[ \begin{bmatrix} 1 & -1 & 1 \ 4 & 2 & 1 \ 4 & -2 & 1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} -6 \ 9 \ -3 \end{bmatrix} \] And the solution for the equation \(AX=B\) is \(X=A^{-1} B=\) \[ \begin{bmatrix} 0 \ 0 \ -6 \end{bmatrix} \]

Step 1: Formulate matrices A and B

The first step is to determine the matrices \(A\) and \(B\) from the given system of equations. Matrix \(A\) is formed by the coefficients of \(x\), \(y\), and \(z\) and Matrix \(B\) contains the constants on the right of the equations. Thus, we have: \[ A=\begin{bmatrix} 1 & -1 & 1 \ 4 & 2 & 1 \ 4 & -2 & 1 \end{bmatrix} \] and \[ B=\begin{bmatrix} -6 \ 9 \ -3 \end{bmatrix} \]

Step 2: Compute the inverse of matrix A

The next step is to compute the inverse of matrix \(A\) which will allow us to solve the system of equations. Using a graphing tool, calculator, or manual computation, we find that the inverse of matrix \(A\) is: \[ A^{-1}=\begin{bmatrix} 0 & 0.5 & -0.5 \ 0 & 0.5 & 0.5 \ 1 & -1 & -1 \end{bmatrix} \]

Step 3: Compute the solution

Now that we have determined the inverse of matrix \(A\), the last step is to compute \(A^{-1} B\) to find the solution vector for the system of equations. Using a graphing tool or calculator, we find: \[ X=A^{-1} B=\begin{bmatrix} 0 \ 0 \ -6 \end{bmatrix} \]

Key Concepts

Systems of Equations

A system of equations is like a set of interconnected puzzles, where the goal is to find out the exact values of the variables involved, which in this case are \( x \), \( y \), and \( z \). In our system, each equation represents a different perspective or rule about these variables. A typical approach is to solve each equation for one variable at a time. However, when working with linear systems, it can be very efficient to use matrix algebra.
By representing systems of equations in matrix form, we simplify the process of finding solutions. The system:

• \( x-y+z = -6 \)
• \( 4x+2y+z = 9 \)
• \( 4x-2y+z = -3 \)

can be translated into the matrix equation \( A X = B \), where \( A \) is the coefficients' matrix, \( X \) is the variable matrix, and \( B \) is the constant matrix. This matrix representation is particularly neat because it transforms the problem into one continuous operation.

Matrix Inverse

Finding the inverse of a matrix is an essential step in solving systems of equations using matrix algebra. The inverse of a matrix \( A \), denoted as \( A^{-1} \), essentially reverses the multiplication operation. When multiplied by the original matrix \( A \), it yields the identity matrix, much like finding reciprocal in arithmetic.

For a 3x3 matrix like the one in our problem, calculating the inverse manually can be complex, but many graphing utilities and calculators easily compute this inverse. The critical property utilized here is:

• \( A A^{-1} = I \)
• \( I \) is the identity matrix.
• \( AX = B \) becomes \( X = A^{-1}B \).

In our example, once we found \( A^{-1} \), we multiplied it by \( B \) to find the solution. Getting the inverse sets up the pathway to resolving values for \( x, \) \( y, \) and \( z \).

Graphing Utility

Graphing utilities are powerful tools that make solving complex mathematical problems quicker and more efficient. By using a graphing calculator or software, you can input matrices and perform operations like finding an inverse and matrix multiplication without manual computation. This is very handy when dealing with larger matrices or when time is of the essence.
These utilities can plot graphs, calculate derivatives, and evaluate integrals too, making them invaluable for students. In this context, they simplify solving our system of equations in matrix form by performing operations like:

• Entering the matrix \( A \) and matrix \( B \).
• Computing \( A^{-1} \).
• Finding the product \( A^{-1}B \) to get the solutions.

Although doing calculations by hand facilitates deeper understanding, using a graphing utility streamlines the process significantly.

Matrix Multiplication

Once the inverse of matrix \( A \) is found, the next step is matrix multiplication to solve the equation \( A^{-1}B \). Understanding matrix multiplication is crucial for manipulating systems of equations efficiently. Each element of the resulting product matrix is derived from the sum of products of corresponding elements of rows from the first matrix (\( A^{-1} \)) and columns from the second matrix (\( B \)).
The rules of matrix multiplication are:

• Rows from the first matrix and columns from the second matrix are multiplied.
• The results are added together to form a new matrix.
• The number of columns in the first matrix should match the number of rows in the second matrix.

By multiplying the inverse \( A^{-1} \) with \( B \), we obtain the solution matrix \( X \), which provides the values of the unknown variables, concluding the sequence of operations needed to solve the system. It’s efficient and significantly reduces the complexity of solving linear equation systems by hand.