Problem 46 - A Graphical Approach to College Algebra

Question

Solve each system by using the matrix inverse method. $$\begin{aligned} 3 x+2 y-w &=0 \\ 2 x+z+2 w &=5 \\ x+2 y-z &=-2 \\ 2 x-y+z+w &=2 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( x = 1 \), \( y = -1 \), \( z = 2 \), \( w = 1 \).

Step 1: Formulate the problem in matrix form

Re-arrange the given linear equations into matrix form. Let the system of equations be represented as \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is the variable matrix, and \( B \) is the constants matrix.The coefficient matrix \( A \) is:\[A = \begin{bmatrix} 3 & 2 & 0 & -1 \ 2 & 0 & 1 & 2 \ 1 & 2 & -1 & 0 \ 2 & -1 & 1 & 1 \end{bmatrix}\]The variable matrix \( X \) is:\[X = \begin{bmatrix} x \ y \ z \ w \end{bmatrix}\]The constants matrix \( B \) is:\[B = \begin{bmatrix} 0 \ 5 \ -2 \ 2 \end{bmatrix}\]

Step 2: Calculate the inverse of the coefficient matrix

Find the inverse of the matrix \( A \), denoted as \( A^{-1} \). This can be done using a matrix calculator or manually applying methods like the row reduction technique. The inverse is calculated as follows:\[A^{-1} = \begin{bmatrix} -3/13 & 4/13 & -2/13 & 2/13 \ 4/13 & 3/13 & 1/13 & -7/13 \ 5/13 & -6/13 & 1/13 & 4/13 \ -8/13 & 5/13 & 2/13 & 1/13 \end{bmatrix}.\]

Step 3: Multiply the inverse of A with B

Using the relationship \( X = A^{-1}B \), calculate the product of \( A^{-1} \) with the matrix \( B \).\[A^{-1}B = \begin{bmatrix} -3/13 & 4/13 & -2/13 & 2/13 \ 4/13 & 3/13 & 1/13 & -7/13 \ 5/13 & -6/13 & 1/13 & 4/13 \ -8/13 & 5/13 & 2/13 & 1/13 \end{bmatrix} \times \begin{bmatrix} 0 \ 5 \ -2 \ 2 \end{bmatrix} = \begin{bmatrix} 1 \ -1 \ 2 \ 1 \end{bmatrix}\]This gives us the values of \( x \), \( y \), \( z \), and \( w \).

Step 4: Interpret the solution

The solution to the system is the values obtained from the product \( A^{-1}B \). Thus the solution is \( x = 1 \), \( y = -1 \), \( z = 2 \), and \( w = 1 \).

Key Concepts

System of Linear Equations

A system of linear equations is a collection of two or more linear equations that we deal with all at once. Solving such systems means finding the values of the variables that satisfy all the given equations simultaneously. Linear equations are equations of the first degree, which means each term is either a constant or the product of a constant and a single variable.
For instance, in the exercise given, we have four equations with four variables:

• 3x + 2y - w = 0
• 2x + z + 2w = 5
• x + 2y - z = -2
• 2x - y + z + w = 2

These equations form a system of linear equations because every equation within it lines up in a straight line when graphed in a multi-dimensional space, in this case, a four-dimensional space. In real-world problems, systems of linear equations often arise in different fields like physics, engineering, economics, and statistics, where they are used to model situations with multiple interacting factors.

Coefficient Matrix

A coefficient matrix is a matrix containing only the coefficients of the variables in a system of linear equations. When we look at the system of linear equations, each equation can be seen as a row in this matrix.
For example, from our problem, the matrix is:\[A = \begin{bmatrix} 3 & 2 & 0 & -1 \ 2 & 0 & 1 & 2 \ 1 & 2 & -1 & 0 \ 2 & -1 & 1 & 1 \end{bmatrix}\]This matrix holds the coefficients in the same order in which the variables appear (x, y, z, w). It is important to remember that any missing variable in the original equations should be considered to have a coefficient of zero in the matrix layout.
Using the coefficient matrix is an organized way of grouping the data so that computational techniques like matrix operations can be easily applied to solve the system. It's the structure that allows us to utilize the beauty of linear algebra to find solutions in an elegant and efficient manner.

Inverse Matrix

An inverse matrix is defined such that when it is multiplied by the original matrix, it yields the identity matrix. In simpler words, if you have a matrix \( A \), its inverse \( A^{-1} \) will make \( A \times A^{-1} = I \), where \( I \) is the identity matrix with ones on its diagonal and zeros elsewhere.
In the context of solving a system of linear equations, the inverse matrix is crucial because it provides a direct method to find the solution. Once we multiply the inverse of the coefficient matrix by the constants on the other side of the equations, we can solve for the variable matrix.

• To find the inverse of a matrix, several methods can be applied such as row reduction, or more advanced techniques if you are using technology.
• In the exercise, the inverse of the coefficient matrix \( A \) was calculated, giving us:\[A^{-1} = \begin{bmatrix} -3/13 & 4/13 & -2/13 & 2/13 \ 4/13 & 3/13 & 1/13 & -7/13 \ 5/13 & -6/13 & 1/13 & 4/13 \ -8/13 & 5/13 & 2/13 & 1/13 \end{bmatrix} \]Once you have this inverse, finding the solution is straightforward by performing matrix multiplication \( A^{-1}B \) to find the values of the variables.

Understanding how the inverse matrix works not only helps solve equations but also ensures we approach these problems with detail and precision, especially in handling real-world data that can be erratic and interdependent.