Problem 36 - Technical Mathematics with Calculus

Question

Solve each system of equations by calculator using the unit matrix method. Three Equations in Three Unknowns. $$\begin{array}{r} \frac{x}{10}+\frac{y}{5}+\frac{z}{20}=\frac{1}{4} \\ x+y+z=6 \\ \frac{x}{3}+\frac{y}{2}+\frac{z}{6}=1 \end{array}$$

Step-by-Step Solution

Answer

To find the solution, calculate matrix A's inverse and multiply it by matrix B. The resulting matrix X provides the values of x, y, and z.

Step 1: Create the Coefficient Matrix

First, form the coefficient matrix A by using the coefficients of variables x, y, and z from each equation. For the given system, the matrix A is: \[ A = \begin{pmatrix} \frac{1}{10} & \frac{1}{5} & \frac{1}{20} \ 1 & 1 & 1 \ \frac{1}{3} & \frac{1}{2} & \frac{1}{6} \end{pmatrix} \]

Step 2: Form the Constant Matrix

Next, construct the constant matrix B using the constants from the right side of the equations. The matrix B is: \[ B = \begin{pmatrix} \frac{1}{4} \ 6 \ 1 \end{pmatrix} \]

Step 3: Use Calculator to Form the Inverse Matrix of A

Using a calculator, find the inverse of matrix A, denote it as \( A^{-1} \). This can usually be done by entering matrix A into the calculator and using the matrix function to calculate the inverse.

Step 4: Multiply the Inverse of Matrix A with Matrix B

Multiply the inverse matrix \( A^{-1} \) by the constant matrix B to find the variable matrix X (which contains the values of x, y, and z). The operation is: \[ X = A^{-1}B \]

Step 5: Interpret the Result

The resulting matrix X will give the values of x, y, and z. Each entry in matrix X corresponds to the value of one of the variables in the order they appear in the equations.

Key Concepts

Coefficient Matrix

Understanding the coefficient matrix is fundamental when solving systems of equations using matrices. In technical mathematics, a system of linear equations can be represented in matrix form. The coefficient matrix, denoted by A, comprises the coefficients of the variables from the equations.

For the exercise above, the system of equations was transformed into a coefficient matrix by collecting the coefficients of x, y, and z. Each row of the matrix corresponds to an equation, and each column corresponds to a variable. Proper formation of the coefficient matrix is crucial as it lays the groundwork for further computations using matrix operations.

Inverse Matrix

Once you have the coefficient matrix, the next step is to find an inverse matrix, represented as \( A^{-1} \). An inverse matrix is essentially a matrix that, when multiplied by the original matrix, yields the identity matrix. The beauty of the inverse lies in its ability to solve linear equations algebraically without needing to do substitution or elimination.

Finding the inverse requires making sure that the determinant of your matrix is not zero, as only non-singular matrices have inverses. Use your calculator as meticulously outlined in Step 3 of the provided solution to accurately compute the inverse matrix for the given system.

Matrix Multiplication

The penultimate step is the multiplication of the inverse matrix with the constant matrix. Matrix multiplication is not the same as multiplying numbers; it involves a series of operations based on the rows of the first matrix and the columns of the second. Accuracy in multiplication is vital—it's how we ensure the solutions to our equations hold water.

In the context of the exercise, multiplying the inverse of the coefficient matrix \( A^{-1} \) with the constant matrix B, will unravel matrix X which contains the desired values for x, y, and z. Remember, the order of multiplication matters in matrix multiplication, so it's always the inverse matrix first, followed by the constant matrix.

Algebraic Methods in Technical Mathematics

Algebraic methods in technical mathematics, particularly matrix algebra, provide a structured way to solve complex systems. This discipline encompasses a variety of operations and concepts—like solving systems using the inverse matrix method as demonstrated. It is a robust framework for solving linear equations, which appear not only in pure mathematics but also in fields like engineering and economics.

To effectively apply these algebraic methods, comprehension of the underlying procedures and their sequential implementation is essential. The process outlined in the solution is a testament to the systematic approach required in algebraic methods to move from a set of equations to a matrix, then to an inverse, and finally to the desired solutions.