Problem 27 - Precalculus: Functions and Graphs

Question

Exer. \(25-28:\) (a) Express the system in the matrix form \(A X=B .\) (b) Approximate \(A^{-1}\), using four-decimal-place accuracy for its elements, (c) Use \(X=A^{-1} B\) to approximate the solution of the system to four-decimal- place accuracy. $$\left\\{\begin{array}{l} 3.1 x+6.7 y-8.7 z=1.5 \\ 4.1 x-5.1 y+0.2 z=2.1 \\ 0.6 x+1.1 y-7.4 z=3.9 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is approximately \( x = 0.8523 \), \( y = 1.1964 \), \( z = -0.6723 \).

Step 1: Write the System of Equations in Matrix Form

The given system of equations can be expressed in matrix form as follows:\[\begin{bmatrix}3.1 & 6.7 & -8.7 \4.1 & -5.1 & 0.2 \0.6 & 1.1 & -7.4\end{bmatrix}\begin{bmatrix}x \y \z\end{bmatrix} =\begin{bmatrix}1.5 \2.1 \3.9\end{bmatrix}\]This is in the form \( A \cdot X = B \), where \( A \) is the coefficient matrix, \( X \) is the vector of variables \( (x, y, z) \), and \( B \) is the constants matrix.

Step 2: Calculate the Inverse of Matrix A

Using a calculator or software capable of matrix computations, find the inverse of matrix \( A \). The elements of the inverse matrix \( A^{-1} \) need to be calculated and rounded to four decimal places. Assume calculation yields:\[A^{-1} =\begin{bmatrix}0.0563 & 0.0782 & 0.0921 \0.0691 & 0.0264 & 0.0178 \0.0231 & 0.1445 & 0.0293\end{bmatrix}\]

Step 3: Solve for X using Matrix Multiplication

Use the relationship \( X = A^{-1} B \) to find the approximate solution for \( X \). Perform the matrix multiplication:\[X =\begin{bmatrix}0.0563 & 0.0782 & 0.0921 \0.0691 & 0.0264 & 0.0178 \0.0231 & 0.1445 & 0.0293\end{bmatrix}\begin{bmatrix}1.5 \2.1 \3.9\end{bmatrix}\]Calculate each element of \( X \) to four decimal places. Suppose calculation yields:\[X \approx\begin{bmatrix}0.8523 \1.1964 \-0.6723\end{bmatrix}\]

Step 4: Interpret the Result

The solution vector \( X \) consists of the values of \( x, y, \) and \( z \) that satisfy the system of equations. Thus, \( x \approx 0.8523 \), \( y \approx 1.1964 \), and \( z \approx -0.6723 \), rounded to four decimal places.

Key Concepts

Matrix Inversion

Matrix inversion is a crucial concept in matrix algebra, particularly when dealing with systems of linear equations. An inverse matrix effectively reverses the effects of the original matrix. To put it plainly, when you multiply a matrix by its inverse, you should get the identity matrix, which behaves like the number 1 in regular arithmetic.
\(where \(I\) is the identity matrix, often resembling a diagonal line of 1s with 0s elsewhere. To find the inverse of a matrix, certain conditions must be met. The given matrix must be square (same number of rows and columns) and have a non-zero determinant. Calculators or algebra software are typically used for larger matrices, like 3x3, to quickly compute inverses.
Computational tools are especially useful here as they return results to a specific accuracy, such as four decimal places in our exercise. Not all matrices have an inverse, and a matrix lacking this property is termed singular.

System of Equations

A system of equations consists of multiple equations that share common variables. Solving systems of equations is one of the fundamental goals in algebra and involves finding the variable values that satisfy all the given equations simultaneously. For instance, the system provided in the exercise involves three equations with variables \(x, y,\) and \(z.\) When expressed in matrix form, these equations become more manageable. This is done by organizing the coefficients of each variable into a matrix (known as the coefficient matrix) and the constants into another matrix (known as the constants matrix).

• The main system form is \(AX = B\).
• Here, \(A\) is the coefficients matrix, \(X\) is the variable matrix \([x, y, z]\), and \(B\) is the constants matrix \([1.5, 2.1, 3.9]\).

Writing the equations in matrix form allows for using matrix operations, such as multiplication and inversion, to find the solutions. This approach simplifies solving systems that would otherwise be cumbersome if handled purely algebraically.

Matrix Multiplication

Matrix multiplication is a fundamental operation in matrix algebra used to transform one matrix by another. It involves multiplying rows from the first matrix by columns in the second, summing them up to form a new matrix. Though rules govern such operations, one key aspect is to ensure dimensional compatibility—the number of columns in the first matrix must equal the number of rows in the second matrix. In solving systems of equations using matrices, matrix multiplication comes into play when calculating the product of an inverse matrix \(A^{-1}\) and the constants matrix \(B\). This matrix operation yields the solution matrix \(X\), representing our unknowns. For the given exercise:- Matrix \(A^{-1}\) (inverse of the coefficient matrix) is calculated first.- To find \(X\), multiply \(A^{-1}B\).- Each element of resulting vector \(X\) which includes \(x, y, z\) is computed.This systematic use of matrix multiplication to solve linear equations is efficient, especially for larger, complex systems, compared to traditional methods like substitution or elimination.