Problem 51 - College Algebra with Modeling and Visualization

Question

Complete the following. (A) Write the system in the form \(A X=B\). (B) Solve the system by finding \(A^{-1}\) and then using the equation \(\boldsymbol{X}=\boldsymbol{A}^{-1} \boldsymbol{B}\). (Hint: Some of your answers from Exercises \(15-28\) may be helpful.) $$ \begin{array}{rr} x+\quad z= & -7 \\ 2 x+y+3 z= & -13 \\ -x+y+z= & -4 \end{array} $$

Step-by-Step Solution

Answer

The solution is \( x = -25 \), \( y = -1 \), \( z = -17 \).

Step 1: Write the System in Matrix Form

Express the system of equations as a matrix equation in the form \( AX = B \).The system is:\[\begin{array}{rcl}x + 0y + z & = & -7, \2x + y + 3z & = & -13, \-x + y + z & = & -4\end{array}\]The matrix \( A \) consists of the coefficients of \( x, y, \) and \( z \) for each equation:\[A = \begin{bmatrix} 1 & 0 & 1 \ 2 & 1 & 3 \ -1 & 1 & 1 \end{bmatrix}\]The variable matrix \( X \) and the constant matrix \( B \) are:\[X = \begin{bmatrix} x \ y \ z \end{bmatrix}, \quad B = \begin{bmatrix} -7 \ -13 \ -4 \end{bmatrix}\]Thus, \(AX = B\) is:\[\begin{bmatrix} 1 & 0 & 1 \ 2 & 1 & 3 \ -1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} -7 \ -13 \ -4 \end{bmatrix}\]

Step 2: Calculate the Inverse of Matrix A

To solve \( X = A^{-1}B \), first find \( A^{-1} \). We'll use the formula for the inverse of a 3x3 matrix:\[A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A)\]Firstly, calculate the determinant of \( A \):\[\text{det}(A) = 1\cdot (1 \cdot 1 - 3 \cdot 1) - 0\cdot(2 \cdot 1 - 3 \cdot (-1)) + 1\cdot(2 \cdot 1 - 1 \cdot (-1))= 1(-2) + 0 + 1 \cdot 3 = 1\]Since the determinant is \(1\), the matrix is invertible. Next, find the adjugate of \( A \) and hence \( A^{-1} \), which will result in:\[A^{-1} = \begin{bmatrix} -2 & 3 & 3 \ 1 & -1 & -2 \ 0 & 1 & 1 \end{bmatrix}\]

Step 3: Solve for X using Inverse of A

Now solve for \( X \) using the formula \( X = A^{-1}B \):\[X = \begin{bmatrix} -2 & 3 & 3 \ 1 & -1 & -2 \ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} -7 \ -13 \ -4 \end{bmatrix}\]Multiply the matrices:\[\begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix}(-2)(-7) + 3(-13) + 3(-4) \1(-7) + (-1)(-13) + (-2)(-4) \0(-7) + 1(-13) + 1(-4)\end{bmatrix} = \begin{bmatrix} -25 \ -1 \ -17 \end{bmatrix}\]Thus, the solution is \( x = -25 \), \( y = -1 \), \( z = -17 \).

Step 4: Conclusion: Solution of the System

We found the values of \( x \), \( y \), and \( z \) that satisfy the given system of equations. The solution is:\[\begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} -25 \ -1 \ -17 \end{bmatrix}\]

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. Typically, the goal is to find the values of these variables that satisfy all equations simultaneously. When written in matrix form, a system like:

• \[ x + 0y + z = -7 \]
• \[ 2x + y + 3z = -13 \]
• \[ -x + y + z = -4 \]

can be expressed as \( AX = B \), where \( A \) is a matrix of coefficients, \( X \) is the matrix of variables, and \( B \) is the constant matrix. Writing systems in this form simplifies the process of finding solutions using matrix operations.

Inverse Matrix

An inverse matrix, denoted as \( A^{-1} \), is a fundamental concept in matrix algebra used to solve equations like \( AX = B \). The inverse of a matrix is similar to the reciprocal in numbers. When multiplied by the original matrix, the inverse results in the identity matrix:

• \[ A imes A^{-1} = I \]

To find an inverse, first ensure that the determinant is not zero. If the determinant is zero, the matrix isn't invertible. For a 3x3 matrix, the inverse can be found using the formula:

• \[ A^{-1} = \frac{1}{\text{det}(A)} \times \text{adj}(A) \]

If you can calculate it, you can then solve for \( X \) using \( A^{-1}B \) with ease.

Matrix Multiplication

Matrix multiplication involves multiplying rows by columns. This operation is a cornerstone of transforming matrices, but it's not as straightforward as multiplying individual numbers. To multiply, follow these steps:

• Align the matrices so that the number of columns in the first matrix equals the number of rows in the second.
• Multiply elements of the row by corresponding elements of the column and sum them.
• Repeat for each row and column combination.

For example, to solve for \( X = A^{-1}B \), you compute the product of the inverse matrix and the constant matrix, which provides the values of variables \( x, y, \) and \( z \).

Determinant Calculation

Determinants are a key part of matrix algebra. They help determine whether a matrix is invertible and feature prominently in the calculation of inverses. For a 3x3 matrix \( A \), the determinant is calculated using the rule:

• \[ \text{det}(A) = a(ei − fh) − b(di − fg) + c(dh − eg) \]

where each letter corresponds to elements of the matrix. In the given exercise, the determinant is calculated as \( 1 \), confirming the matrix is invertible. Understanding determinants can simplify matrix equations and guide whether solutions can be found using inverses.