Problem 70 - Precalculus

Question

apply matrix algebra to solve the system of linear equations. $$\begin{aligned}&x+2 y-3 z=1\\\&x-y-z=3\\\&x \quad-4 z=0\end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = 1\), \(y = -\frac{11}{7}\), \(z = -\frac{3}{7}\).

Step 1: Write the System in Matrix Form

The system of equations can be represented in matrix form as \(AX = B\), where \[A = \begin{bmatrix} 1 & 2 & -3 \ 1 & -1 & -1 \ 1 & 0 & -4 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \ z \end{bmatrix}, \quad B = \begin{bmatrix} 1 \ 3 \ 0 \end{bmatrix} \].

Step 2: Find the Inverse of Matrix A

Calculate the inverse of matrix \(A\) if it exists. The inverse of a 3x3 matrix is computed using the formula involving the determinant and adjugate of the matrix. If \(A^{-1}\) exists, calculate it to use in solving the system.

Step 3: Calculate the Determinant of A

The determinant of matrix \(A\) is found as \[ \text{det}(A) = 1(-1)(-4) + 2(-1)(1) + (-3)(1)(0) - ((-3)(-1)(1) + 2(0)(1) + 1(-4)(-1)) = -7 \]. The determinant is \(-7\), which is non-zero, so \(A^{-1}\) exists.

Step 4: Compute the Inverse of Matrix A

Use the adjugate and determinant to find \(A^{-1}\). Compute \[ A^{-1} = \frac{1}{-7}\begin{bmatrix} 4 & -4 & -1 \ 1 & 4 & 5 \ 1 & 2 & -3 \end{bmatrix} \].

Step 5: Solve for X Using Matrix Inverse

Multiply both sides of the matrix equation \(AX = B\) by \(A^{-1}\) to find \(X\): \[ X = A^{-1}B = \frac{1}{-7}\begin{bmatrix} 4 & -4 & -1 \ 1 & 4 & 5 \ 1 & 2 & -3 \end{bmatrix} \begin{bmatrix} 1 \ 3 \ 0 \end{bmatrix} = \begin{bmatrix} \frac{7}{7} \ \frac{-11}{7} \ \frac{-3}{7} \end{bmatrix} \].

Step 6: Verify the Solution

Substitute \(x = 1\), \(y = -\frac{11}{7}\), and \(z = -\frac{3}{7}\) back into the original equations to ensure they satisfy the system, confirming the solution is correct.

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more algebraic equations involving the same set of variables. In the given exercise, we have three equations with three variables: \(x\), \(y\), and \(z\). Systems like these are fundamental in mathematics and appear frequently in various fields of study. To solve such systems, we aim to find the values of the variables that satisfy all the equations in the system simultaneously. Converting this system into matrix form helps simplify and streamline the process of finding the solution.

When dealing with a system of equations:

• You can represent them using matrices, transforming the equations into a format that is easier to manipulate algebraically.
• The coefficients of the variables form a matrix, the variables themselves form a vector, and the constants on the right side form another vector.

This transformation allows us to use powerful matrix algebra methods to solve the system, such as finding a matrix inverse.

Matrix Inverse

The inverse of a matrix, similar to the reciprocal of a number, is a matrix that, when multiplied with the original matrix, yields the identity matrix. Not all matrices have inverses. A matrix must be square (xndimension) and have a non-zero determinant to possess an inverse.

The inverse of a matrix \(A\) is denoted as \(A^{-1}\). If it exists:

• Multiplying \(A\) by \(A^{-1}\) results in the identity matrix \(I\), where each diagonal element is 1, and all other elements are 0.
• The identity matrix serves as the multiplicative identity in matrix algebra, much like 1 does for real numbers.

Finding the inverse involves calculating the determinant and the adjugate (or adjoint) of the matrix. The inverse is calculated as \(A^{-1} = \frac{1}{ ext{det}(A)} \, \text{adj}(A)\). It's important to calculate the inverse properly to solve linear systems efficiently.

Determinant of a Matrix

The determinant is a special number that can be calculated from a square matrix. It provides important properties of the matrix, such as whether the matrix has an inverse.

For a 3x3 matrix with elements \(a, b, c\), etc., the determinant is calculated using the formula:\[\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\]The determinant offers insights:

• If the determinant is zero, the matrix does not have an inverse, and thus, a unique solution for the system of equations cannot be found.
• A non-zero determinant confirms that the matrix is invertible.

In the exercise, the determinant was calculated to be \(-7\), confirming that the matrix has an inverse.

Solving Linear Systems

Solving linear systems algebraically can be straightforward with enough practice. The exercise demonstrates using the matrix inverse method. Here's how it works:

First, represent the system of linear equations as a matrix equation \(AX = B\). To find the solution vector \(X\):

• Calculate the matrix inverse \(A^{-1}\).
• Multiply both sides of the matrix equation by \(A^{-1}\) to isolate \(X\): \(X = A^{-1}B\).

This process effectively transforms the problem of solving a system of equations into a problem of performing matrix multiplication. Once you have \(X\), substitute these solutions back into the original equations to verify. Remember that having an invertible matrix (non-zero determinant) is crucial to using this method effectively. This approach is valuable in disciplines like engineering and computer science, where systems with many variables must be solved efficiently.