Problem 68 - Precalculus

Question

apply matrix algebra to solve the system of linear equations. $$x-y+z=0$$ $$x+y+z=2$$ $$-x+2 y-3 z=1$$

Step-by-Step Solution

Answer

The solution to the system is \( x = 1 \), \( y = 0 \), \( z = 1 \).

Step 1: Set Up the Coefficient Matrix

Identify the coefficients for each variable from the equations and construct the corresponding coefficient matrix. The system of equations can be written in matrix form as: \[ \begin{bmatrix} 1 & -1 & 1 \ 1 & 1 & 1 \ -1 & 2 & -3 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 0 \ 2 \ 1 \end{bmatrix} \] The matrix of coefficients is: \[ A = \begin{bmatrix} 1 & -1 & 1 \ 1 & 1 & 1 \ -1 & 2 & -3 \end{bmatrix} \] and the vector of constants is: \[ B = \begin{bmatrix} 0 \ 2 \ 1 \end{bmatrix} \]

Step 2: Find the Inverse of Coefficient Matrix

To solve the system using matrix algebra, calculate the inverse of matrix \( A \), denoted as \( A^{-1} \). Use the formula for finding the inverse if a 3x3 matrix inverse doesn't already exist. If \( \det(A) eq 0 \), the inverse exists.

Step 3: Apply the Formula for Solving the System

Once \( A^{-1} \) is determined, solve for \( \begin{bmatrix} x \ y \ z \end{bmatrix} \) using the formula:\[ \begin{bmatrix} x \ y \ z \end{bmatrix} = A^{-1}B \] This involves matrix multiplication of \( A^{-1} \) and \( B \) to find the values of \( x \), \( y \), and \( z \).

Step 4: Calculate and Verify the Result

Perform the multiplication \( A^{-1} B \) and find the values for \( x \), \( y \), and \( z \): \[ \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} \] Verify that substituting these values into the original equations yields consistent results, confirming the solution's validity.

Key Concepts

System of Linear Equations

A system of linear equations consists of multiple linear equations that share common variables. Each equation represents a straight line, and the solution to the system is the point or points where these lines intersect. In this context, we are dealing with three equations:

• \(x - y + z = 0\)
• \(x + y + z = 2\)
• \(-x + 2y - 3z = 1\)

These equations represent planes in space. To solve the system means finding a set of values for the variables \(x\), \(y\), and \(z\) that satisfy all the equations simultaneously. Understanding the intersection of these lines in space is crucial as it visually represents how the solutions relate to one another.

Coefficient Matrix

The coefficient matrix is derived from the coefficients of the variables in the system of linear equations. For the given system, the coefficient matrix is:\[ A = \begin{bmatrix} 1 & -1 & 1 \ 1 & 1 & 1 \ -1 & 2 & -3 \end{bmatrix} \]This matrix organizes the coefficients of \(x\), \(y\), and \(z\) from each equation into rows within the matrix. Arranging these coefficients in a matrix form simplifies solving the system using matrix algebra techniques. Additionally, solving for the variables often requires manipulation of this matrix, such as finding its inverse or transforming it into a different form like row-echelon form.

Matrix Inverse

The inverse of a matrix is akin to the reciprocal of a number in scalar arithmetic. Calculating the inverse of a square matrix, denoted as \(A^{-1}\), is critical when using matrices to solve systems of linear equations. For a matrix to have an inverse, its determinant must be non-zero.

• If the determinant \(\det(A) eq 0\), the matrix is invertible.
• If \(\det(A) = 0\), the matrix does not have an inverse, indicating the system might be dependent or inconsistent.

In our context, after confirming the invertibility of matrix \(A\), its inverse \(A^{-1}\) can be calculated. Using \(A^{-1}\), we can find the solution to the equations through matrix multiplication, effectively reversing the transformation encoded by matrix \(A\). This process demonstrates the power of linear algebra in analyzing and solving complex equation systems.

Matrix Multiplication

Matrix multiplication is the process of multiplying rows of one matrix by columns of another. It is a fundamental operation in matrix algebra used to solve systems of equations and perform various transformations in data.To solve a system of equations using matrices, once the inverse of the coefficient matrix (\(A^{-1}\)) is found, it is multiplied by the vector of constants (\(B\)) to solve for the variable matrix \(\begin{bmatrix} x \ y \ z \end{bmatrix}\). The resulting product gives the values of the variables that satisfy the system:\[\begin{bmatrix} x \ y \ z \end{bmatrix} = A^{-1}B\]This computation, represented as \[A^{-1} B\], involves taking the dot product of rows of \(A^{-1}\) with the column of \(B\). This calculation yields a new vector containing the solution set for the variables \(x\), \(y\), and \(z\). Through this approach, matrix multiplication efficiently provides a powerful analytical tool to handle complex linear systems.