Question

Rewrite the system of equations in matrix form. Find the solution to the linear system by simultaneously manipulating the equations and the matrix. $$ \begin{array}{l} 2 x+4 y=10 \\ -x+y=4 \end{array} $$

Short answer

The solution is \( x = -1, y = 3 \).

Step-by-step solution

Write the System of Equations

We have the following system of equations: 1) \( 2x + 4y = 10 \) 2) \( -x + y = 4 \)

Express in Matrix Equation Form

Rewriting the system in matrix form, we have: \[\begin{bmatrix}2 & 4 \-1 & 1\end{bmatrix}\begin{bmatrix}x \y\end{bmatrix} = \begin{bmatrix}10 \4\end{bmatrix}\] This results in the matrix form \( Ax = b \), where \( A \) is the coefficient matrix, \( x \) is the variable vector, and \( b \) is the constant vector.

Solve Using Row Operations

We will use Gaussian elimination to manipulate the matrix until it is in row-echelon form. Begin by adding row 1 multiplied by 0.5 to row 2.The matrix becomes:\[\begin{bmatrix}2 & 4 \0 & 3\end{bmatrix}\begin{bmatrix}x \y\end{bmatrix} = \begin{bmatrix}10 \9\end{bmatrix}\]

Back Substitution

Substitute back to find values of variables:From the second row, we have: \( 3y = 9 \), thus \( y = 3 \).Substitute \( y = 3 \) into the first equation:\( 2x + 4(3) = 10 \) which simplifies to \( 2x + 12 = 10 \). Simplifying gives \( 2x = -2 \), so \( x = -1 \).

Solution Verification

Substitute \( x = -1 \) and \( y = 3 \) back into the original equations to verify:For the first equation, \( 2(-1) + 4(3) = -2 + 12 = 10 \) (True).For the second equation, \( -(-1) + 3 = 1 + 3 = 4 \) (True).Both equations are satisfied, confirming the solution as \( x = -1, y = 3 \).

Key concepts

Linear Equations

Linear equations are equations of the first degree, meaning they involve variables raised only to the power of one. In this particular exercise, we have two linear equations: \( 2x + 4y = 10 \) and \( -x + y = 4 \). These equations describe straight lines in a two-dimensional space when plotted on a graph.

Linear equations are useful in finding relationships between variables. Often, they are used in real-world contexts, such as calculating costs, distances, or other measurements that involve a constant rate of change. Understanding how to manipulate and solve these equations is a pivotal skill in algebra and mathematics in general.

• Each equation can be represented graphically as a line.
• The solution to a system of linear equations is the point where all equations intersect.
• Linear systems can have one solution (intersect at a single point), no solution (parallel lines), or infinitely many solutions (same line).

Understanding linear equations lays the foundation for more complex algebraic concepts.

Matrix Form

Matrix form allows for the compact representation of systems of linear equations. By rewriting the system \( 2x + 4y = 10 \) and \( -x + y = 4 \) in matrix form, we create a more structured way of handling the equations. This is done using:

• The coefficient matrix \( A = \begin{bmatrix} 2 & 4 \ -1 & 1 \end{bmatrix} \).
• The variable vector \( x = \begin{bmatrix} x \ y \end{bmatrix} \).
• The constant vector \( b = \begin{bmatrix} 10 \ 4 \end{bmatrix} \).

In matrix form, the system is represented as \( Ax = b \). This form is beneficial for both symbolic and numerical calculations. It serves as the foundation for various algebraic operations and for applying methods such as Gaussian elimination to solve systems of equations. By transforming linear equations into matrix form, we standardize the process and simplify it for computational solutions. It allows easy manipulation through various matrix operations.

Gaussian Elimination

Gaussian elimination is a method used to solve systems of linear equations by transforming the coefficient matrix into a simpler form called row-echelon form. This simplification makes it easier to derive the solution through back substitution.

The steps of Gaussian elimination involve using row operations to form zeros below the main diagonal of the matrix. This particular exercise uses such row operations:

• Adding the first row multiplied by 0.5 to the second row.
• Resulting in the new matrix \( \begin{bmatrix} 2 & 4 \ 0 & 3 \end{bmatrix} \).

• These operations do not change the solution to the equations, they only simplify the form.
• The ultimate goal is to turn the matrix into an easier structure to work with, particularly row-echelon form.

By the end of Gaussian elimination, the matrix is set up for straightforward back substitution, allowing you to easily find the variable values.

Row-Echelon Form

Row-echelon form is a special type of matrix used in solving linear systems. Its form is triangular, with zeros below the diagonal. The transformation to a row-echelon form simplifies the process of finding solutions using back substitution.

For the given exercise, the matrix becomes \( \begin{bmatrix} 2 & 4 \ 0 & 3 \end{bmatrix} \) through Gaussian elimination. This form is particularly useful because:

• The lower portion of non-diagonal elements are zeros. Thus, solving from bottom to top (backwards) becomes feasible.
• The solution becomes clearer, as each step of substitution resolves one variable.
• It significantly reduces complexity and calculation time.

Once in row-echelon form, begin back substitution from the last row up to the first. From the last row in this matrix, solve for \( y \) and use its value to find \( x \) in the first row. This streamlined structure is perfect for solving these kinds of systems and is widely used in both theoretical and applied mathematics.