Question

Rewrite the system of equations in matrix form. Find the solution to the linear system by simultaneously manipulating the equations and the matrix. $$ \begin{array}{l} x+y=3 \\ 2 x-3 y=1 \end{array} $$

Short answer

The solution is \( x = 2 \) and \( y = 1 \).

Step-by-step solution

Convert the System into Matrix Form

First, identify the coefficients of the variables and the constants from the given equations. The system of equations is \( x + y = 3 \) and \( 2x - 3y = 1 \). The coefficients of \( x \) are \( 1 \) and \( 2 \), and for \( y \) are \( 1 \) and \(-3\). The constants on the right side are \( 3 \) and \( 1 \). Define matrix \( A \) for coefficients, \( X \) for variables, and \( B \) for constants: \[ A = \begin{bmatrix} 1 & 1 \ 2 & -3 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \end{bmatrix}, \quad B = \begin{bmatrix} 3 \ 1 \end{bmatrix} \] Thus, the matrix equation is: \[ AX = B \]

Express the System with Augmented Matrix

Combine the coefficient matrix with the constants by forming an augmented matrix. Write it as follows: \[ \left[ \begin{array}{cc|c}1 & 1 & 3 \2 & -3 & 1\end{array} \right] \]

Apply Row Reduction (Gaussian Elimination)

To solve, apply Gaussian elimination to the augmented matrix. Start by eliminating \( x \) from the second row. You can replace the second row with \(-2\) times the first row added to the second row: 1. Multiply the first row by \(-2\): \(-2) \times (1, 1 \mid 3) = (-2, -2 \mid -6) \) 2. Add this to the second row: \( (2, -3 \mid 1) + (-2, -2 \mid -6) = (0, -5 \mid -5) \) The new matrix is: \[ \left[ \begin{array}{cc|c}1 & 1 & 3 \0 & -5 & -5\end{array} \right] \]

Solve for Remaining Variables

Next, solve the matrix system. From the transformed matrix, the second equation is: \(-5y = -5 \), which simplifies to \( y = 1 \). Substitute \( y = 1 \) back into the first original equation: \( x + 1 = 3 \). Thus, \( x = 2 \).

Verify the Solution

Substitute \( x = 2 \) and \( y = 1 \) back into the original equations to verify: 1. First equation: \( x + y = 3 \): \( 2 + 1 = 3 \), which holds true. 2. Second equation: \( 2x - 3y = 1 \): \( 2(2) - 3(1) = 4 - 3 = 1 \), which is also true. This confirms our solution.

Key concepts

Matrix Form

Writing a system of linear equations in matrix form is a helpful method for dealing with multiple equations at once. Imagine you have a system like:

• Equation 1: \( x + y = 3 \)
• Equation 2: \( 2x - 3y = 1 \)

You start by taking the coefficients of the variables from these equations to form a matrix. This is matrix \( A \). Next, you stack your variables \( x \) and \( y \) into a column matrix, often called matrix \( X \). Finally, you also make a matrix \( B \) for the constants alone. In the end, these matrices combine in the equation \( AX = B \). This simplifies solving systems of equations with methods that work well with matrices.

Gaussian Elimination

Gaussian Elimination is a systematic method used to solve systems of linear equations. It involves two main stages: the forward elimination phase and the back substitution phase. The goal is to transform the original system of equations into an upper triangular matrix.

• **Forward Elimination**: Reduce the system step by step. You eliminate the variables in a column-wise manner, from top to bottom of your matrix.
• **Back Substitution**: Once in upper triangular form, solve for variables starting from the last equation back to the first.

This method helps in managing large systems and can be implemented easily through a sequence of algorithmic steps that leverage simple row operations on matrices.

Augmented Matrix

An augmented matrix merges the original matrices for coefficients and constants into one matrix. It serves as a practical step in matrix operations, especially when you want to perform row operations directly rather than dealing separately with equations. For example, consider the system of equations:

• Equation 1: \( x + y = 3 \)
• Equation 2: \( 2x - 3y = 1 \)

Its augmented matrix is written as:\[\left[\begin{array}{cc|c}1 & 1 & 3 \2 & -3 & 1\end{array}\right]\]Here, the vertical line separates the coefficients from the constants and allows easier application of row operations during Gaussian elimination.

Row Reduction

Row reduction, essential to Gaussian Elimination, involves performing a sequence of row operations to simplify an augmented matrix. There are three types of row operations you can utilize:

• Swapping two rows
• Multiplying a row by a nonzero scalar
• Adding a multiple of one row to another

These operations aim to move towards an upper triangular form, making it feasible to solve each equation step by step, starting from the bottom. In our example, you'd adjust the second row so that it works towards eliminating the off-diagonal elements step by step, performing calculations until achieving a zero below the leading coefficient (pivot) in the first column. This structured reduction is at the heart of solving linear equations effectively using matrices.