Problem 62 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use matrices to solve the system of equations, if possible. Use Gaussian elimination with back-substitution. $$\left\\{\begin{aligned} x+2 y &=0 \\ x+y &=6 \\ 3 x-2 y &=8 \end{aligned}\right.$$

Step-by-Step Solution

Answer

Therefore, the values of \(x\) and \(y\) that satisfy the system of equations are \(x=-12\) and \(y=-6\).

Step 1: Create the Augmented Matrix

The first step is to express the system as an augmented matrix. Each row of this matrix represents an equation and each column represents a coefficient of one of the variables. So the augmented matrix is \[\left[\begin{array}{cc|c} 1 & 2 & 0 \ 1 & 1 & 6 \ 3 & -2 & 8 \end{array}\right]\]

Step 2: Perform Row Operations

Then perform row operations to get the matrix into Row Echelon Form. It means there should be a leading 1 in the first row, and if possible, zeros below it. To achieve that, subtract row 1 from row 2 and subtract 3 times row 1 from row 3 to get, \[\left[\begin{array}{cc|c} 1 & 2 & 0 \ 0 & -1 & 6 \ 0 & -8 & 8 \end{array}\right]\] In the next step, multiply row 2 by -1 and add 8 times row 2 to row 3 to eliminate the -8 in the last row, we get \[\left[\begin{array}{cc|c} 1 & 2 & 0 \ 0 & 1 & -6 \ 0 & 0 & 0 \end{array}\right]\]

Step 3: Back-substitution

Finally, with the last row indicating 0=0, we can disregard it. Then from the second row, we can get \(y=-6\). Substituting \(y\) into the first row we get \(x=2*(-6)= -12\).

Key Concepts

Gaussian elimination

When faced with a system of linear equations, one efficient way to find a solution is by using Gaussian elimination. This method transforms the system into a simpler form, which can then be solved by a process known as back-substitution. The goal is to convert the system into an upper triangular matrix, with zeros below the leading diagonal, making it straightforward to extract the values of the unknowns.

In our example with the equations involving variables x and y, we begin by setting up the equations in matrix form. This matrix is then manipulated through a series of row operations, to form a matrix where the bottom rows become zeroes, or where the last row represents a true statement such as 0=0, indicating that the previous rows sufficiently determine the solution. A significant benefit of Gaussian elimination is its ability to identify systems with no solution or an infinite number of solutions, with minimal computation.

Row operations

Row operations are the tools we use to manipulate matrices during the Gaussian elimination process. There are three fundamental row operations:

• Swap the positions of two rows
• Multiply a row by a nonzero scalar
• Add or subtract the multiple of one row to another row

These operations are used to simplify the augmented matrix step by step while maintaining the equivalence of the system. The objective is to create zeros beneath the pivot position, which is the leading non-zero element in any row. By applying these row operations strategically, we can advance the matrix towards row echelon form. It's much like a game of strategy, where each move is calculated to simplify the system without altering the precise balance of the equations.

Augmented matrix

An augmented matrix is a compact way to represent a system of linear equations. It combines the coefficients of the variables and the constants from the right side of the equations into a single matrix that encapsulates all the necessary information. The line in the matrix separates the coefficients from the constants, echoing the equal sign in the original equations.

Constructing an augmented matrix is the first step in solving a system using Gaussian elimination. For instance, the system of equations given in the exercise is translated into an augmented matrix that visually organizes and prepares the data for the row operations that follow. Each row in the matrix corresponds to an equation, and each column corresponds to a different variable, except for the final column, which holds the constant terms. Once in augmented matrix form, a clear path unfolds for manipulating the equations as a single unit, simplifying both notation and solution steps.

Back-substitution

Once we've reduced the augmented matrix to row echelon form through Gaussian elimination, we can find the solution to the system by back-substitution. Starting from the last non-zero row, we read off the value of one variable, which we then substitute into the above rows, working our way backward.

In the example given, after the row operations, we arrive at a scenario where the system tells us directly that the value of y is -6. This newfound value can now be used in the first row to uncover the value of x. Back-substitution is the final leg of the journey, where we unravel each variable's value one by one, climbing up the row echelon form. It's like finishing a puzzle - as pieces fall into place, the picture (in this case, the solution to the system of equations) becomes clear.