Problem 62 - Precalculus with Limits

Question

In Exercises 57-64, (a) write the system of linear equations as a matrix equation, \(AX\ =\ B\), and (b) use Gauss-Jordan elimination on the augmented matrix \([A\ \vdots\ B]\) to solve for the matrix \(X\). \(\begin{cases} x_1 + x_2 - 3x_3= -1 \\ -x_1 + 2x_2 = 1 \\ x_1 - x_2 + x_3 = 2 \end{cases}\)

Step-by-Step Solution

Answer

The solution to the system represented by our matrix equation is \(X = \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix}\).

Step 1: Write the Matrix Equation

Represent the system of equations in matrix form. Here, we have \(A = \begin{bmatrix} 1 & 1 & -3 \ -1 & 2 & 0 \ 1 & -1 & 1\end{bmatrix}\), \(X = \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix}\) and \(B = \begin{bmatrix} -1 \ 1 \ 2 \end{bmatrix}\). So, the matrix equation will be \(AX = B\).

Step 2: Form the Augmented Matrix

Next, form the augmented matrix \([A | B]\) by appending the B matrix to A. So we have \([A | B] = \begin{bmatrix} 1 & 1 & -3 & | & -1 \ -1 & 2 & 0 & | & 1 \ 1 & -1 & 1 & | & 2 \end{bmatrix}\)

Step 3: Perform Gauss-Jordan Elimination

We need to perform row operations to convert the matrix to row reduced echelon form (RREF). The final RREF matrix is \(\begin{bmatrix} 1 & 0 & 0 & | & 1 \ 0 & 1 & 0 & | & -1 \ 0 & 0 & 1 & | & 1 \end{bmatrix}\), achieved by swapping rows, multiplying rows by scalars, and adding multiples of one row to another.

Step 4: Extract the Solution

The resulting matrix tells us that \(x_1 = 1\), \(x_2 = -1\) and \(x_3 = 1\). We can represent that as a matrix: \(X = \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix}\).

Key Concepts

System of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. In our example, the system consists of three equations:

• \(x_1 + x_2 - 3x_3= -1\)
• \(-x_1 + 2x_2 = 1\)
• \(x_1 - x_2 + x_3 = 2\)

The goal is to find the values of the variables \(x_1, x_2,\) and \(x_3\) that satisfy all three equations simultaneously. Solving such systems can be done using methods like substitution, elimination, or matrix operations. When dealing with larger systems, or systems that can be represented in a more standardized format, using matrices is particularly useful. This allows us to apply linear algebra techniques to find a solution efficiently.

Augmented Matrix

An augmented matrix is an essential tool in solving systems of linear equations using matrices. This special type of matrix combines both the coefficient and the constant terms from the equations into a single matrix. For example, from our original system of equations, we form the matrices:

• Coefficient matrix \(A = \begin{bmatrix} 1 & 1 & -3 \ -1 & 2 & 0 \ 1 & -1 & 1 \end{bmatrix}\)
• Constant matrix \(B = \begin{bmatrix} -1 \ 1 \ 2 \end{bmatrix}\)

The augmented matrix \([A | B]\) is then created by appending matrix \(B\) to matrix \(A\):\[[A | B] = \begin{bmatrix} 1 & 1 & -3 & | & -1 \ -1 & 2 & 0 & | & 1 \ 1 & -1 & 1 & | & 2 \end{bmatrix}\]This formation makes it easier to apply row operations needed for methods such as Gauss-Jordan elimination. In essence, the augmented matrix holds all necessary data to solve for the unknowns simultaneously.

Row Reduced Echelon Form

Row reduced echelon form (RREF) is the result of applying certain row operations to simplify an augmented matrix. The goal is to transform the matrix in such a way that the system of equations it represents becomes trivially solvable. When a matrix is in RREF, each leading entry in a row is '1', all elements below and above these entries are '0', and each leading '1' is to the right of the leading '1' in the previous row. To achieve this form, you can perform operations such as:

• Swapping rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting a multiple of one row from another

Using Gauss-Jordan elimination, the augmented matrix from our example problem was transformed into:\[\begin{bmatrix} 1 & 0 & 0 & | & 1 \ 0 & 1 & 0 & | & -1 \ 0 & 0 & 1 & | & 1 \end{bmatrix}\]This matrix directly gives the solutions to the variables: \(x_1 = 1\), \(x_2 = -1\), and \(x_3 = 1\). Using RREF ensures a clear and straightforward path to finding solutions to more complex systems of linear equations.