Problem 26 - Algebra and Trigonometry

Question

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{aligned} x &-3 z=-2 \\ 2 x+2 y+z &=4 \\ 3 x+y-2 z &=5 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solutions to the system of equations are: \(x = 1\), \(y = -0.2\), and \(z = 1.2\)

Step 1: Setup the Augmented Matrix

To set up the augmented matrix, write down the coefficients of each variable in each equation and the constant that each equation is equal to. For the given system of equations the augmented matrix is: \[\begin{pmatrix} 1 & 0 & -3 & -2 \ 2 & 2 & 1 & 4 \ 3 & 1 & -2 & 5 \end{pmatrix} \]

Step 2: Perform Gaussian Elimination

The aim here is to create a matrix that has a triangular form, i.e., all entries below the main diagonal are zero. Begin by swapping row 1 and row 2, and then add (-2) times the new row 1 to row 2, and (-3) times the new row 1 to row 3. The new matrix becomes:\[\begin{pmatrix} 2 & 2 & 1 & 4 \ 0 & -2 & -5 & -10 \ 0 & -5 & -5 & -7 \end{pmatrix}\]Next, swap the second row and third row, then add times (5/2) the second row to third row to create a zero in the third row's second column. The result is:\[\begin{pmatrix} 2 & 2 & 1 & 4 \ 0 & -5 & -5 & -7 \ 0 & 0 & -5 & -6 \end{pmatrix} \]

Step 3: Perform Back-substitution

Perform back-substitution to solve for the variables. This involves dividing each row by the coefficient of the leading variable (to make it equal to 1) and then substituting back into the previous rows. Hence, the final matrix is:\[\begin{pmatrix} 1 & 1 & 0.5 & 2 \ 0 & 1 & 1 & 1.4 \ 0 & 0 & 1 & 1.2 \end{pmatrix} \]Therefore, from the final matrix, the solutions to the system of equations are: \(x = 1\), \(y = -0.2\), and \(z = 1.2\).

Key Concepts

Matrix Algebra

Matrix algebra involves using matrices to solve systems of linear equations. This process is efficient, especially when dealing with complex systems involving several variables. A matrix is essentially a grid of numbers that represents equations. In the given exercise, we start by forming an augmented matrix using the coefficients from the system of equations. This matrix helps us methodologically organize and visualize the data.

In matrix algebra, the augmented matrix is key to performing operations like Gaussian and Gauss-Jordan elimination. It's set up by writing down each equation’s coefficients, including the constants on the right side of the equations. For example, the system:

• \( x - 3z = -2 \)
• \( 2x + 2y + z = 4 \)
• \( 3x + y - 2z = 5 \)

can be represented as the matrix:\[\begin{pmatrix}1 & 0 & -3 & -2 \2 & 2 & 1 & 4 \3 & 1 & -2 & 5\end{pmatrix}\] This matrix embodies the structure needed to carry out further elimination methods.

Back-Substitution

Back-substitution is a critical step when solving a system of equations using Gaussian elimination. Once a matrix has been transformed into an upper triangular form, back-substitution is used to find the values of unknown variables.

When a matrix looks mainly like this:

• All entries below its main diagonal are zeros.
• The remaining entries are either zeros or form a descending "staircase" from left to right.

Back-substitution enables us to start solving from the bottom row of the matrix. Let's consider the final upper triangular matrix obtained:\[\begin{pmatrix}2 & 2 & 1 & 4 \0 & -5 & -5 & -7 \0 & 0 & -5 & -6\end{pmatrix}\] Here, you would start from the third equation, solving for \(z\) first, and then substitute this value back up through the preceding equations to find \(y\) and \(x\). This approach allows for efficient computation, unraveling one variable at a time until the entire system is resolved.

Gauss-Jordan Elimination

Gauss-Jordan elimination is a refinement of Gaussian elimination that simplifies any matrix to reduced row-echelon form (RREF). Unlike Gaussian elimination, which provides an upper triangular matrix, Gauss-Jordan strives for an even simpler matrix where every column containing a leading 1 has zeros in all other positions.

Transforming a matrix using Gauss-Jordan elimination involves similar steps:

• Reduce matrix to upper triangular form.
• Normalize each leading coefficient to 1 by division.
• Eliminate above and below each leading 1, ensuring that the rest of the column is zero.

With the given example, after performing Gaussian elimination, additional operations are performed to simplify the matrix into:\[\begin{pmatrix}1 & 1 & 0.5 & 2 \0 & 1 & 1 & 1.4 \0 & 0 & 1 & 1.2\end{pmatrix}\]From here, each variable can be directly read off without needing back-substitution. This makes Gauss-Jordan elimination a powerful method for achieving a straightforward solution, especially when programming automated systems to solve linear equations directly.