Problem 54 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use the matrix capabilities of a graphing utility to solve (if possible) the system of linear equations. $$\left\\{\begin{array}{l} 2 x+3 y+5 z=4 \\ 3 x+5 y-9 z=7 \\ 5 x+9 y+17 z=13 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution for the system of equations is \(x\), \(y\), and \(z\) given by the back-substitution in step 4.

Step 1: Create an Augmented Matrix

First transform the system of equations into an augmented matrix. From the given system, the augmented matrix becomes \(\begin{bmatrix} 2 & 3 & 5 & 4 \ 3 & 5 & -9 & 7 \ 5 & 9 & 17 & 13 \end{bmatrix}\)

Step 2: Simplify Augmented Matrix to Row-Echelon Form

Now, simplify the matrix into row-echelon form using elementary row operations. Subtract the appropriate multiples of the first row from the other two to eliminate \(x\) in the second and third rows.

Step 3: Continue Simplifying Matrix

Next, use similar operations to eliminate \(y\) in the third row. Now, the variables can be solved by back-substitution.

Step 4: Solve for \(x\), \(y\), and \(z\)

From the simplified matrix, one can directly read off the solutions or explicitly solve for the variables \(x\), \(y\), and \(z\). The last equation from the simplified matrix will give the value of \(z\). Substituting this value in the second equation will yield the value for \(y\), and these two values substituted into the first equation will yield the value for \(x\).

Key Concepts

Augmented Matrix

An augmented matrix is an essential tool to transform a system of linear equations into a more manageable form. When given a system of equations such as: \[ \begin{array}{l} 2x + 3y + 5z = 4 \ 3x + 5y - 9z = 7 \ 5x + 9y + 17z = 13 \end{array} \] we translate it into a matrix form without variables. This matrix includes coefficients of the variables and constants from the equations. Constructing an augmented matrix allows us to handle equations numerically and apply systematic operations. For the given problem, the augmented matrix looks like this: \[ \begin{bmatrix} 2 & 3 & 5 & 4 \ 3 & 5 & -9 & 7 \ 5 & 9 & 17 & 13 \end{bmatrix} \] where the last column represents the constants from each equation. By structuring it this way, we create a format suitable for further operations that simplify finding solutions.

Matrix Row-Echelon Form

Turning an augmented matrix into a row-echelon form is a two-step magic trick in solving systems of equations. Simply put, this form resembles a staircase pattern or triangular form where every leading entry (the first non-zero number from the left) in a row is to the right of the leading entry of the row above. This structure makes it easier to solve variables step by step, starting from the last row.
In our example, once we have our augmented matrix, we perform a sequence of operations to zero out coefficients below the first row's leading entry, starting from the second. The process continues down the matrix until it reaches the desired echelon form.
This form usually looks like this: \[ \begin{bmatrix} a & b & c & d \ 0 & e & f & g \ 0 & 0 & h & i \end{bmatrix} \]
Ensuring the rows are staggered in this manner, we enable solving the system using back-substitution.

Elementary Row Operations

Elementary row operations are the steps we take to manipulate our augmented matrix into a row-echelon form. These operations are like the magic moves that change the matrix without altering the solutions to the equations it represents. They include:

• **Row Switching:** Swapping two rows if it's advantageous to do so.
• **Row Multiplication:** Multiplying all entries of a row by a non-zero scalar to make numbers simpler or to prepare for row addition.
• **Row Addition:** Adding or subtracting one row from another to eliminate variables, particularly aimed at making certain numbers become zero.

In our problem, we could subtract a multiple of the first row from the second and third rows to zero out the first variable's coefficients below the diagonal. Each step must be chosen carefully to steadily work towards the triangle-like row-echelon structure, and ultimately to solve the system through back-substitution.