Problem 26 - Precalculus: Mathematics for Calculus

Question

A matrix is given in row-echelon form. (a) Write the system of equations for which the given matrix is the augmented matrix. (b) Use back-substitution to solve the system. $$\left[\begin{array}{rrrr} 1 & 1 & -3 & 8 \\ 0 & 1 & -3 & 5 \\ 0 & 0 & 1 & -1 \end{array}\right]$$

Step-by-Step Solution

Answer

The solutions are \( x = 3 \), \( y = 2 \), and \( z = -1 \).

Step 1: Interpret the Matrix as a System of Equations

The given matrix represents the system of equations, where each row corresponds to an equation. The entries in the last column are the constants on the right side of the equation.- First row: \( x + y - 3z = 8 \)- Second row: \( y - 3z = 5 \)- Third row: \( z = -1 \)

Step 2: Substitute the Known Value of z

From the third equation, we know that \( z = -1 \). Substitute \( z = -1 \) into the second equation:\[ y - 3(-1) = 5 \]Simplify to find \( y \):\[ y + 3 = 5 \] \[ y = 2 \]

Step 3: Substitute Known Values to Find x

Now that we know \( z = -1 \) and \( y = 2 \), substitute these into the first equation:\[ x + y - 3z = 8 \]\[ x + 2 - 3(-1) = 8 \]Simplify to solve for \( x \):\[ x + 2 + 3 = 8 \]\[ x + 5 = 8 \]\[ x = 3 \]

Key Concepts

Back-Substitution

Back-substitution is a process used to solve systems of equations that have been transformed into a row-echelon form, usually for linear equations. The row-echelon form makes it easier to find solutions by focusing on the last row first, which often has the simplest equation. We solve this equation for one variable and then substitute this value back into the previous equations to solve for the remaining variables. This step-by-step process continues backward until all variables have been determined.
In our example, we begin with the third equation, which directly gives us the value of the variable \( z \) as \( -1 \).
Then, using this known value, we move to the second row, substitute \( z \) and solve for \( y \). Finally, with both \( y \) and \( z \) resolved, we substitute back into the first equation to find \( x \). This backward approach sequentially narrows down and solves each variable efficiently.

Augmented Matrix

An augmented matrix is a key element in solving linear systems of equations. It combines the coefficients of variables and constants from a system of equations into a compact, structured form. This representation simplifies both the process of solving the equations and aids in visualizing them. In an augmented matrix, each row corresponds to an equation, with the last column representing the constant terms of the equations.
Take the given row-echelon form matrix:

• First row represents: \( x + y - 3z = 8 \)
• Second row becomes: \( y - 3z = 5 \)
• Third row simplifies to: \( z = -1 \)

These rows, therefore, map directly to the system of equations that can be interpreted and solved using back-substitution. The power of the augmented matrix lies in its ability to concisely package complex information for easier manipulation and solving.

System of Equations

A system of equations consists of multiple equations that share common variables. The goal is to find values for these variables that satisfy all equations simultaneously. Systems can vary in complexity, ranging from two linear equations with two unknowns to much larger systems with several equations and unknowns.
In a practical scenario, these systems often relate to real-world problems requiring precise solutions. For example, systems of equations can help determine equilibrium points in physics, balanced financial portfolios, or even logistical planning in operations research.
In the given exercise, the system of equations represented by the provided augmented matrix includes three linear equations involving variables \( x \), \( y \), and \( z \). The challenge is finding values for these variables that satisfy all three equations, which was effectively done using methods like substitution once transformed to row-echelon form through an augmented matrix.