Problem 52 - Algebra and Trigonometry Real Mathematics, Real People

Question

Write the system of linear equations represented by the augmented matrix. Then use back-substitution to find the solution. (Use the variables \(x, y,\) and \(z,\) if applicable.) $$\left[\begin{array}{ccccc} 1 & 0 & -2 & \vdots & -7 \\ 0 & 1 & 1 & \vdots & 9 \\ 0 & 0 & 1 & \vdots & -3 \end{array}\right]$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = -13\), \(y = 12\), \(z = -3\)

Step 1: Convert the matrix to a system of equations

The given matrix represents the following system of linear equations: \[\begin{align*} x - 2z &= -7 \ y + z &= 9 \ z &= -3 \end{align*}\]Each row of the matrix corresponds to one equation with the numbers on the left-hand-side of the column-bar being the coefficients of the variables \(x, y, z\) (in this order) and the number on the right-hand side of the bar being the constant term of the equation.

Step 2: Use back-substitution to solve for the variables

We start from the last equation, substituting the value of \(z = -3\) into the second equation. This gives us\[\begin{align*} y + (-3) &= 9 \ y &= 9 + 3 \ y &= 12 \end{align*}\]After we have found \(y = 12\), we substitute the value of \(z\) and \(y\) into the first equation to find the value of \(x\):\[\begin{align*} x - 2(-3) &= -7 \ x + 6 &= -7 \ x &= -7 - 6 \ x &= -13 \end{align*}\]

Key Concepts

Augmented Matrix

An augmented matrix is a powerful tool used in linear algebra to solve systems of equations. It's a rectangular array of numbers organized by rows and columns. Each row corresponds to an equation and each column to a coefficient or constant of the system.

The matrix in the exercise is augmented because it includes both the coefficients of the variables and the constants from the right-hand side of the equations. The vertical dot (\(\vdots\)) within the matrix separates the coefficients from these constants:

• The first part on the left includes the coefficients of variables \(x, y,\) and \(z.\)
• The numbers on the right, after the dot, are the constants of each equation.

Transforming a system of equations into an augmented matrix allows us to manipulate it more conveniently with linear algebra techniques. This in turn, makes it easier to find solutions.

Back-Substitution

Back-substitution is a method used to solve a system of equations that has been transformed into a row-echelon form or an upper triangular form. When you have a system in these forms, it becomes possible to find the solution starting from the last equation and moving upwards.

In our example, we start with the third row of the augmented matrix, which is an equation in terms of \(z:\) \[ z = -3 \]Once \(z\) is known, you substitute its value into the second equation (second row):

• Replace \(z\) in the equation \( y + z = 9 \)
• This becomes \( y + (-3) = 9 \)
• Solve for \(y,\) finding that \( y = 12 \).

With \(y\) and \(z\) determined, substitute these back into the first equation (first row), \( x - 2z = -7 \), allowing us to solve for \(x.\)
This technique of starting from the bottom and working upwards makes it more efficient to find the values of all variables, following a structured path to the solution.

System of Equations

A system of equations involves multiple linear equations with shared variables. Solving these systems means finding values for the variables that satisfy all given equations simultaneously. In our scenario:

• The system consists of three equations.
• Each equation corresponds to a row in our augmented matrix.

These equations can be written as:

\[ \begin{align*} x - 2z &= -7 \ y + z &= 9 \ z &= -3 \end{align*} \]
Each equation adds a layer of constraints, narrowing down the possible solutions to where all conditions meet.

When solving, you often encounter various methods like substitution, elimination, and matrix manipulation. In this case, the matrix approach offers a structured and efficient way of tackling the equations, especially when paired with techniques like back-substitution.