Problem 16 - Precalculus: Functions and Graphs

Question

Use matrices to solve the system. $$\left\\{\begin{aligned} 5 x+2 y-z &=10 \\ y+z &=-3 \end{aligned}\right.$$

Step-by-Step Solution

Answer

Example solution: \( x = 3.2, y = -3, z = 0 \).

Step 1: Write the System in Matrix Form

The given system of equations can be represented in matrix form as \( A\mathbf{x} = \mathbf{b} \), where \( A \) is the coefficient matrix, \( \mathbf{x} \) is the variable vector, and \( \mathbf{b} \) is the constant vector. The equations are:1. \( 5x + 2y - z = 10 \)2. \( y + z = -3 \)So, \( A = \begin{bmatrix} 5 & 2 & -1 \ 0 & 1 & 1 \end{bmatrix} \), \( \mathbf{x} = \begin{bmatrix} x \ y \ z \end{bmatrix} \), and \( \mathbf{b} = \begin{bmatrix} 10 \ -3 \end{bmatrix} \).

Step 2: Augment the Matrix

Combine the coefficient matrix \( A \) and the constant vector \( \mathbf{b} \) to form an augmented matrix \( [A|\mathbf{b}] \).The augmented matrix is:\[ \begin{bmatrix} 5 & 2 & -1 & | & 10 \ 0 & 1 & 1 & | & -3 \end{bmatrix} \]

Step 3: Employ Row Reduction Techniques

Use row operations to reduce the augmented matrix to row-echelon form or reduced row-echelon form. 1. Start with the existing matrix: \[ \begin{bmatrix} 5 & 2 & -1 & | & 10 \ 0 & 1 & 1 & | & -3 \end{bmatrix} \]2. Row 2 is already in a convenient form for \( y \), so no modification is needed here.3. Substitute the value of \( y \) from Row 2 into Row 1. - From Row 2, \( y = -3 - z \). - Substituting in Row 1: \( 5x + 2(-3 - z) - z = 10 \). - Simplify: \( 5x - 6 - 2z - z = 10 \) gives \( 5x - 3z = 16 \).

Step 4: Solve for Remaining Variables

Now solve the set of simplified equations.1. From the simplified row 1: \( 5x - 3z = 16 \) - Rearrange to express \( x \) in terms of \( z \): \( x = \frac{16 + 3z}{5} \)2. From row 2: \( y = -3 - z \) 3. Use a specific value of \( z \) (e.g., \( z = 0 \)) to find one possible solution quickly. - If \( z = 0 \): - \( y = -3 \) - \( x = \frac{16 + 3(0)}{5} = \frac{16}{5} = 3.2 \)

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more equations with multiple variables. These equations are called "linear" because their graphs produce straight lines. By solving a system of equations, we determine the values of the variables that satisfy all the equations simultaneously.
In the given system:

• The first equation is \(5x + 2y - z = 10\).
• The second equation is \(y + z = -3\).

Here, \(x\), \(y\), and \(z\) are the variables, and these two equations must be satisfied together to find their corresponding values. This forms the backbone of solving any linear system.

Row Reduction Techniques

Row reduction techniques are used to simplify matrices into a form that makes it easier to solve linear systems.
There are different forms that a matrix can be reduced to, such as row-echelon form or reduced row-echelon form. Here, we use elementary row operations which include:

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

For the given problem:
We start with the matrix:
\[\begin{bmatrix}5 & 2 & -1 & | & 10 \0 & 1 & 1 & | & -3\end{bmatrix} \]We notice that the second row is already convenient for solving \(y\). Thus, we substitute the expression for \(y\) from this row into the first row, facilitating finding \(x\) and \(z\). This highlights the efficiency of row reduction techniques.

Augmented Matrix

An augmented matrix combines the coefficient matrix and the constant vector of the system of equations into a single matrix. This is a critical part of the process when using matrices to solve systems, as it simplifies handling the equations all at once.
For instance, for our equations:

• The coefficient matrix \(A\) for the variables \([x, y, z]\) is \(\begin{bmatrix}5 & 2 & -1 \0 & 1 & 1\end{bmatrix} \).
• The constant vector \(\mathbf{b}\) is \(\begin{bmatrix}10 \-3\end{bmatrix} \).

When combined, they create the augmented matrix:\[\begin{bmatrix}5 & 2 & -1 & | & 10 \0 & 1 & 1 & | & -3\end{bmatrix}\]By using the augmented matrix, we perform computations in a more structured way and apply row reductions efficiently.

Variable Substitution in Matrices

Variable substitution in matrices involves expressing one variable in terms of others to reduce the complexity of the equations. It plays a vital role in the process of solving linear systems by helping to decompose more complex equations.
In the given exercise, after forming and reducing the augmented matrix, we use:

• From the second row: \(y = -3 - z\).
• Substitute \(y\) from the second equation into the first equation.
• This substitution simplifies the equations, aiding in quickly finding solutions for \(x\) and \(z\).

This technique is especially useful when one equation is simpler than others, as seen here where \(y\) was directly expressed and substituted, leading to simplified calculations.