Question

Rewrite the system of equations in matrix form. Find the solution to the linear system by simultaneously manipulating the equations and the matrix. $$ \begin{aligned} -5 x_{2}+2 x_{3} &=-11 \\ x_{1}\quad+2 x_{3} &=15 \\ -3 x_{2}+x_{3} &=-8 \end{aligned} $$

Short answer

The solution is \(x_1 = 31\), \(x_2 = 1\), \(x_3 = -8\).

Step-by-step solution

Write the System as Augmented Matrix

First, rewrite the given system of equations in matrix form. For the system:\[\begin{align*}-5x_2 + 2x_3 &= -11 \ x_1 + 2x_3 &= 15 \ -3x_2 + x_3 &= -8 \end{align*}\]We identify the coefficients of each variable:\[\begin{bmatrix} 0 & -5 & 2 \ 1 & 0 & 2 \ 0 & -3 & 1 \end{bmatrix}\begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} -11 \ 15 \ -8 \end{bmatrix}\]

Use Row Operations to Solve the System

To find the solution, perform operations to achieve the reduced row-echelon form:1. Start with swapping Row 2 and Row 1 to have a non-zero leading term in the first row.2. Apply row operations like addition/multiplication to get leading 1s and eliminate other entries in each column.Using operations, we end with:\[\begin{bmatrix} 1 & 0 & 2 \ 0 & 1 & -\frac{7}{3} \ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} 15 \ 3 \ -8 \end{bmatrix}\]

Read the Solution from the Matrix

From the reduced matrix, we have:- \(x_1 + 2x_3 = 15\)- \(x_2 - \frac{7}{3}x_3 = 3\)- \(x_3 = -8\)Substitute \(x_3 = -8\) into the other equations to find \(x_1\) and \(x_2\).

Substitute and Solve for Each Variable

1. Substitute \(x_3 = -8\) into the equation for \(x_1\):\[x_1 + 2(-8) = 15 \implies x_1 = 31\]2. Substitute \(x_3 = -8\) into the equation for \(x_2\):\[x_2 - \frac{7}{3}(-8) = 3 \implies x_2 = \frac{3}{3} = 1\]The solution to the system is \(x_1 = 31\), \(x_2 = 1\), and \(x_3 = -8\).

Key concepts

Linear Equations

Linear equations are the foundation of algebra and are equations of the first degree. This means that the variables in linear equations are not raised to any power other than one. They can be seen as straight lines when graphed on a coordinate plane. In matrix algebra, systems of linear equations are often used to represent complex problems in various fields such as physics and engineering.

A typical system of linear equations looks like this:

• \( ax + by + cz = d \)
• \( ex + fy + gz = h \)
• \( ix + jy + kz = l \)

Here, \(a, b, c, e, f, g, i, j,\) and \(k\) are coefficients, \(x, y, z\) are variables, and \(d, h, l\) are constants. Solving these systems means finding the values of the variables that make all the equations true simultaneously.

Linear equations can represent anything from simple everyday problems to complex engineering challenges. Understanding them is key to diving into more advanced mathematical concepts.

Augmented Matrix

An augmented matrix is a tool used to represent a system of linear equations in a more compact form. Instead of writing out each equation, we place numbers into a matrix, which consists of rows and columns that correspond to the coefficients and constants of each equation.

For example, if we have the system:

• \(-5x_2 + 2x_3 = -11 \)
• \(x_1 + 2x_3 = 15 \)
• \(-3x_2 + x_3 = -8 \)

We can express this as an augmented matrix:

• \[ \begin{bmatrix} 0 & -5 & 2 & | & -11 \ 1 & 0 & 2 & | & 15 \ 0 & -3 & 1 & | & -8 \end{bmatrix} \]

The line separates the coefficients of the variables from the constants of the equations. This format allows us to use matrix techniques to find solutions, simplifying the process. Using an augmented matrix is especially useful when dealing with larger systems of equations.

Row Operations

Row operations are techniques used on matrices to simplify and solve systems of linear equations. They help convert a matrix into a form that is easier to interpret and solve, such as reduced row-echelon form. In essence, they are the steps through which we manipulate rows to achieve desired outcomes, like having leading 1s in a row or zeroes below leading coefficients.

There are three primary types of row operations:

• **Swapping two rows**: Useful when a row with a zero leading term can be shifted for a non-zero one.
• **Multiplying a row by a non-zero constant**: This helps in scaling the row to achieve the desired leading coefficient.
• **Adding or subtracting a multiple of one row to another row**: Aids in eliminating unwanted coefficients.

Applying these operations correctly will transform your matrix step by step until it arrives at reduced row-echelon form. This approach provides a clear path to the solution of the system. Understanding row operations is fundamental in manipulating and solving matrices.

Reduced Row-Echelon Form

The reduced row-echelon form (RREF) of a matrix is a simplified version of the matrix that makes it very easy to read off the solutions to the original system of linear equations. It's the end goal of performing row operations on an augmented matrix.

A matrix in RREF has certain characteristics:

• Every leading entry in a row is 1.
• Leading 1s are the only non-zero entries in their respective columns.
• Leading 1s appear to the right of leading 1s in previous rows.
• Rows containing only zeros, if any, are at the bottom of the matrix.

In the context of our example system, achieving the RREF allowed us to easily substitute values back into the equations to find solutions like \(x_3 = -8\), then back-track to determine \(x_1\) and \(x_2\). Once you have a matrix in RREF, the system of equations can be solved quickly and efficiently. RREF simplifies finding the solution and is a powerful tool in matrix algebra.