Question

Rewrite the system of equations in matrix form. Find the solution to the linear system by simultaneously manipulating the equations and the matrix. $$ \begin{aligned} -5 x_{1}+\quad 2 x_{3} &=14 \\ x_{2} &=1 \\ -3 x_{1}+\quad x_{3} &=8 \end{aligned} $$

Short answer

The solution is \(x_1 = -2\), \(x_2 = 1\), \(x_3 = 2\).

Step-by-step solution

Write the System in Augmented Matrix Form

To rewrite the system of equations in matrix form, we first identify the coefficients of each variable in each equation. Recognizing missing variables as having a coefficient of zero, the given system is:Equations:1. \(-5x_1 + 0x_2 + 2x_3 = 14\)2. \(0x_1 + 1x_2 + 0x_3 = 1\)3. \(-3x_1 + 0x_2 + 1x_3 = 8\)Augmented matrix:\[\begin{bmatrix}-5 & 0 & 2 & | & 14 \0 & 1 & 0 & | & 1 \-3 & 0 & 1 & | & 8\end{bmatrix}\]

Use Row Operations to Simplify the Matrix

Apply row operations to simplify the matrix to row-echelon form:1. Swap Row 2 and Row 3 to bring the equation with \(x_2\) to the bottom for simplicity:\[\begin{bmatrix}-5 & 0 & 2 & | & 14 \-3 & 0 & 1 & | & 8 \0 & 1 & 0 & | & 1\end{bmatrix}\]2. Multiply Row 1 by \(1/(-5)\) to make the leading coefficient 1:\[\begin{bmatrix}1 & 0 & -2/5 & | & -14/5 \-3 & 0 & 1 & | & 8 \0 & 1 & 0 & | & 1\end{bmatrix}\]3. Add 3 times Row 1 to Row 2 to eliminate the first element in Row 2:\[\begin{bmatrix}1 & 0 & -2/5 & | & -14/5 \0 & 0 & 1/5 & | & 2/5 \0 & 1 & 0 & | & 1\end{bmatrix}\]

Solve the Matrix Form of the System

From the simplified matrix, we can read the equations directly:1. \(x_1 - \frac{2}{5}x_3 = -\frac{14}{5}\)2. \(\frac{1}{5}x_3 = \frac{2}{5}\)3. \(x_2 = 1\)From the second equation, solve for \(x_3\):\(x_3 = 2\).Substitute \(x_3 = 2\) into the first equation to solve for \(x_1\):\[x_1 - \frac{2}{5}(2) = -\frac{14}{5}\]\[x_1 - \frac{4}{5} = -\frac{14}{5}\]\[x_1 = -\frac{14}{5} + \frac{4}{5}\]\[x_1 = -2\]

Verify the Solution

Substitute \(x_1 = -2\), \(x_2 = 1\), and \(x_3 = 2\) back into the original equations to verify:1. \(-5(-2) + 0 + 2(2) = 14\) \[10 + 4 = 14\] True2. \(1 = 1\) True3. \(-3(-2) + 0 + 2 = 8\) \[6 + 2 = 8\] TrueSince these verify, the solution is correct.

Key concepts

Matrix Form

When working with linear systems, arranging the system into a matrix form can make the problem easier to solve. Essentially, a matrix form is a compact way to express a set of equations by listing the coefficients of the variables in a rectangular array. Each row of this matrix represents an equation, while each column corresponds to the coefficients of one of the variables.

• To begin, identify the coefficient of each variable in your equations. Don't forget that missing variables have a coefficient of zero.
• For example, given the equations: \(-5x_1 + 2x_3 = 14\) \(x_2 = 1\) \(-3x_1 + x_3 = 8\) we organize them into a matrix based on the coefficients.
• Remember: the arrangement brings uniformity and clarity, allowing for more organized operations as we proceed to solve the equations.

Augmented Matrix

The augmented matrix is an essential concept when dealing with systems of linear equations. It combines the coefficient matrix and the constant matrix into one, indicating that the system is balanced.
When creating an augmented matrix:

• First, write the coefficients of the variables in a neat matrix form.
• Then, add a dashed line to separate it from the constants on the right side of the equations. This single matrix representation simplifies handling complex systems.
• For example, the given linear system transforms into: \[ \begin{bmatrix} -5 & 0 & 2 & | & 14 \ 0 & 1 & 0 & | & 1 \ -3 & 0 & 1 & | & 8 \end{bmatrix} \]

Why go to all this effort? Simply put, an augmented matrix streamlines calculations and operations, making it easier to visualize and manipulate the system as a whole.

Row Operations

Row operations are critical tools to manipulate matrices and are used to simplify them into a form that aids in determining the solutions of the system. These operations include swapping rows, scaling rows by a non-zero constant, and adding or subtracting scaled rows.
During the row operations:

• Swap rows if needed to position an equation in a way that's easier to solve, as was done by swapping rows 2 and 3.
• Scale rows to make coefficients of leading variables equal to one, simplifying equations—illustrated by dividing the first row by \(-5\).
• Add or subtract rows to eliminate coefficients from below the leading variable, making it easier to deduce values. For example, multiply the first row by 3 and add it to the second row to null the coefficient below.

Row operations allow us to transform the matrix into a row-echelon or even reduced row-echelon form, which directly showcases the information needed to find the solutions.

Solution of Linear Systems

The solution of a linear system involves finding the values of the variables that fulfill all equations involved. Once a matrix is simplified using row operations, it becomes straightforward to interpret the results from this simplified form.
How do you extract these solutions?

• From the simplified matrix, rewrite the equivalent linear equations.
• Solve for each variable from bottom to top, a method called back substitution, as demonstrated by solving for \(x_3\) and substituting into previous equations to find \(x_1\).
• Verify the solution by substituting back into the original equations. This step ensures the derived values truly satisfy the original system.

Through a combination of augmented matrices, row operations, and simplification, the solution of a linear system becomes much more accessible and less prone to error.