Introduction to Matrices: Videos & Practice Problems

Understanding matrices is essential for solving systems of equations. A matrix organizes coefficients into a grid format, with row operations like swapping, multiplying, and adding rows to manipulate the data. The goal is to achieve row echelon form, characterized by ones along the diagonal and zeros below, or reduced row echelon form, which also requires zeros above the diagonal. This process, known as Gaussian elimination, simplifies solving equations, allowing for straightforward back substitution to find variable values.

Introduction to Matrices

In mathematics, a matrix is a structured way to organize numbers into a grid format, consisting of rows and columns. Understanding matrices begins with recognizing their dimensions, which are defined by the number of rows and columns they contain. For instance, a matrix with 2 rows and 3 columns is referred to as a 2 by 3 matrix.

When dealing with systems of equations, matrices provide a compact representation of the coefficients and constants involved. This representation is known as an augmented matrix. In an augmented matrix, the coefficients of the variables are arranged in rows corresponding to each equation, while the constants are placed in an additional column, separated by a vertical line that signifies the equal sign.

To convert a system of equations into an augmented matrix, one must extract the coefficients of each variable and the constants from the equations. For example, consider the system:

• 2x - 3y + z = -4
• 6x + 3y = 13
• y - z = 8

In this case, the first equation contributes the coefficients 2, -3, and 1 for x, y, and z, respectively, with -4 as the constant. The second equation contributes 6, 3, and 0 (since there is no z term), with 13 as the constant. The third equation contributes 0 for x, 1 for y, and -1 for z, with 8 as the constant. Thus, the augmented matrix for this system would be:

\[\begin{bmatrix}2 & -3 & 1 & | & -4 \\6 & 3 & 0 & | & 13 \\0 & 1 & -1 & | & 8\end{bmatrix}\]

This matrix effectively summarizes the information from the system of equations, allowing for easier manipulation and analysis. As you progress in your studies, you will learn various operations that can be performed on matrices, enhancing your ability to solve systems of equations efficiently.

Write the equations in standard form, then represent the system using an augmented matrix.
$3x+5y-9=0$
$8x=-4y+3$

• A.
• B.
• C.
• D.

Answer: (C)

Write the system of equations represented by the augmented matrix shown.

• A. $x+2y+3z=5;5y+4z=1;4x+7y=12$
• B. $x+2y+3x=5;5x+4y=1;4x+7y=12$
• C. $x+2y+3x=5;5y+4z=1;4y+7z=12$
• D. $x+2y+3x=-5;5y+4z=-1;4x+7y=-12$

Answer: (A)

Performing Row Operations on Matrices

In the study of matrices, understanding row operations is essential as they provide a systematic way to manipulate and solve systems of equations. A matrix serves as a compact representation of a system of equations, and just like equations, we can perform specific operations on the rows of a matrix. There are three primary row operations to master: swapping rows, multiplying a row by a non-zero number, and adding a multiple of one row to another.

The first operation, swapping rows, is straightforward. It involves exchanging the positions of two rows within the matrix. For example, if we have rows represented as r₁ and r₂, after swapping, r₁ becomes r₂ and vice versa. This operation is denoted in textbooks with a notation that distinguishes the old row (small r) from the new row (capital R).

The second operation is multiplying a row by a non-zero number. This operation allows us to scale a row, which is particularly useful in adjusting coefficients to facilitate elimination. For instance, if we multiply r₂ by 2, the new row R₂ will contain all elements of r₂ multiplied by 2. This maintains the equivalence of the system represented by the matrix.

The third operation involves adding a multiple of one row to another. This is a crucial step in the elimination process. For example, if we take r₂ and add it to a multiple of r₃, we rewrite r₂ as the sum of its original values and the scaled values from r₃. This operation is vital for eliminating variables and simplifying the system.

To illustrate these operations, consider a matrix with rows r₁, r₂, and r₃. If we swap r₂ and r₃, we simply exchange their positions. If we multiply r₁ by 1/2, we adjust all its coefficients accordingly. Finally, when adding a multiple of r₃ to r₂, we calculate the new values by performing the addition for each corresponding element.

These row operations are foundational in matrix algebra and are particularly useful in solving linear systems using methods such as Gaussian elimination. Mastery of these operations allows for efficient manipulation of matrices, leading to solutions of complex systems of equations.

Perform the indicated Row Operation.



SWAP $R_1\leftrightarrow R_2$

• A.
• B.
• C.
• D.

Answer: (A)

Perform the indicated Row Operation.



ADD $R_1+2\cdot R_3\rightarrow R_1$

• A.
• B.
• C.
• D.

Answer: (B)

Solving Systems of Equations - Matrices (Row-Echelon Form)

In solving systems of equations, one effective method involves transforming the system into a matrix and applying row operations to achieve a specific structure known as row echelon form. This form is characterized by having ones along the diagonal and zeros below it, which simplifies the process of solving for the variables.

To illustrate this, consider a system of equations such as:

-x + 2y = 4

x + 7y = 14

First, we convert this system into an augmented matrix, which consists of the coefficients of the variables and the constants:

\[ \begin{bmatrix} -1 & 2 & | & 4 \\ 1 & 7 & | & 14 \end{bmatrix} \]

The goal is to manipulate this matrix using row operations—specifically, swapping rows, multiplying rows by non-zero constants, and adding rows—to achieve the desired row echelon form. The first step is to create a leading one in the top left corner. In this case, we can swap the two rows to place a one in the first position:

\[ \begin{bmatrix} 1 & 7 & | & 14 \\ -1 & 2 & | & 4 \end{bmatrix} \]

Next, we focus on eliminating the entry below the leading one. To do this, we add the first row to the second row:

\[ \begin{bmatrix} 1 & 7 & | & 14 \\ 0 & 9 & | & 18 \end{bmatrix} \]

Now, we need to convert the second row's leading coefficient into a one. This is achieved by multiplying the entire second row by \(\frac{1}{9}\):

\[ \begin{bmatrix} 1 & 7 & | & 14 \\ 0 & 1 & | & 2 \end{bmatrix} \]

At this point, we have achieved row echelon form, with ones along the diagonal and zeros below. The next step is to convert this matrix back into a system of equations:

x + 7y = 14

y = 2

From the second equation, we can directly solve for \(y\). Substituting \(y = 2\) back into the first equation allows us to solve for \(x\):

x + 7(2) = 14

x + 14 = 14

x = 0

This method of solving systems of equations is often referred to as Gaussian elimination. It emphasizes the importance of systematically applying row operations to simplify the matrix, ultimately leading to a straightforward solution through back substitution.

Example 1

In solving a system of equations using row operations, the goal is to transform the corresponding matrix into row echelon form. This form is characterized by having all leading coefficients (the first non-zero number from the left in a non-zero row) as 1s along the diagonal, with all zeros below these leading 1s. The elements above the diagonal can be any values.

To begin, convert the system of equations into an augmented matrix by extracting the coefficients of the variables. For example, if the system is represented as:

\[\begin{align*}1x + 3y + 4z &= 2 \\2x + 5y + 7z &= 9 \\4x + 8y + 10z &= 14\end{align*}\]

the corresponding augmented matrix would be:

\[\begin{bmatrix}1 & 3 & 4 & | & 2 \\2 & 5 & 7 & | & 9 \\4 & 8 & 10 & | & 14\end{bmatrix}\]

Next, apply row operations to achieve the desired form. The operations include:

• Swapping two rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting a multiple of one row to another row

For instance, to eliminate the entries below the leading 1 in the first column, you can manipulate the second and third rows. If you want to make the second row's first entry zero, you can add a multiple of the first row to the second row. Continuing this process systematically will help you achieve zeros below the leading 1s.

Once the matrix is in row echelon form, you can back substitute to find the values of the variables. For example, if the final matrix looks like this:

\[\begin{bmatrix}1 & 3 & 4 & | & 2 \\0 & 1 & 1 & | & -5 \\0 & 0 & 1 & | & 7\end{bmatrix}\]

From the last row, you can directly read off that \( z = 7 \). Substitute \( z \) into the second row to find \( y \):

\[y + 7 = -5 \implies y = -12\end{p}

Finally, substitute both \( y \) and \( z \) back into the first equation to solve for \( x \):

\[x + 3(-12) + 4(7) = 2 \implies x - 36 + 28 = 2 \implies x = 10\end{p}

Thus, the solution to the system of equations is \( x = 10 \), \( y = -12 \), and \( z = 7 \). This methodical approach using row operations is essential for solving systems of equations efficiently.

Example 2

To solve a system of equations involving three variables (x, y, z), we can utilize matrices and convert them into row echelon form. This process involves organizing the coefficients of the equations into a matrix and performing row operations to achieve a specific format where the leading coefficients (or pivots) are 1, and all entries below these pivots are 0.

Start by constructing the augmented matrix from the system of equations. For example, if the equations are represented as:

1. \(2x + 4y + 6z = 24\)

2. \(x + 5y + 12z = 60\)

3. \(3x + 6y + 15z = 20\)

the corresponding augmented matrix would be:

\[\begin{bmatrix}2 & 4 & 6 & | & 24 \\1 & 5 & 12 & | & 60 \\3 & 6 & 15 & | & 20\end{bmatrix}\]

Next, perform row operations to manipulate the matrix. The goal is to create leading 1s in the diagonal and zeros below them. For instance, you can swap rows to position a row with a leading 1 at the top. After swapping, you might have:

\[\begin{bmatrix}1 & 5 & 12 & | & 60 \\2 & 4 & 6 & | & 24 \\3 & 6 & 15 & | & 20\end{bmatrix}\]

Then, eliminate the entries below the leading 1 by adding or subtracting multiples of the first row from the others. For example, to eliminate the 2 in the second row, you can replace it with:

Row 2 = Row 2 - 2 * Row 1

Continuing this process will lead to a matrix that looks like:

\[\begin{bmatrix}1 & 5 & 12 & | & 60 \\0 & -6 & -18 & | & -96 \\0 & -9 & -21 & | & -160\end{bmatrix}\]

Next, focus on the second row to create a leading 1 by multiplying the entire row by \(-\frac{1}{6}\). This gives:

\[\begin{bmatrix}1 & 5 & 12 & | & 60 \\0 & 1 & 3 & | & 16 \\0 & -9 & -21 & | & -160\end{bmatrix}\]

Now, eliminate the entry below the leading 1 in the second row by adding a multiple of the second row to the third row. This will yield:

\[\begin{bmatrix}1 & 5 & 12 & | & 60 \\0 & 1 & 3 & | & 16 \\0 & 0 & 6 & | & 144\end{bmatrix}\]

Finally, to create a leading 1 in the third row, divide the entire row by 6:

\[\begin{bmatrix}1 & 5 & 12 & | & 60 \\0 & 1 & 3 & | & 16 \\0 & 0 & 1 & | & 24\end{bmatrix}\]

Now that the matrix is in row echelon form, you can convert it back into a system of equations:

1. \(x + 5y + 12z = 60\)

2. \(y + 3z = 16\)

3. \(z = 24\)

Substituting \(z\) back into the second equation gives:

\(y + 3(24) = 16\), leading to \(y = 16 - 72 = -56\).

Finally, substitute both \(y\) and \(z\) back into the first equation to solve for \(x\):

\(x + 5(-56) + 12(24) = 60\), which simplifies to find \(x\).

This systematic approach allows for the efficient solving of systems of equations using matrices, providing a clear pathway to the solution through organized steps and operations.

Solving Systems of Equations - Matrices (Reduced Row-Echelon Form)

In solving systems of equations, two important forms of matrices are the row echelon form and the reduced row echelon form. Both methods utilize similar row operations, but they aim for different configurations in the matrix. In row echelon form, the goal is to have ones along the diagonal and zeros below it, while in reduced row echelon form, the matrix must have ones along the diagonal and zeros both above and below the diagonal.

The primary advantage of achieving reduced row echelon form is that it simplifies the process of converting the matrix back into a system of equations. Once in this form, the solution can be directly read from the matrix without further manipulation, unlike in row echelon form, where additional steps such as substitution are necessary to find the variable values.

To illustrate this, consider a matrix that has already been transformed into row echelon form. For example, if we have a matrix represented as:

\[\begin{bmatrix}1 & 7 & 14 \\0 & 1 & 2\end{bmatrix}\]

To convert this into reduced row echelon form, we need to eliminate the 7 above the second row's leading 1. This can be achieved by performing a row operation that involves adding a multiple of the second row to the first row. Specifically, we can multiply the second row by -7 and add it to the first row:

\[\text{New Row 1} = \text{Row 1} + (-7) \times \text{Row 2}\]

This operation results in:

\[\begin{bmatrix}1 & 0 & 0 \\0 & 1 & 2\end{bmatrix}\]

Now, this matrix is in reduced row echelon form. When converting back to equations, we can easily read the solutions: from the first row, we see that \(x = 0\), and from the second row, we find \(y = 2\). This method not only provides the same solutions as the row echelon form but does so with less subsequent work.

In summary, while both row echelon and reduced row echelon forms are useful for solving systems of equations, the reduced row echelon form streamlines the process of finding solutions. This method is often referred to as Gauss-Jordan elimination, distinguishing it from the Gaussian elimination associated with row echelon form.

Solve the system of equations by using row operations to write a matrix in REDUCED row-echelon form.
$4x+2y+3z=6$
$x+y+z=3$$5x+y+2z=5$

• A. $x=0,y=-3,z=4$
• B. $x=-\frac14,y=-\frac{19}{4},z=\frac{11}{2}$
• C. $x=1,y=4,z=-2$
• D. $x=\frac12,y=\frac12,z=1$

Answer: (C)

Here’s what students ask on this topic:

What is a matrix and how is it used to represent a system of equations?

A matrix is a rectangular array of numbers arranged in rows and columns. It is used to represent a system of equations by organizing the coefficients of the variables into a grid format. For example, the system of equations:

$\left\{\begin{array}{c}{x}^1+y^1=4\\ x^2+y^2=5\end{array}\right\}$

can be represented as the augmented matrix:

$\left[\begin{array}{cccc}1& 1& |& 4\\ 2& 1& |& 5\end{array}\right]$

This matrix format simplifies the process of solving the system using row operations.

What are the basic row operations that can be performed on a matrix?

The three basic row operations that can be performed on a matrix are:

1. Swapping Rows: Interchanging the positions of two rows.
2. Multiplying a Row by a Non-Zero Scalar: Multiplying all elements of a row by a non-zero constant.
3. Adding a Multiple of One Row to Another: Adding a multiple of one row to another row to create a new row.

These operations are used to manipulate the matrix into a desired form, such as row echelon form or reduced row echelon form, to simplify solving the system of equations.

What is row echelon form and how is it used to solve systems of equations?

Row echelon form (REF) is a form of a matrix where all non-zero rows are above any rows of all zeros, and the leading entry of each non-zero row is to the right of the leading entry of the previous row. Additionally, the leading entry in each non-zero row is 1, and all entries in the column below a leading 1 are zeros.

To solve a system of equations using REF, you perform row operations to transform the matrix into REF. Once in REF, you can use back substitution to solve for the variables. This method simplifies the process of solving complex systems of equations.

What is the difference between row echelon form and reduced row echelon form?

Row echelon form (REF) and reduced row echelon form (RREF) are both forms of matrices used to solve systems of equations, but they have key differences:

• Row Echelon Form (REF): In REF, the matrix has ones along the diagonal and zeros below the diagonal. The leading entry in each non-zero row is 1, and all entries below the leading 1 are zeros.
• Reduced Row Echelon Form (RREF): In RREF, the matrix has ones along the diagonal and zeros both below and above the diagonal. This form is more restrictive and often requires additional row operations to achieve.

RREF provides a more straightforward solution as it directly gives the values of the variables without needing further back substitution.

How do you perform Gaussian elimination to solve a system of equations?

Gaussian elimination is a method used to solve systems of linear equations by transforming the system's augmented matrix into row echelon form (REF) using row operations. The steps are:

1. Form the Augmented Matrix: Write the system of equations as an augmented matrix.
2. Apply Row Operations: Use row operations (swapping, multiplying, adding) to transform the matrix into REF. The goal is to get ones along the diagonal and zeros below.
3. Back Substitution: Once in REF, convert the matrix back into equations and solve for the variables starting from the bottom row upwards.

This method simplifies the process of solving complex systems by breaking it down into manageable steps.