Matrices

Section 1.2 Matrices

In this section we will introduce matrices, one of the main computational tools in linear algebra. We will learn how to translate the information about a linear system to a matrix and then manipulate that matrix to solve the original system.

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Some of the matrix manipulations later in this section may seem to come from nowhere. We intend this first example to motivate these upcoming operations.

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Example 1.2.1.

Consider the following system of two linear equations in two variables:

\begin{align*} 3x+3y \amp =-3\\ 2x-y \amp =7 \text{.} \end{align*}

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We first multiply both sides of the top equation by \(\frac{1}{3}\) in order to get “easier” coefficients on our variables. After taking this action, we have the following equivalent linear system:

\begin{align*} x+y \amp =-1\\ 2x-y \amp =7\text{.} \end{align*}

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We can now use the first equation to eliminate one of the variables in the second equation. If we add \(-2\) times the first equation to the second equation, we’ll only have the \(y\) variable left. The second equation is transformed, resulting in this equivalent linear system:

\begin{align} x+y \amp =-1\tag{1.2}\\ -3y \amp =9\text{.}\tag{1.3} \end{align}

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We can now multiply both sides of Equation (1.3) by \(-\frac{1}{3}\) to find that \(y=-3\text{.}\) Plugging this value in for \(y\) in Equation (1.2) quickly gives us \(x=2\text{.}\) We have solved the linear system at this point, and we have determined that the only solution to the system is \((2,-3)\text{.}\) (Readers should check that this is in fact a solution by plugging these values into the original linear equations.)

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Example 1.2.1 is much longer than it needs to be, and at this point some readers may wonder what they’ve gotten themselves into—the mathematics so far (such as it is) is far from advanced. The real point of this example is to make explicit the operations used to solve a simple linear system. Once we have better notation and terminology, solving linear systems will be much faster (especially with the aid of technology). This is where matrices come in so handy.

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Definition 1.2.2.

A matrix is a rectangular array of numbers. If \(m\) and \(n\) are natural numbers, then an \(m\times n\) matrix is one with \(m\) rows and \(n\) columns. The entries in a matrix are referred to by their row and column numbers, so entry \((i,j)\) is the number in the \(i\)th row and \(j\)th column. (Row numbers increase from the top of the matrix down, and column numbers increase from the left of the matrix to the right.)

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The main diagonal of a matrix refers to those entries on the \((i,i)\)-diagonal of the matrix—starting at the upper left and going down to the right. In other words, an entry is on the main diagonal if and only if its row and column numbers are the same.

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A submatrix of a matrix refers to the matrix that remains after removing one or more rows and/or columns from a matrix.

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Every system of linear equations generates two important matrices—the coefficient matrix and the augmented matrix.

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Definition 1.2.3.

Given a system of \(m\) linear equations in \(n\) variables (hereafter, we will call this an \(m\times n\) linear system),

\begin{align*} a_{11}x_1 + \cdots + a_{1n}x_n \amp =b_1\\ \vdots \hspace{.2in} \amp\phantom{=} \hspace{8pt} \vdots\\ a_{m1}x_1 + \cdots + a_{mn}x_n \amp =b_m\text{,} \end{align*}

the coefficient matrix of the system is

\begin{equation*} \begin{bmatrix} a_{11} \amp \cdots \amp a_{1n} \\ \vdots \amp \amp \vdots \\ a_{m1} \amp \cdots \amp a_{mn} \end{bmatrix}\text{,} \end{equation*}

and the augmented matrix of the system is

\begin{equation*} \left[\begin{array}{@{}ccc|c@{}} a_{11} \amp \cdots \amp a_{1n} \amp b_1 \\ \vdots \amp \amp \vdots \amp \vdots \\ a_{m1} \amp \cdots \amp a_{mn} \amp b_m \end{array}\right]\text{.} \end{equation*}

We form the augmented matrix by “augmenting” the coefficient matrix with the column of constants.

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Note that the number of equations in the linear system corresponds to the number of rows of both the coefficient and augmented matrices, and the number of variables in the linear system corresponds to the number of columns in the coefficient matrix. (The augmented matrix of a linear system has one more column than the number of variables.)

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Example 1.2.4.

For the linear system

\begin{align*} 2x_1 - 7x_2 + x_3 \amp =-8\\ -x_1 + 4x_3 \amp =-2\text{,} \end{align*}

the coefficient and augmented matrices are, respectively,

\begin{equation*} \begin{bmatrix} 2 \amp -7 \amp 1 \\ -1 \amp 0 \amp 4 \end{bmatrix} \hspace{.3in} \text{and} \hspace{.3in} \left[\begin{array}{@{}ccc|c@{}} 2 \amp -7 \amp 1 \amp -8 \\ -1 \amp 0 \amp 4 \amp -2 \end{array}\right]\text{.} \end{equation*}

Note the \(0\) in position \((2,2)\) as it corresponds to the absence of an \(x_2\) term in the second equation of the linear system.

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We must get comfortable switching between linear systems and their associated matrices. In particular, we need to understand why specific forms of matrices are especially useful when solving linear systems.

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Example 1.2.5.

Consider the following matrix as the augmented matrix for a linear system:

\begin{equation*} \left[ \begin{array}{@{}ccc|c@{}} 1 \amp 0 \amp 0 \amp 7 \\ 0 \amp 1 \amp 0 \amp -2 \\ 0 \amp 0 \amp 1 \amp \frac{2}{3} \end{array}\right]\text{.} \end{equation*}

In one sense, this is the best possible augmented matrix we could have for a \(3\times 3\) linear system, because the solutions to the system are obvious. Translating this matrix back to equation form gives us \(x=7\text{,}\) \(y=-2\text{,}\) and \(z=\frac{2}{3}\text{.}\)

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Almost no linear systems will come to us with an augmented matrix as simple as the one in Example 1.2.5. (This is mostly because such a linear system is, well, boring. It takes no work to solve a system like this.) Our goal, however, is to take any given linear system and manipulate its augmented matrix to be as close to this sort of matrix as we can get.

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As we work with augmented matrices, we are restricted in the arithmetic we perform on them because, most of all, we want to preserve the solution sets of the corresponding linear systems. In our next definition, we describe the three “legal” ways we have to manipulate a matrix in this fashion.

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Definition 1.2.6.

The following operations on a matrix are called elementary row operations.

Add a multiple of one row to another row, replacing that second row with the result. (We will call this the replace row operation.)
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Multiply every entry in a row by a nonzero constant. (We will call this the scale row operation.)
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Switch the location of any two rows in the matrix. (We will call this the switch row operation.)
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Two matrices are called row equivalent if there is a sequence of elementary row operations that transforms one matrix into the other.

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When defining “row equivalent” in the previous definition, careful readers will note one implied fact that must be checked. The word “equivalent” brings to mind an equivalence relation, which includes the property of the relation being symmetric. Therefore, the definition implies that all of the elementary row operations are “reversible”—that is, for each elementary row operation, there is an elementary row operation that reverses the change that was just made. This is something to prove! (See Exercise 1.2.9.)

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Theorem 1.2.7.

Suppose that \(A\) and \(B\) are augmented matrices corresponding to systems of linear equations. Then if \(A\) and \(B\) are row equivalent, the linear systems to which they correspond are also equivalent.

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Proof.

We will prove this statement directly. Because elementary row operations only involve one or two rows of a matrix at a time, it is sufficient to prove this theorem for systems of two linear equations.

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We suppose that we have the following \(2\times n\) linear system:

\begin{align*} a_{11}x_1 + \cdots + a_{1n}x_n \amp =b_1\\ a_{21}x_1 + \cdots + a_{2n}x_n \amp =b_2\text{,} \end{align*}

which produces the following augmented matrix:

\begin{equation*} A = \left[\begin{array}{@{}ccc|c@{}} a_{11} \amp \cdots \amp a_{1n} \amp b_1 \\ a_{21} \amp \cdots \amp a_{2n} \amp b_2 \end{array}\right]\text{.} \end{equation*}

Further, we suppose that \((c_1,\ldots,c_n)\) is a solution to the linear system. If we apply the switch row operation to \(A\text{,}\) this corresponds to writing the second equation above the first in the linear system. It is immediate that \((c_1,\ldots,c_n)\) is still a solution to this system.

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If we apply the scale row operation to \(A\text{,}\) multiplying row 1 (without loss of generality) by a nonzero constant \(d\text{,}\) then we have the matrix

\begin{equation*} B = \left[\begin{array}{@{}ccc|c@{}} da_{11} \amp \cdots \amp da_{1n} \amp db_1 \\ a_{21} \amp \cdots \amp a_{2n} \amp b_2 \end{array}\right]\text{.} \end{equation*}

We must show that \((c_1,\ldots,c_n)\) is a solution to the corresponding linear system.

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It is obvious that \((c_1,\ldots,c_n)\) satisfies the second equation in this new linear system since that equation is unchanged. If \((c_1,\ldots,c_n)\) satisfied the first equation of the original system, then

\begin{equation*} a_{11}c_1 + \cdots + a_{1n}c_n = b_1\text{.} \end{equation*}

We can now show that \((c_1,\ldots,c_n)\) satisfies the first equation of the second system by substitution:

\begin{align*} da_{11}c_1 + \cdots + da_{1n}c_n \amp = d(a_{11}c_1 + \cdots + a_{1n}c_n) \\ \amp = d(b_1)\text{.} \end{align*}

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We must now show that the replace row operation preserves solutions. We let \(k\) be a nonzero constant and we replace (without loss of generality) the second row of \(A\) with the old second row plus \(k\) times the first row. Here is the resulting matrix \(C\text{:}\)

\begin{equation*} C = \left[\begin{array}{@{}ccc|c@{}} a_{11} \amp \cdots \amp a_{1n} \amp b_1 \\ ka_{11}+a_{21} \amp \cdots \amp ka_{1n}+a_{2n} \amp kb_1+b_2 \end{array}\right]\text{.} \end{equation*}

In the linear system which corresponds to \(C\text{,}\) the first equation is unchanged from the first equation in the linear system that corresponds to \(A\text{,}\) so we only need to be concerned with the second equation. By virtue of the fact that \((c_1,\ldots,c_n)\) satisfied both equations of the first system, we know that

\begin{equation*} a_{11}c_1 + \cdots + a_{1n}c_n = b_1 \end{equation*}

and

\begin{equation*} a_{21}c_1 + \cdots + a_{2n}c_n = b_2\text{.} \end{equation*}

We can now use this in the following calculation:

\begin{equation*} \begin{split} (ka_{11}+a_{21})c_1 + \cdots \amp + (ka_{1n}+a_{2n})c_n \\ \amp = (ka_{11}c_1 + \cdots +ka_{1n}c_n) + (a_{21}c_1 + \cdots + a_{2n}c_n) \\ \amp = k(a_{11}c_1 + \cdots +a_{1n}c_n) + (a_{21}c_1 + \cdots + a_{2n}c_n) \\ \amp = kb_1 + b_2. \end{split} \end{equation*}

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There is technically one more collection of facts to verify before this proof is complete. We have shown that, for all three elementary row operations, if \((c_1,\ldots,c_n)\) is a solution to the linear system corresponding to \(A\) then it will also be a solution to the linear system corresponding to \(B\) (where \(B\) is the result of applying one elementary row operation to \(A\)). But “equivalent” linear systems means that the solution sets are equal as sets. This means that we must assume we have a solution for the linear system corresponding to the matrix \(B\) and show that it is a solution for the linear system corresponding to \(A\text{.}\) We claim that this concern can be dispensed with by invoking another result in this section. Connecting the last dots of this proof is left to the reader in Exercise 1.2.12.

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Roughly speaking, we want to use elementary row operations to transform the augmented matrix for a linear system into a matrix which has 1s along the main diagonal and 0s as the other entries in those columns. This is not always possible, and we will describe the situation more precisely below, but here is an example to illustrate the process.

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Example 1.2.8.

We start by considering the following matrix as the augmented matrix of a linear system:

\begin{equation*} \left[\begin{array}{@{}ccc|c@{}} 2 \amp 2 \amp -1 \amp 8 \\ -3 \amp -2 \amp 2 \amp -12 \\ 5 \amp 0 \amp 4 \amp 11 \end{array}\right]\text{.} \end{equation*}

We first scale the first row by \(\frac{1}{2}\) to produce a 1 in the \((1,1)\) entry:

\begin{equation*} \left[\begin{array}{@{}ccc|c@{}} 1 \amp 1 \amp -\frac{1}{2} \amp 4 \\[3pt] -3 \amp -2 \amp 2 \amp -12 \\ 5 \amp 0 \amp 4 \amp 11 \end{array}\right]\text{.} \end{equation*}

We then replace the second row with the sum of the second row and three times the first row:

\begin{equation*} \left[\begin{array}{@{}ccc|c@{}} 1 \amp 1 \amp -\frac{1}{2} \amp 4 \\[3pt] 0 \amp 1 \amp \frac{1}{2} \amp 0 \\[3pt] 5 \amp 0 \amp 4 \amp 11 \end{array}\right]\text{.} \end{equation*}

We will again use the 1 in the \((1,1)\) entry to “eliminate” the 5 in the \((3,1)\) entry. We replace the third row with \(-5\) times the first row plus the third:

\begin{equation*} \left[\begin{array}{@{}ccc|c@{}} 1 \amp 1 \amp -\frac{1}{2} \amp 4 \\[3pt] 0 \amp 1 \amp \frac{1}{2} \amp 0 \\[3pt] 0 \amp -5 \amp \frac{13}{2} \amp -9 \end{array}\right]\text{.} \end{equation*}

Now that we have “cleared out” the entries under the \((1,1)\) entry, we can do the same for the 1 in the \((2,2)\) entry. (In future examples we may need to scale first to have a \(1\) here.) We replace the third row with 5 times the second row plus the third:

\begin{equation*} \left[\begin{array}{@{}ccc|c@{}} 1 \amp 1 \amp -\frac{1}{2} \amp 4 \\[3pt] 0 \amp 1 \amp \frac{1}{2} \amp 0 \\[3pt] 0 \amp 0 \amp 9 \amp -9 \end{array}\right]\text{.} \end{equation*}

We can now scale the third row by \(\frac{1}{9}\) to produce 1s along the main diagonal:

\begin{equation*} \left[\begin{array}{@{}ccc|c@{}} 1 \amp 1 \amp -\frac{1}{2} \amp 4 \\[3pt] 0 \amp 1 \amp \frac{1}{2} \amp 0 \\[3pt] 0 \amp 0 \amp 1 \amp -1 \end{array}\right]\text{.} \end{equation*}

We are halfway done, as we have 0s below the main diagonal. We now need to use elementary row operations to produce 0s above the main diagonal. We first replace the second row with \(-\frac{1}{2}\) times the third row plus the second:

\begin{equation*} \left[\begin{array}{@{}ccc|c@{}} 1 \amp 1 \amp -\frac{1}{2} \amp 4 \\[3pt] 0 \amp 1 \amp 0 \amp \frac{1}{2} \\[3pt] 0 \amp 0 \amp 1 \amp -1 \end{array}\right]\text{.} \end{equation*}

We can again use the 1 in the \((3,3)\) position to produce a 0 in the \((1,3)\) position. We replace the first row with \(\frac{1}{2}\) times the third row plus the first:

\begin{equation*} \left[\begin{array}{@{}ccc|c@{}} 1 \amp 1 \amp 0 \amp \frac{7}{2} \\[3pt] 0 \amp 1 \amp 0 \amp \frac{1}{2} \\[3pt] 0 \amp 0 \amp 1 \amp -1 \end{array}\right]\text{.} \end{equation*}

Finally, we use the 1 in the \((2,2)\) entry to produce a 0 in the \((1,2)\) entry. We replace the first row with \(-1\) times the second row plus the first:

\begin{equation*} \left[\begin{array}{@{}ccc|c@{}} 1 \amp 0 \amp 0 \amp 3 \\ 0 \amp 1 \amp 0 \amp \frac{1}{2} \\[3pt] 0 \amp 0 \amp 1 \amp -1 \end{array}\right]\text{.} \end{equation*}

We now have the matrix in the form we wanted, because we can read off the solution: \(x=3\text{,}\) \(y=\frac{1}{2}\text{,}\) and \(z=-1\text{.}\)

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We will now define and standardize this form of the augmented matrix that is so helpful in solving the related linear system. In what follows, by “nonzero row (column)” we mean a row (column) with at least one nonzero entry, and by the “leading entry” of a row we mean the row’s leftmost nonzero entry.

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Definition 1.2.9.

A matrix \(A\) is in row-echelon form (REF) if all of the following hold:

all rows of all zeros are collected at the bottom of the matrix;

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each leading entry of a row is in a column to the right of the column of the leading entry for the row above it; and

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all entries in a column below the leading entry of a row are zeros.

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If a matrix \(A\) is in row-echelon form and also satisfies the following two conditions, it is in reduced row-echelon form (RREF):

each leading entry in a nonzero row is 1; and

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each leading 1 is the only nonzero entry in its column.

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Note 1.2.10.

The reader should observe that every matrix which is in RREF is automatically in REF, but the converse is not true.

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These definitions are a mouthful, but they are useful. Here is an example showing some matrices that do and do not meet these criteria.

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Example 1.2.11.

The following two matrices are in REF but not RREF:

\begin{equation*} \begin{bmatrix} 2 \amp -1 \amp 3 \amp 0 \\ 0 \amp 5 \amp -3 \amp 7 \\ 0 \amp 0 \amp 2 \amp -1 \\ 0 \amp 0 \amp 0 \amp 0 \end{bmatrix} \hspace{12pt} \begin{bmatrix} -4 \amp 9 \amp 6 \\ 0 \amp 12 \amp -10 \end{bmatrix}\text{.} \end{equation*}

The following two matrices are in RREF:

\begin{equation*} \begin{bmatrix} 1 \amp 0 \amp 4 \amp 5 \\ 0 \amp 1 \amp 2 \amp -3 \\ 0 \amp 0 \amp 0 \amp 0 \end{bmatrix} \hspace{12pt} \begin{bmatrix} 1 \amp -7 \amp 0 \amp 2 \\ 0 \amp 0 \amp 1 \amp -5 \end{bmatrix}\text{.} \end{equation*}

Neither of the following matrices are in REF or RREF:

\begin{equation*} \begin{bmatrix} 0 \amp 2 \amp 5 \amp -8 \\ 0 \amp -1 \amp 7 \amp 7 \\ 4 \amp 2 \amp 1 \amp 0 \end{bmatrix} \hspace{12pt} \begin{bmatrix} 2 \amp 3 \amp -2 \amp 5 \\ 0 \amp -1 \amp 3 \amp 6 \\ -4 \amp -3 \amp 0 \amp 12 \end{bmatrix}\text{.} \end{equation*}

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Definition 1.2.12.

When a matrix \(A\) is in row-echelon form, then the leading entry in each row is called a pivot. The location of this entry within the matrix is called a pivot position. Finally, any column containing a pivot is called a pivot column.

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One of the reasons we have Definition 1.2.9 is because (we will prove this below) every matrix can be put into RREF. What follows here is a description of the algorithm used to transform every matrix into RREF. This process is usually called “Gaussian elimination” or “Gauss-Jordan elimination.”

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Algorithm 1.2.13. The Row Reduction Algorithm.

This row reduction algorithm consists of five steps. The first four (collectively) produce a matrix in row-echelon form; after the fifth step the matrix will be in reduced row-echelon form.

Start with the leftmost nonzero column. This will be a pivot column with the pivot position at the top.

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Select a nonzero entry in this column and use the switch operation to move it to the top of the column (if necessary).

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Use elementary row operations to create zeros below this pivot.

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Ignore the row with the pivot just considered. Apply steps 1–3 to the submatrix that remains. Repeat this process until all nonzero rows have been handled.

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Select the rightmost pivot and use the scale operation to make it a 1. Now use elementary row operations to create 0s above this pivot. Move upward and to the left, repeating this process for all remaining pivots.

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Example 1.2.14.

We consider the following matrix \(A\) and put it into reduced row-echelon form:

\begin{equation*} A = \left[\begin{array}{@{}cccc|c@{}} -2 \amp -2 \amp 6 \amp 14 \amp 4 \\ 2 \amp 3 \amp -4 \amp -4 \amp -1 \\ -3 \amp -5 \amp 4 \amp -2 \amp -2 \end{array}\right]\text{.} \end{equation*}

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The first column is nonzero, and we will leave the current top row in its place. Though it is not necessary to create a \(1\) in the pivot position at this point in the process, it is often useful to do so. (We are not violating Algorithm 1.2.13 by producing a leading 1 this early in the process, but this is a step of row reduction that many, including this author, prefer to perform now to make future steps less painful.) We multiply the first row by \(-\frac{1}{2}\) to achieve this:

\begin{equation*} \left[\begin{array}{@{}cccc|c@{}} 1 \amp 1 \amp -3 \amp -7 \amp -2 \\ 2 \amp 3 \amp -4 \amp -4 \amp -1 \\ -3 \amp -5 \amp 4 \amp -2 \amp -2 \end{array}\right]\text{.} \end{equation*}

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We now use the \(1\) in the \((1,1)\) position to create zeros in the column below it. We add \(-2\) times row 1 to row 2 and we add \(3\) times row 1 to row 3. Since there is no “interaction” between these operations, we will perform them at the same time, though the reader should certainly take one operation at a time if this combination raises one’s blood pressure:

\begin{equation*} \left[\begin{array}{@{}cccc|c@{}} 1 \amp 1 \amp -3 \amp -7 \amp -2 \\ 0 \amp 1 \amp 2 \amp 10 \amp 3 \\ 0 \amp -2 \amp -5 \amp -23 \amp -8 \end{array}\right]\text{.} \end{equation*}

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According to the algorithm, we now ignore row 1 and repeat the process for the remaining matrix. There is already a \(1\) at the “top” of (this portion of) the second column, so we use that entry to create zeros below it. We add twice the second row to the third row:

\begin{equation*} \left[\begin{array}{@{}cccc|c@{}} 1 \amp 1 \amp -3 \amp -7 \amp -2 \\ 0 \amp 1 \amp 2 \amp 10 \amp 3 \\ 0 \amp 0 \amp -1 \amp -3 \amp -2 \end{array}\right]\text{.} \end{equation*}

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The matrix is in row-echelon form now, so we proceed to step 5 of the algorithm. We multiply the third row by \(-1\) to produce a \(1\) in the \((3,3)\) position:

\begin{equation*} \left[\begin{array}{@{}cccc|c@{}} 1 \amp 1 \amp -3 \amp -7 \amp -2 \\ 0 \amp 1 \amp 2 \amp 10 \amp 3 \\ 0 \amp 0 \amp 1 \amp 3 \amp 2 \end{array}\right]\text{.} \end{equation*}

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We now use the \(1\) we just created to produce zeros in the column above it. We add \(-2\) times the third row to the second, and we add \(3\) times the third row to the first:

\begin{equation*} \left[\begin{array}{@{}cccc|c@{}} 1 \amp 1 \amp 0 \amp 2 \amp 4 \\ 0 \amp 1 \amp 0 \amp 4 \amp -1 \\ 0 \amp 0 \amp 1 \amp 3 \amp 2 \end{array}\right]\text{.} \end{equation*}

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The final step is to use the pivot in position \((2,2)\) to create a \(0\) above it. We add \(-1\) times the second row to the first:

\begin{equation} \left[\begin{array}{@{}cccc|c@{}} 1 \amp 0 \amp 0 \amp -2 \amp 5 \\ 0 \amp 1 \amp 0 \amp 4 \amp -1 \\ 0 \amp 0 \amp 1 \amp 3 \amp 2 \end{array}\right]\text{.}\tag{1.4} \end{equation}

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The process is now complete, so the matrix in (1.4) is the result of reducing \(A\) to RREF.

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We note that this example is different from Example 1.2.8 in an important way. The matrix in Example 1.2.14 corresponds to a system with three equations and four variables, while the matrix in Example 1.2.8 corresponds to a system with three equations and three variables. We only mention this to point out that the size of the original matrix puts some restrictions on the possibilities for its RREF, and the reader should be aware of this when completing the reading questions and the exercises at the end of this section.

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Reading Questions Reading Questions

1.

Write down the coefficient matrix and the augmented matrix for the following linear system:

\begin{align*} 2x_1-3x_2+ \tfrac{1}{2}x_3 \amp =7\\ -5x_2+x_3-x_1 \amp =-4 \\ 6x_2+9x_1 \amp =-1\text{.} \end{align*}

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2.

Although the reduced row-echelon form of a matrix is unique (we will prove this soon), the row-echelon form of a matrix is not unique. For the following matrix \(A\text{,}\) write two distinct row-echelon forms:

\begin{equation*} A = \left[\begin{array}{@{}cc|c@{}} 2 \amp -3 \amp 4 \\ -1 \amp 2 \amp 0 \end{array}\right]\text{.} \end{equation*}

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3.

Put the following matrix into reduced row-echelon form. Record each step along the way (both the elementary row operations used and the matrices obtained as a result):

\begin{equation*} A = \left[\begin{array}{@{}cc|c@{}} 2 \amp -3 \amp 5 \\ -1 \amp 2 \amp -4 \\ 5 \amp -9 \amp 8 \end{array}\right]\text{.} \end{equation*}

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Exercises Exercises

1.

Write the augmented matrix that corresponds to the following linear system:

\begin{align*} 2x_2 - 4x_1 + \sqrt{2}x_4 \amp = 8\\ 3x_3 - 19x_2 \amp = -1\\ 0.5x_4 - 3x_3 + x_1 \amp = 0\text{.} \end{align*}

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2.

Write the linear system that corresponds to the following matrix, assuming this is the augmented matrix of the system:

\begin{equation*} \left[\begin{array}{@{}ccc|c@{}} 1 \amp 0 \amp -3 \amp 4 \\ 0 \amp -2 \amp 1 \amp -5 \\ 0 \amp 0 \amp 0 \amp 0 \end{array}\right]\text{.} \end{equation*}

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3.

For each of the following, describe the elementary row operation that was used to transform the matrix on the left into the matrix on the right. Be specific in your description of the operation. (You should not just answer “scale” or “switch”, etc.)

\(\displaystyle \begin{bmatrix} 2 \amp 1 \amp -3 \\ -1 \amp 4 \amp 5 \\ -2 \amp 0 \amp 2 \end{bmatrix} \longrightarrow \begin{bmatrix} 2 \amp 1 \amp -3 \\ 1 \amp 5 \amp 2 \\ -2 \amp 0 \amp 2 \end{bmatrix}\)
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\(\displaystyle \begin{bmatrix} 2 \amp 1 \amp -3 \\ -1 \amp 4 \amp 5 \\ -2 \amp 0 \amp 2 \end{bmatrix} \longrightarrow \begin{bmatrix} -1 \amp 4 \amp 5 \\ 2 \amp 1 \amp -3 \\ -2 \amp 0 \amp 2 \end{bmatrix}\)
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\(\displaystyle \begin{bmatrix} 2 \amp 1 \amp -3 \\ -1 \amp 4 \amp 5 \\ -2 \amp 0 \amp 2 \end{bmatrix} \longrightarrow \begin{bmatrix} 2 \amp 1 \amp -3 \\ -1 \amp 4 \amp 5 \\ -1 \amp 0 \amp 1 \end{bmatrix}\)
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4.

For each of the following matrices, determine if the matrix is in row-echelon form, reduced row-echelon form, both, or neither.

\(\displaystyle \begin{bmatrix} 2 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \\ 0 \amp -1 \amp 1 \end{bmatrix}\)
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\(\displaystyle \begin{bmatrix} 0 \amp 2 \amp 3 \\ 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \end{bmatrix}\)
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\(\displaystyle \begin{bmatrix} -1 \amp 2 \\ 0 \amp 2 \\ 0 \amp 0 \end{bmatrix}\)
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\(\displaystyle \begin{bmatrix} 1 \amp 0 \amp 7 \\ 0 \amp 1 \amp -4 \end{bmatrix}\)
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Answer.

Neither
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REF
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REF
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Both
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5.

Consider the following matrix \(A\text{:}\)

\begin{equation*} A = \begin{bmatrix} -2 \amp 3 \\ -3 \amp 5 \end{bmatrix}\text{.} \end{equation*}

Find two distinct matrices \(B\) and \(C\) which are row equivalent to \(A\) and are in row echelon form. (There are many correct answers!)

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Answer.

Two possibilities (there are many!):

\begin{equation*} B = \begin{bmatrix} -2 \amp 3 \\ 0 \amp 1/2 \end{bmatrix}, \hspace{6pt} \text{and} \hspace{6pt} C = \begin{bmatrix} 1 \amp -3/2 \\ 0 \amp 1/2 \end{bmatrix}\text{.} \end{equation*}

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6.

Use Algorithm 1.2.13 to put each of these matrices into RREF.

\(\displaystyle \begin{bmatrix} -3 \amp 4 \\ 1 \amp 2 \end{bmatrix}\)
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\(\displaystyle \begin{bmatrix} 2 \amp 4 \amp 5 \\ -1 \amp 3 \amp -4 \end{bmatrix}\)
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\(\displaystyle \begin{bmatrix} 4 \amp 8 \amp 8 \\ 4 \amp -1 \amp -8 \\ 0 \amp 8 \amp -7 \end{bmatrix}\)
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\(\displaystyle \begin{bmatrix} -3 \amp -8 \amp -1 \\ -4 \amp -3 \amp -3 \\ 2 \amp 13 \amp -1 \end{bmatrix}\)
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Answer.

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\(\displaystyle \begin{bmatrix} 1 \amp 0 \amp 31/10 \\ 0 \amp 1 \amp -3/10 \end{bmatrix}\)
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\(\displaystyle \begin{bmatrix} 1 \amp 0 \amp 21/23 \\ 0 \amp 1 \amp -5/23 \\ 0 \amp 0 \amp 0 \end{bmatrix}\)
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7.

For each of the following matrix sizes, list all of the possible RREFs for matrices of that size. Use the symbols \(\blacksquare\) for a pivot, \(\ast\) for an unspecified number, and \(0\) for a zero entry.

\(\displaystyle 2\times 2\)
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\(\displaystyle 2\times 3\)
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\(\displaystyle 3\times 2\)
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Answer.

There are 4 possible RREFs.
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There are 7 possible RREFs.
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There are 4 possible RREFs.
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Writing Exercises

8.

Suppose \(A\) is a \(3\times 4\) matrix. What is the maximum number of pivots in any RREF of \(A\text{?}\) Explain.
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Suppose \(B\) is a \(6\times 4\) matrix. What is the minimum number of rows of zeros in any RREF of \(B\text{?}\) Explain.
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9.

Prove that each of the elementary row operations is reversible. In other words, if an elementary row operation was used to transform a matrix \(A\) into a matrix \(B\text{,}\) prove that there is another elementary row operation (of the same type) which will transform \(B\) back into \(A\text{.}\)

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10.

Recall that an equivalence relation is reflexive, symmetric, and transitive. Now, fix the integers \(m\) and \(n\) and consider row equivalence as a relation on all \(m\times n\) matrices. Prove that this is an equivalence relation. (Hint: another exercise in this section may be helpful in your argument.)

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11.

Prove or disprove: The following two matrices are row equivalent. (Hint: another exercise in this section may be helpful in your argument.)

\begin{equation*} A = \begin{bmatrix} 9 \amp -9 \amp -10 \\ -1 \amp -1 \amp 9 \\ -5 \amp -2 \amp -1 \end{bmatrix} \hspace{.3in} B = \begin{bmatrix} 0 \amp 3 \amp -2 \\ 7 \amp -2 \amp 1 \\ 7 \amp 6 \amp -8 \end{bmatrix} \end{equation*}

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12.

Complete the proof of Theorem 1.2.7. This can be accomplished by proving the statement in the final paragraph of the given proof.

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