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Chapter 3: Problem 34

Prove Theorem \(3.16:\) Let \(e\) be an elementary row operation and let \(E\) be the corresponding \(m\)-square elementary matrix; that is, \(E=e(I) .\) Then \(e(A)=E A\), where \(A\) is any \(m \times n\) matrix. Let \(R_{i}\) be the row \(i\) of \(A\); we denote this by writing \(A=\left[R_{1}, \ldots, R_{m}\right] .\) If \(B\) is a matrix for which \(A B\) is defined then \(A B=\left[R_{1} B, \ldots, R_{m} B\right] .\) We also let $$ e_{i}=(0, \ldots, 0, \hat{1}, 0, \ldots, 0), \quad^{-}=i $$ Here \(^{\wedge}=i\) means 1 is the ith entry. One can show (Problem 2.45) that \(e_{i} A=R_{i}\). We also note that \(I=\left[e_{1}, e_{2}, \ldots, e_{m}\right]\) is the identity matrix. (i) Let \(e\) be the elementary row operation "Interchange rows \(R_{i}\) and \(R_{j}, "\) Then, for \(^{-}=i\) and \(^{2}=j\), $$ E=e(I)=\left[e_{1}, \ldots, \widehat{e}_{j}, \ldots, \widehat{\widehat{e}_{i}}, \ldots, e_{m}\right] $$ and $$ e(A)=\left[R_{1}, \ldots, \widehat{R}_{j}, \ldots, \widehat{R}_{i}, \ldots, R_{m}\right] $$ Thus, $$ E A=[e_{1} A, \ldots, \widehat{e_{j} A}, \ldots, \overbrace{\widehat{e_{i} A}}, \ldots, e_{m} A]=\left[R_{1}, \ldots, \widehat{R}_{j}, \ldots, \widehat{\widehat{R}}_{i}, \ldots, R_{m}\right]=e(A) $$ (ii) Let \(e\) be the elementary row operation "Replace \(R_{i}\) by \(k R_{i}(k \neq 0) .\) " Then, for \(^{\wedge}=i\), $$ E=e(I)=\left[e_{1}, \ldots, \hat{k e}_{i}, \ldots, e_{m}\right] $$ and $$ e(A)=\left[R_{1}, \ldots, \widehat{k R}_{i}, \ldots, R_{m}\right] $$ Thus, $$ E A=\left[e_{1} A, \ldots, \widehat{k e_{i} A}, \ldots, e_{m} A\right]=\left[R_{1}, \ldots, \widehat{k R}_{i}, \ldots, R_{m}\right]=e(A) $$ (iii) Let \(e\) be the elementary row operation "Replace \(R_{i}\) by \(k R_{j}+R_{i} "\) Then, for \(^{\wedge}=i\) $$ E=e(I)=[e_{1}, \ldots, k e_{j} \overbrace{+e}_{i}, \ldots, e_{m}] $$ and $$ e(A)=\left[R_{1}, \ldots, k R_{j}+R_{i}, \ldots, R_{m}\right] $$ Using \(\left(k e_{j}+e_{i}\right) A=k\left(e_{j} A\right)+e_{i} A=k R_{j}+R_{i}\), we have $$ \begin{aligned} E A &=\left[e_{1} A, \quad \ldots, \quad\left(k e_{j}+e_{i}\right) A,\right.&\left.\ldots, \quad e_{m} A\right] \\ &=\left[\begin{array}{llll} R_{1}, & \ldots, & k R_{j}+R_{i}, & \ldots & R_{m} \end{array}\right]=e(A) \end{aligned} $$

Short Answer

Expert verified
In summary, for any elementary row operation 'e' and its corresponding elementary matrix 'E', applying the operation to matrix 'A' can also be achieved by multiplying 'E' by 'A'. We demonstrated this property for three cases: interchanging rows, multiplying a row by a scalar, and adding a multiple of one row to another row. Hence, we have proved that e(A) = EA for all cases.

Step by step solution

01

Case 1: Interchanging Rows

Interchange rows i and j of the matrix 'A'. The elementary matrix 'E' will have its i-th and j-th rows interchanged. Thus, when multiplying 'E' by 'A', the same operation will be applied to 'A' and we will have e(A) = EA.
02

Case 2: Multiplying a Row by a Scalar

Replace the i-th row of matrix 'A' with a multiple of the same row (e.g., k * row i). The elementary matrix 'E' will have its i-th row multiplied by the same scalar k. When multiplying 'E' by 'A', the same operation will be applied to 'A'. Hence, e(A)=EA.
03

Case 3: Add a Multiple of One Row to Another Row

Replace row i with a multiple of another row plus itself (e.g., k * row j + row i) in matrix 'A'. The elementary matrix 'E' will have its i-th row replaced with (k * e_j + e_i). When multiplying 'E' by 'A', the same operation will be applied to 'A'. Therefore, e(A) = EA.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elementary Matrices
In linear algebra, an elementary matrix is one that is achieved by performing a single elementary row operation on an identity matrix. These operations include row switching, row multiplication, and row addition. Elementary matrices are significant because they are the building blocks for more complex matrix operations. They can also be used to represent the steps needed to transform a given matrix into its row echelon form or reduced row echelon form.

Furthermore, elementary matrices are helpful in understanding the structure of linear transformations. Since they correspond to basic row operations, any sequence of such operations needed to solve a set of linear equations, or to carry out matrix inversion, can be represented as a product of elementary matrices. Therefore, understanding the concept of elementary matrices is crucial to grasp more advanced algebraic procedures.
Matrix Multiplication
The process of matrix multiplication is central in linear algebra and it's how elementary row operations are applied to matrices. Unlike element-wise multiplication, matrix multiplication involves taking the dot product of rows and columns. Specifically, to multiply two matrices, the number of columns in the first matrix must match the number of rows in the second matrix.

When you multiply an elementary matrix by some matrix A, you are effectively performing the elementary row operation represented by the elementary matrix on A. This multiplication results in a new matrix where the specified operation has been applied. It鈥檚 important to realize that matrix multiplication is not commutative, meaning that the order in which you multiply matrices matters.
Linear Algebra Theorems
Theorems in linear algebra serve as the foundation for understanding and proving various properties about vectors, matrices, and systems of linear equations. One important theorem, as highlighted in the exercise, is the relationship between elementary row operations and elementary matrices. This theorem establishes that performing an elementary row operation on a matrix is equivalent to multiplying the matrix by the corresponding elementary matrix on the left.

Another central theorem in linear algebra is the Invertible Matrix Theorem, which connects the concepts of linear independence, spans, column space, rank, determinants, and eigenvalues. Familiarity with these theorems not only aids in solving matrix equations but also in understanding the behavior of linear systems and transformations.
Row Echelon Form
The row echelon form of a matrix is a form where all non-zero rows (rows with at least one non-zero element) are above any rows of all zeroes, and the leading coefficient (the first non-zero number from the left, also called the pivot) of a nonzero row is always to the right of the leading coefficient of the row above it.

A matrix is in row echelon form when it meets these criteria, which simplifies the process of solving linear equation systems. The continuing process to achieve reduced row echelon form (RREF) makes the system even simpler, often allowing the solutions to be read directly from the matrix. Understanding how to manipulate a matrix to reach this form is often a crucial step in solving linear equations; elementary row operations are key to achieving it.

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Most popular questions from this chapter

Reduce each of the following matrices to row canonical form: (a) \(A=\left[\begin{array}{rrrrr}2 & 2 & -1 & 6 & 4 \\ 4 & 4 & 1 & 10 & 13 \\\ 8 & 8 & -1 & 26 & 23\end{array}\right]\) (b) \(\quad B=\left[\begin{array}{rrr}5 & -9 & 6 \\ 0 & 2 & 3 \\ 0 & 0 & 7\end{array}\right]\) (a) First reduce \(A\) to echelon form by applying the operations "Replace \(R_{2}\) by \(-2 R_{1}+R_{2} "\) and "Replace \(R_{3}\) by \(-4 R_{1}+R_{3}\)," and then applying the operation "Replace \(R_{3}\) by \(-R_{2}+R_{3} .\) " These operations yield $$ A \sim\left[\begin{array}{rrrrr} 2 & 2 & -1 & 6 & 4 \\ 0 & 0 & 3 & -2 & 5 \\ 0 & 0 & 3 & 2 & 7 \end{array}\right] \sim\left[\begin{array}{rrrrr} 2 & 2 & -1 & 6 & 4 \\ 0 & 0 & 3 & -2 & 5 \\ 0 & 0 & 0 & 4 & 2 \end{array}\right] $$ Now use back-substitution on the echelon matrix to obtain the row canonical form of \(A\). Specifically, first multiply \(R_{3}\) by \(\frac{1}{4}\) to obtain the pivot \(a_{34}=1\), and then apply the operations "Replace \(R_{2}\) by \(2 R_{3}+R_{2}^{\prime \prime}\) and "Replace \(R_{1}\) by \(-6 R_{3}+R_{1} . "\) These operations yield $$ A \sim\left[\begin{array}{rrrrr} 2 & 2 & -1 & 6 & 4 \\ 0 & 0 & 3 & -2 & 5 \\ 0 & 0 & 0 & 1 & \frac{1}{2} \end{array}\right] \sim\left[\begin{array}{llrll} 2 & 2 & -1 & 0 & 1 \\ 0 & 0 & 3 & 0 & 6 \\ 0 & 0 & 0 & 1 & \frac{1}{2} \end{array}\right] $$ Now multiply \(R_{2}\) by \(\frac{1}{3}\), making the pivot \(a_{23}=1\), and then apply "Replace \(R_{1}\) by \(R_{2}+R_{1}\)," yielding $$ A \sim\left[\begin{array}{llrll} 2 & 2 & -1 & 0 & 1 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & \frac{1}{2} \end{array}\right] \sim\left[\begin{array}{lllll} 2 & 2 & 0 & 0 & 3 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & \frac{1}{2} \end{array}\right] $$ Finally, multiply \(R_{1}\) by \(\frac{1}{2}\), so the pivot \(a_{11}=1 .\) Thus, we obtain the following row canonical form of \(A\) : $$ A \sim\left[\begin{array}{lllll} 1 & 1 & 0 & 0 & \frac{3}{2} \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & \frac{1}{2} \end{array}\right] $$ (b) Because \(B\) is in echelon form, use back-substitution to obtain $$ B \sim\left[\begin{array}{lll} 5 & -9 & 6 \\ 0 & 2 & 3 \\ 0 & 0 & 1 \end{array}\right] \sim\left[\begin{array}{lll} 5 & -9 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{array}\right] \sim\left[\begin{array}{lll} 5 & -9 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \sim\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \sim\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] $$ The last matrix, which is the identity matrix \(I\), is the row canonical form of \(B\). (This is expected, because \(B\) is invertible, and so its row canonical form must be \(I\).)

Write \(v\) as a linear combination of \(u_{1}, u_{2}, u_{3}\), where (a) \(v=(3,10,7)\) and \(u_{1}=(1,3,-2), u_{2}=(1,4,2), u_{3}=(2,8,1)\); b) \(v=(2,7,10)\) and \(u_{1}=(1,2,3), u_{2}=(1,3,5), u_{3}=(1,5,9)\) c) \(v=(1,5,4)\) and \(u_{1}=(1,3,-2), u_{2}=(2,7,-1), u_{3}=(1,6,7)\). Find the equivalent system of linear equations by writing \(v=x u_{1}+y u_{2}+z u_{3} .\) Alternatively, use the ugmented matrix \(M\) of the equivalent system, where \(M=\left[u_{1}, u_{2}, u_{3}, v\right] .\) (Here \(u_{1}, u_{2}, u_{3}, v\) are the columns (M.) (a) The vector equation \(v=x u_{1}+y u_{2}+z u_{3}\) for the given vectors is as follows: $$ \left[\begin{array}{r} 3 \\ 10 \\ 7 \end{array}\right]=x\left[\begin{array}{r} 1 \\ 3 \\ -2 \end{array}\right]+y\left[\begin{array}{l} 1 \\ 4 \\ 2 \end{array}\right]+z\left[\begin{array}{l} 2 \\ 8 \\ 1 \end{array}\right]=\left[\begin{array}{r} x+y+2 z \\ 3 x+4 y+8 z \\ -2 x+2 y+z \end{array}\right] $$ Form the equivalent system of linear equations by setting corresponding entries equal to each other, and then reduce the system to echelon form: \(\begin{array}{rlrlr}x+y+2 z & =3 & x+y+2 z=3 & x+y+2 z=3 \\ 3 x+4 y+8 z & =10 & \text { or } & y+2 z=1 & \text { or } & y+2 z=1 \\ -2 x+2 y+z & =7 & 4 y+5 z=13 & -3 z=9\end{array}\) The system is in triangular form. Back-substitution yields the unique solution \(x=2, y=7, z=-3\). Thus, \(v=2 u_{1}+7 u_{2}-3 u_{3}\). Alternatively, form the augmented matrix \(M=\left[u_{1}, u_{2}, u_{3}, v\right]\) of the equivalent system, and reduce \(M\) to echelon form: $$ M=\left[\begin{array}{rrrr} 1 & 1 & 2 & 3 \\ 3 & 4 & 8 & 10 \\ -2 & 2 & 1 & 7 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & 1 & 2 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & 4 & 5 & 13 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & 1 & 2 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & -3 & 9 \end{array}\right] $$ The last matrix corresponds to a triangular system that has a unique solution. Back-substitution yields the solution \(x=2, y=7, z=-3 .\) Thus, \(v=2 u_{1}+7 u_{2}-3 u_{3}\). b) Form the augmented matrix \(M=\left[u_{1}, u_{2}, u_{3}, v\right]\) of the equivalent system, and reduce \(M\) to the echelon form: $$ M=\left[\begin{array}{lllr} 1 & 1 & 1 & 2 \\ 2 & 3 & 5 & 7 \\ 3 & 5 & 9 & 10 \end{array}\right] \sim\left[\begin{array}{llll} 1 & 1 & 1 & 2 \\ 0 & 1 & 3 & 3 \\ 0 & 2 & 6 & 4 \end{array}\right] \sim\left[\begin{array}{cccc} 1 & 1 & 1 & 2 \\ 0 & 1 & 3 & 3 \\ 0 & 0 & 0 & -2 \end{array}\right] $$ The third row corresponds to the degenerate equation \(0 x+0 y+0 z=-2\), which has no solution. Thus, the system also has no solution, and \(v\) cannot be written as a linear combination of \(u_{1}, u_{2}, u_{3}\). c) Form the augmented matrix \(M=\left[u_{1}, u_{2}, u_{3}, v\right]\) of the equivalent system, and reduce \(M\) to echelon form: $$ M=\left[\begin{array}{rrrr} 1 & 2 & 1 & 1 \\ 3 & 7 & 6 & 5 \\ -2 & -1 & 7 & 4 \end{array}\right] \sim\left[\begin{array}{llll} 1 & 2 & 1 & 1 \\ 0 & 1 & 3 & 2 \\ 0 & 3 & 9 & 6 \end{array}\right] \sim\left[\begin{array}{llll} 1 & 2 & 1 & 1 \\ 0 & 1 & 3 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ The last matrix corresponds to the following system with free variable \(z\) : $$ \begin{array}{r} x+2 y+z=1 \\ y+3 z=2 \end{array} $$ Thus, \(v\) can be written as a linear combination of \(u_{1}, u_{2}, u_{3}\) in many ways. For example, let the free variable \(z=1\), and, by back- substitution, we get \(y=-2\) and \(x=2\). Thus, \(v=2 u_{1}-2 u_{2}+u_{3}\).

Determine whether each of the following equations is linear: (a) \(5 x+7 y-8 y z=16\) (b) \(x+\pi y+e z=\log 5\) (c) \(3 x+k y-8 z=16\)

Find the inverse of each of the following \(n \times n\) matrices: (a) \(A\) has l's on the diagonal and superdiagonal (entries directly above the diagonal) and 0 's elsewhere. (b) \(B\) has l's on and above the diagonal, and 0 's below the diagonal.

Consider the system $$\begin{array}{c} x+a y=4 \\ a x+9 y=b \end{array}$$ (a) For which values of \(a\) does the system have a unique solution? (b) Find those pairs of values \((a, b)\) for which the system has more than one solution.
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