Skip to main content
🔗

Subsection 3.2 Elementary matrices

🔗

Definition 3.2.56. Elementary matrices.

🔗
A elementary matrix, denoted by Ei, derives from applying one elementary row operation to the identity matrix of the same size.
Below is a list of elementary row operations and their corresponding elementary matrix.
  • R1→12R1E=[1200010001].
  • R2→R2−R1E=[100−110001].
  • R3→R3+2R2E=[1000010002100001].
  • R2↔R4E=P24=[1000000100100100].
🔗

Subsubsection 3.2.1 Using elementary matrices

🔗
Let A be a m×n matrix and let E be an m×m elementary matrix. The matrix multiplication EA is the matrix that results when the same row operation is performed on A as that performed to produce the elementary matrix E.
Let
A=[25102−5121105],B=[11052−5122510],C=[11050−1022510].
  1. Find the elementary matrix E1 such that E1A=B.
    B results from interchanging rows 1 and 3 on A, then the elementary matrix is
    E1=[001010100].
    so that
    E1A=[001010100][25102−5121105]=[11052−5122510].
  2. Find the elementary matrix E2 such that E2B=C.
    C results from the row operation R2→R2−R3 applied to B, then the elementary matrix is
    E2=[10001−1001].
    so that
    E2B=[10001−1001][11052−5122510]=[11050−1022510].
🔗

Subsubsection 3.2.2 Row reduction with elementary matrices

🔗
🔗
The row reduction of a matrix A to row-echelon form can be obtained by successive multiplication on the left by elementary matrices,
AREF=ErEr−1⋯E3E2E1A.
🔗
We denote the product of all elementary matrices as the elimination matrix, E,
E=ErEr−1⋯E3E2E1⇒AREF=EA.
Let
A=[25102−5121105],B=[11052−5122510],C=[11050−1022510].
Find the elementary matrix E3 such that E3A=C.
From the previous example we have
B=E1A,withE1=[001010100].
And C from B as,
C=E2B,withE2=[10001−1001].
This implies that one can obtain C from A as
C=E2B=E2(E1A)=E2E1A=[10001−1001][001010100][25102−5121105]=[001−110100][25102−5121105]=[11050−1022510]
Find ARREF where
A=[25102−5121105].
We will take matrices E1 and E2 from the previous example, so that, so far, our elimination matrix is
E=E2E1=[001−110100],
and the current matrix to be reduced is
[11050−1022510],
we continue with the reduction:
[11050−1022510]R3→R3−2R1[11050−1020−150],E3=[100010−201]E=E3E2E1=[100010−201][001−110100]=[001−11010−2]
[11050−10224−1]R1→R1+R2[1070−1020−150],E4=[110010001]E=E4E3E2E1=[110010001][001−11010−2]=[−111−11010−2]
[1850−10224−1]R2→−110R2[10701−150−150],E5=[1000−1100001]E=E5⋯E1=[1000−1100001][−111−11010−2]=[−111110−110010−2]
[11050−10224−1]R3→R3+15R2[10701−1500−3],E6=[1000100151]E=E6⋯E1=[1000100151][−111110−110010−2]=[−111110−110052−32−2]
[11050−10224−1]R3→−13R3[10701−15001],E7=[10001000−13]E=E7⋯E1=[10001000−13][−111110−110052−32−2]=[−111110−1100−561223]
[11050−10224−1]R1→R1−7R3[10001−15001],E8=[10−7010001]E=E8⋯E1=[10−7010001][−111110−1100−561223]=[296−52−113110−1100−561223]
[11050−10224−1]R2→R2+15R3[100010001],E9=[1000115001]E=E9⋯E1=[1000115001][296−52−113110−1100−561223]=[296−52−113−1150215−561223]
We can verify the elimination matrix by performing the following operation
E⋅A=[296−52−113−1150215−561223][25102−5121105]=[296⋅2+−52⋅2+−113⋅1296⋅5+−52⋅(−5)+−113⋅10296⋅10+−52⋅12+−113⋅5−115⋅2+0⋅2+215⋅1−115⋅5+0⋅(−5)+215⋅10−115⋅10+0⋅12+215⋅5−52⋅2+12⋅2+23⋅1−52⋅5+12⋅(−5)+23⋅10−52⋅10+12⋅12+23⋅5]=[100010001]
🔗

Subsubsection 3.2.3 Solving systems of equations with elimination matrices

🔗
🔗
Assume that we have a system of equations represented as a matrix equation
Ax=b.
🔗
Next, assume that we find an elimination matrix, E such that,
E⋅A=I.
🔗
Now, if we multiply both sides of the matrix equation by E we obtain,
Ax=bEAx=EbIx=Ebx=Eb.
🔗
This implies that for this system of equations the solution can be found by multiplying the elimination matrix to the right-hand-side vector.
🔗
🔗
Consider the following system of equations,
2x1+2x2−2x3=41x1+3x2−2x3=52x1+7x2+2x3=9
🔗
The matrix representation of the system is
[12−1130272][x1x2x3]=[459],
🔗
and augmented matrix,
[12−1413052729]
🔗
We can find the elimination matrix for the augmented matrix as follows,
[12−1413052729]R2→R2−R1[12−1401112729]E1=[100−110001]
[12−1413052729]R3→R3−2R1[12−1401110341]E2=[100010−201]
[12−1413052729]R3→R1−2R2[10−3201110341]E3=[1−20010001]
[12−1413052729]R3→R3−3R2[10−320111001−2]E4=[1000100−31]
[12−1413052729]R1→R1+3R3[100−40111001−2]E5=[103010001]
[12−1413052729]R2→R2−R3[100−40103001−2]E6=[10001−1001]
🔗
The resulting elimination matrix is
E=E6⋅E5⋅E4⋅E3⋅E2⋅E1=[6−113−24−11−31]
<Prev^TopNext>