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Elementary Matrices

Matrix Inverse AlgorithmPrevious Section

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Elementary Matrices
An elementary matrix is a square matrix that is equivalent to the identity matrix after one elementary row operation.
Idea: take the identity matrix and perform one ERO on it. The result is an elementary matrix.
Examples
E1=[0110]E_1=\left[\begin{array}{rr}
0&1\\1&0
\end{array}\right]E1​=[01​10​]⟶\longrightarrow⟶ perform  R1↔R2R_1\leftrightarrow R_2R1​↔R2​ on III
E2=[5001]E_2=\left[\begin{array}{rr}
5&0\\0&1
\end{array}\right]E2​=[50​01​]⟶\longrightarrow⟶ perform R1 → 5R1R_1\  \rightarrow\  5R_1R1​ → 5R1​ on III
E3=[1071]E_3=
\left[\begin{array}{rr}
1&0\\
7&1
\end{array}\right]E3​=[17​01​]⟶\longrightarrow ⟶ perform R2 → R2+7R1R_2\  \rightarrow\  R_2 + 7R_1R2​ → R2​+7R1​ on III
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Elementary Matrices and Matrix Multiplication
When a matrix AAA is left multiplied by an elementary matrix EEE, the result is identical to performing the row operation that created EEE on AAA.
Think of elementary matrices as "encoding" the row operation that created them.
Examples
Let A=[123000]A=\left[\begin{array}{rrr}
1&2&3\\0&0&0
\end{array}\right]A=[10​20​30​].
Let E1=[0110]E_1=\left[\begin{array}{rr}
0&1\\1&0
\end{array}\right]E1​=[01​10​] (R1↔R2R_1 \leftrightarrow R_2R1​↔R2​ on III). Find E1AE_1AE1​A.

E1A=[0110][123000]=[000123]E_1A=\left[\begin{array}{rr}
0&1\\1&0
\end{array}\right]\left[\begin{array}{rrr}
1&2&3\\0&0&0
\end{array}\right]
=
\left[\begin{array}{rrr}
0&0&0\\
1&2&3
\end{array}\right]E1​A=[01​10​][10​20​30​]=[01​02​03​]
The row operation that created E1E_1E1​ (swapping rows) is performed on AAA.

Let E2=[5001]E_2=\left[\begin{array}{rr}
5&0\\0&1
\end{array}\right]E2​=[50​01​] (5R15R_15R1​ on III). Find E2AE_2AE2​A.

E2A=[5001][123000]=[51015000]E_2A=\left[\begin{array}{rr}
5&0\\0&1
\end{array}\right]\left[\begin{array}{rrr}
1&2&3\\0&0&0
\end{array}\right]
=
\left[\begin{array}{rrr}
5&10&15\\0&0&0
\end{array}\right]E2​A=[50​01​][10​20​30​]=[50​100​150​]

Let E3=[1071]E_3=\left[\begin{array}{rr}
1&0\\7&1
\end{array}\right]E3​=[17​01​] (R2+7R1R_2 + 7R_1R2​+7R1​ on III). Find E3AE_3 AE3​A.

E3A=[1071][123000]=[12371421]E_3A=\left[\begin{array}{rr}
1&0\\7&1
\end{array}\right]\left[\begin{array}{rrr}
1&2&3\\0&0&0
\end{array}\right]
=
\left[\begin{array}{rrr}
1&2&3\\7&14&21
\end{array}\right]E3​A=[17​01​][10​20​30​]=[17​214​321​]

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Elementary Matrices and the Matrix Inverse Algorithm
Elementary matrices are the reason the matrix inverse algorithm works.
We can "encode" the sequence of row operations that changes AAA into III as a product of elementary matrices.
A→row operationsI  ⟹  (Ek⋯E3E2E1)⏟A−1A=IA\colorTwo{\xrightarrow[\text{row operations}]{}} I 
\quad \implies\quad
\underbrace{\colorTwo{(E_k\cdots E_3E_2E_1)}}_{A^{-1}}A=IArow operations​I⟹A−1(Ek​⋯E3​E2​E1​)​​A=I
So applying the same sequence of operations to III gives:
 (Ek⋯E3E2E1)⏟A−1I  =  A−1I  =  A−1\underbrace{\colorTwo{(E_k\cdots E_3E_2E_1)}}_{A^{-1}}I \ \ =\ \ A^{-1}I \ \ =\ \  A^{-1}A−1(Ek​⋯E3​E2​E1​)​​I  =  A−1I  =  A−1

Wize Tip
Remember to always multiply each new elementary matrix on the left (E1E_1E1​ should be closest to AAA).

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Inverse of Elementary Matrices
Every elementary matrix EEE is invertible; the "opposite" ERO reduces EEE to III.
The inverse of an elementary matrix is an elementary matrix.
Using these facts along with the sequence that produces  A−1=Ek⋯E3E2E1A^{-1} =\colorTwo{E_k\cdots E_3E_2E_1}
A−1=Ek​⋯E3​E2​E1​, we can conclude:
A=(A−1)−1  ⟹  A=(Ek⋯E3E2E1)−1  ⟹  A=E1−1E2−1E3−1⋯Ek−1\begin{aligned}
A&= (A^{-1})^{-1}\\[0.5em]

\implies 
A&=\colorTwo{(E_k\cdots E_3E_2E_1)}^{-1}\\[0.5em]

\implies
A&=E_1^{-1}E_2^{-1}E_3^{-1}\cdots E_k^{-1}
\end{aligned}A⟹A⟹A​=(A−1)−1=(Ek​⋯E3​E2​E1​)−1=E1−1​E2−1​E3−1​⋯Ek−1​​
Wize Concept
If AAA is invertible, then it can be written as a product of elementary matrices.

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Example: Elementary Matrices
Find an elementary matrix EEE such that A=EBA=EBA=EB, given:

A=[246−1−1−11−13]A=\left[\begin{array}{rrr}
2&4&6\\
-1&-1&-1\\
1&-1&3
\end{array}\right]A=​2−11​4−1−1​6−13​​  and  B=[246−1−1−1315]B=\left[\begin{array}{rrr}
2&4&6\\
-1&-1&-1\\
3&1&5
\end{array}\right]B=​2−13​4−11​6−15​​

Think: what row operation can we perform on BBB to turn it into AAA?
Notice that rows 1 and 2 don't change; only row 3 is different.
  ⟹  \implies⟹ The row operation must be performed on row 3.
Therefore, the rows were not swapped, and we can see that the entries in row 3 were not multiplied by a constant.
We can try R3 → R3−R1R_3 \ \rightarrow\ R_3 - R_1R3​ → R3​−R1​, but this does not work out for the second and third entries of the row.
The only other operation that could have been performed is: R3 → R3+2R2R_3 \ \rightarrow \ R_3 + 2R_2R3​ → R3​+2R2​.
Let's perform this operation on III to create the elementary matrix:
[100010001]R3+2R2⟶E=[100010021]\left[\begin{array}{rrr}
1&0&0\\
0&1&0\\
0&0&1
\end{array}\right]

\begin{array}{l}
\\
\\
R_3 + 2R_2\\
\end{array}

\quad \longrightarrow \quad

\boxed{
E=\left[\begin{array}{rrr}
1&0&0\\
0&1&0\\
0&2&1
\end{array}\right]
}​100​010​001​​R3​+2R2​​⟶E=​100​012​001​​​

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Example: Elementary Matrices and Inverses
Let M=[012−8]M=\left[\begin{array}{rr}
0&1\\
2&-8
\end{array}\right]M=[02​1−8​].
MMM can be reduced to III by performing the following EROs (in this order):
R1↔R2R_1\leftrightarrow R_2R1​↔R2​
R1 → 12R1R_1 \ \rightarrow \ \dfrac{1}{2}R_1R1​ → 21​R1​
R1 → R1+4R2R_1 \ \rightarrow \ R_1 + 4R_2R1​ → R1​+4R2​
Using these row operations, find the elementary matrices E1,E2,E3E_1, E_2, E_3E1​,E2​,E3​ and their inverses such that M=E1−1E2−1E3−1M=E_1^{-1}E_2^{-1}E_3^{-1}M=E1−1​E2−1​E3−1​.

I=[1001]R1↔R2⟶E1=[0110]I=\left[\begin{array}{rr}
1&0\\
0&1
\end{array}\right]

\begin{array}{l}
R_1 \leftrightarrow R_2\\
\\
\end{array}

\quad \longrightarrow \quad

E_1=\left[\begin{array}{rr}
0&1\\
1&0
\end{array}\right]
\qquadI=[10​01​]R1​↔R2​​⟶E1​=[01​10​]
I=[1001]12R1⟶E2=[12001]I=\left[\begin{array}{rr}
1&0\\
0&1
\end{array}\right]

\begin{array}{l}
\dfrac{1}{2}R_1\\
\\
\end{array}

\quad \longrightarrow \quad

E_2=\left[\begin{array}{rr}
\dfrac{1}{2}&0\\[1em]
0&1
\end{array}\right]I=[10​01​]21​R1​​⟶E2​=​21​0​01​​
I=[1001]R1+4R2⟶E3=[1401]I=\left[\begin{array}{rr}
1&0\\
0&1
\end{array}\right]

\begin{array}{l}
R_1 + 4 R_2\\
\\
\end{array}

\quad \longrightarrow \quad

E_3=\left[\begin{array}{rr}
1&4\\
0&1
\end{array}\right]
\qquadI=[10​01​]R1​+4R2​​⟶E3​=[10​41​]
So (E3E2E1)M=I  ⟹  M−1=E3E2E1  ⟹  M=(M−1)−1=(E3E2E1)−1  ⟹  M=E1−1E2−1E3−1\begin{aligned}

(E_3 E_2 E_1)M = I \quad
\implies \quad&
M^{-1} = E_3 E_2 E_1\\[0.5em]

\quad \implies \quad&
M = (M^{-1})^{-1} = (E_3 E_2 E_1)^{-1}\\[0.5em]

\quad \implies \quad&
M=E_1^{-1}E_2^{-1}E_3^{-1}

\end{aligned}(E3​E2​E1​)M=I⟹⟹⟹​M−1=E3​E2​E1​M=(M−1)−1=(E3​E2​E1​)−1M=E1−1​E2−1​E3−1​​
Finding Inverses of Elementary Matrices
We can find the inverse of an elementary matrix by performing the opposite operation on III:
E1E_1E1​ simply swapped rows, so swapping them again "undoes" the operation:
I=[1001]R1↔R2⟶E1−1=[0110]I=\left[\begin{array}{rr}
1&0\\
0&1
\end{array}\right]

\begin{array}{l}
R_1 \leftrightarrow R_2\\
\\
\end{array}

\quad \longrightarrow \quad

E_1^{-1}=\left[\begin{array}{rr}
0&1\\
1&0
\end{array}\right]
\qquadI=[10​01​]R1​↔R2​​⟶E1−1​=[01​10​]
E2E_2E2​ multiplied Row 1 by 12\dfrac{1}{2}21​, so we can undo this by multiplying Row 1 by 222:
I=[1001]2R1⟶E2−1=[2001]I=\left[\begin{array}{rr}
1&0\\
0&1
\end{array}\right]

\begin{array}{l}
2R_1\\
\\
\end{array}

\quad \longrightarrow \quad

E_2^{-1}=\left[\begin{array}{rr}
2&0\\
0&1
\end{array}\right]I=[10​01​]2R1​​⟶E2−1​=[20​01​]
E3E_3E3​ added 4 times Row 2 to Row 1, so we can undo this by subtracting 4 times Row 2 from Row 1:
I=[1001]R1−4R2⟶E3−1=[1−401]I=\left[\begin{array}{rr}
1&0\\
0&1
\end{array}\right]

\begin{array}{l}
R_1 - 4 R_2\\
\\
\end{array}

\quad \longrightarrow \quad

E_3^{-1}=\left[\begin{array}{rr}
1&-4\\
0&1
\end{array}\right]I=[10​01​]R1​−4R2​​⟶E3−1​=[10​−41​]
∴ M = E1−1E2−1E3−1 = [0110][2001][1−401]\therefore\ M\ =\ E_1^{-1}E_2^{-1}E_3^{-1} \ =\ 
\left[\begin{array}{rr}
0&1\\
1&0
\end{array}\right]
\left[\begin{array}{rr}
2&0\\
0&1
\end{array}\right]
\left[\begin{array}{rr}
1&-4\\
0&1
\end{array}\right]∴ M = E1−1​E2−1​E3−1​ = [01​10​][20​01​][10​−41​]
Exercise: check this yourself!

Find an elementary matrix EEE such that EB=CEB=CEB=C, given:

B=[11300364]B=\left[\begin{array}{rr}
1&1\\
3&0\\
0&3\\
6&4
\end{array}\right]B=​1306​1034​​  and  C=[101300364]C=\left[\begin{array}{rr}
10&1\\
3&0\\
0&3\\
6&4
\end{array}\right]C=​10306​1034​​
[Enter your answer for EEE by filling in the missing entries in the matrix.]


		0	0
0	0	1	0
0	0	0	1

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Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Let A=[3−11−1]A=\left[\begin{array}{rr}
3&-1\\
1&-1
\end{array}
\right]
A=[31​−1−1​].

A) Using the matrix inverse algorithm, find A−1A^{-1}A−1.
B) Write A−1A^{-1}A−1 as a product of elementary matrices.

I have answered this question

Extra Practice

Write down matrix A−1A^{-1}A−1 as the product of elementary matrices if:
A=[01211021−1]A=\begin{bmatrix}0&1&2\\1&1&0\\2&1&-1\end{bmatrix}A=​012​111​20−1​​

Write down matrix A−1A^{-1}A−1 as the product of elementary matrices if:
A=[01211021−1]A=\begin{bmatrix}0&1&2\\1&1&0\\2&1&-1\end{bmatrix}A=​012​111​20−1​​

Elementary matrices

Which of the following are elementary matrices?
A=[10001000−3]B=[0001001001001000]A = \begin{bmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & -3\end{bmatrix} \quad B = \begin{bmatrix} 0 & 0 & 0 & 1\\0 & 0 & 1 & 0\\0 & 1 & 0 & 0\\1 & 0 & 0 & 0\end{bmatrix}A=​100​010​00−3​​B=​0001​0010​0100​1000​​
C=[001010100]D=[10−71]C = \begin{bmatrix} 0 & 0 & 1\\0 & 1 & 0\\1 & 0 & 0\end{bmatrix} \quad D = \begin{bmatrix} 1 & 0 \\-7 & 1 \end{bmatrix}C=​001​010​100​​D=[1−7​01​]

Example: Elementary Matrix

Find an elementary matrix EEE such that A=EBA=EBA=EB if:
A=[246−1−1−11−13]A=\left[\begin{array}{rrr}
2&4&6\\
-1&-1&-1\\
1&-1&3
\end{array}\right]A=​2−11​4−1−1​6−13​​ and B=[246−1−1−1315]B=\left[\begin{array}{rrr}
2&4&6\\
-1&-1&-1\\
3&1&5
\end{array}\right]B=​2−13​4−11​6−15​​

Which of the following are elementary matrices?
A=[10001000−3]B=[0001001001001000]A = \begin{bmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & -3\end{bmatrix} \quad B = \begin{bmatrix} 0 & 0 & 0 & 1\\0 & 0 & 1 & 0\\0 & 1 & 0 & 0\\1 & 0 & 0 & 0\end{bmatrix}A=​100​010​00−3​​B=​0001​0010​0100​1000​​
C=[001010100]D=[10−71]C = \begin{bmatrix} 0 & 0 & 1\\0 & 1 & 0\\1 & 0 & 0\end{bmatrix} \quad D = \begin{bmatrix} 1 & 0 \\-7 & 1 \end{bmatrix}C=​001​010​100​​D=[1−7​01​]

Practice Question: Matrix of an ERO

Let EEE be the elementary matrix obtained from I2I_2I2​ by multiplying R2 by 1/3.  Let BBB be any 2×22\times22×2 matrix.  Show that EBEBEB is the result of multiplying R2 of BBB by 1/3.  Find the inverse of EEE and show that it is an elementary matrix.

Matrix Multiplication

Express the following matrix A‾\bcb{ \A } A​ as a product of five (or fewer) elementary matrices); repeat for A‾−1\bcb{ \minv{\A} }A​−1.
A‾=[−300201010]\bcb{
\A=\begin{bmatrix}
-3&0&0\\2&0&1\\0&1&0
\end{bmatrix}
}A​=​−320​001​010​​

Practice: Elementary Matrix

Find an elementary matrix EEE such that EB=CEB=CEB=C if:
B=[11300364]B=\left[\begin{array}{rr}
1&1\\
3&0\\
0&3\\
6&4
\end{array}\right]B=​1306​1034​​ and C=[101300364]C=\left[\begin{array}{rr}
10&1\\
3&0\\
0&3\\
6&4
\end{array}\right]C=​10306​1034​​

Matrix Multiplication

e.g. Express the following matrix A‾\underline{A}A​ as a product of five (or fewer) elementary matrices); repeat for  ⁣A‾−1 ⁣\!\underline{A}^{−1}\!A​−1,

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