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3
Elementary Matrices and Calculating Inverses
Definition. An n × n matrix E is an elementary matrix if it can be obtained from the identity matrix In
by a single elementary row operation. For an elementary row operation ρ we write Eρ for the corresponding
matrix.
There are three types of elementary matrices, corresponding to the three types of row operations. They
look like:
Eri ↔rj =
Eri →αri =
Eri →ri +λrj =
Theorem 3.1. Let A and B be m × n matrices and suppose B is obtained from A by a row operation ρ.
Then B = Eρ A.
Proof. Check by matrix multiplication.
Corollary 3.2. Suppose A can be transformed into B by applying a sequences of elementary row operations
ρ1 , ρ2 , . . . , ρk . Then we have
B = Eρk . . . Eρ2 Eρ1 A.
Proof. By induction on k. The base case k = 1 is exactly Theorem 3.1.
Now let k ≥ 2 and suppose for induction that the statement holds for smaller k. Let A′ be the matrix
obtained from A by applying the transformation ρ1 . Then by Theorem 3.1, A′ = Eρ1 A. Now B is obtained
from A′ by the sequence of (k − 1) elementary row operations ρ2 , . . . , ρk , so by the inductive hypothesis,
B = Eρk . . . Eρ2 A′ = Eρk . . . Eρ2 (Eρ1 A) = Eρk . . . Eρ2 Eρ1 A.
Theorem 3.3. Let E be an elementary n × n matrix. Then E is invertible and E −1 is also an elementary
matrix.
Proof. It is easy to check (exercise!) using the definition of the inverse that:
• the inverse of Eri ↔rj is Eri ↔rj ;
• the inverse of Eri →αri is Eri → 1 ri ;
α
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• the inverse of Eri →ri +λrj is Eri →rj −λrj .
Elementary matrices allow us to provide the promised proof that elementary row operations on an
augmented matrix preserve the solution set.
Theorem 2.1. Suppose M and N are the augmented matrices of two system of equations. If M is row
equivalent to N then the two systems have the same solution set.
Proof. Let M = (A | B) and N = (C | D). Then the two systems of equations can be written in matrix form
(see Section 2.4) as AX = B and CX = D where X is the column matrix whose entries are the variables.
Since M and N are equivalent by a sequences of row operations, A and C are equivalent by the same
sequence of row operations, and so are B and D. So by Corollary 3.2 there are elementary matrices
E1 , . . . , Ek such that
C = Ek . . . E1 A and D = Ek . . . E1 B.
Now suppose Z is column matrix which is a solution to AX = B, that is, AZ = B. Then we have
CZ = (Ek . . . E1 A)Z = (Ek . . . E1 )(AZ) = (Ek . . . E1 )B = D
so Z is also a solution to CX = D.
The converse (showing that a solution to CX = D is also a solution to AX = B) is very similar
(exercise!).
3.1
Invertibility and Systems of Equations
Theorem 3.4. For A an n × n matrix, the following are equivalent:
(i) A is invertible;
(ii) AX = 0n×1 has only the trivial solution X = 0n×1 ;
(iii) the reduced row echelon form of A is In ;
(iv) A is row equivalent to In ;
(v) A can be written as a product of elementary matrices.
Proof. We prove (i) =⇒ (ii) =⇒ (iii) =⇒ (iv) =⇒ (v) =⇒ (i).
(i) =⇒ (ii). Suppose A is invertible. If AX = 0n×1 then
X = In X = (A−1 A)X = A−1 0n×1 = 0n×1
so 0n×1 is the only solution to AX = 0n×1 .
(ii) =⇒ (iii). Suppose (ii) holds. Let M be the reduced row echelon form of A. Since M is row equivalent
to A, it follows from Theorem 2.1 that the equation M X = 0n×1 has only one solution. Since M is square
and in reduced row echelon form, it is either the identity matrix or has a zero row at the bottom. Having
a zero row at the bottom would give lots of solutions to M X = 0n×1 (exercise: why? think about
(M | 0n×1 ) and backtracking), so it must be that M = In .
(iii) =⇒ (iv). By definition, A is row equivalent to its reduced row echelon form.
(iv) =⇒ (v). If A is row equivalent to In then by Corollary 3.2 we have
A = Ek . . . E1 In = Ek . . . E1
for some elementary matrices E1 , . . . , Ek .
(v) =⇒ (i). By Theorem 3.3 elementary matrices are invertible, and by (an inductive argument using)
Lemma 1.3(ii) a product of invertible matrices is invertible.
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3.2
Calculating Inverses
Suppose A is an invertible n × n matrix. By Theorem 3.4 there is a sequence of row operations ρ1 , . . . , ρk
transforming A to In . By Corollary 3.2 this means
In = Eρk . . . Eρ1 A
Multiplying both sides on the right by A−1 we get
A−1 = Eρk . . . Eρ1 AA−1 = Eρk . . . Eρ1 In .
But this means (by Corollary 3.2 again) that applying the sequence of row operations ρ1 , . . . , ρk to In gives
A−1 . This observation gives us an efficient way to find the inverse A−1 of an invertible matrix A:
(i) use Gauss-Jordan elimination to transform A into reduced row echelon form (which by Theorem 3.4
must turn out to be In — if it isn’t then A isn’t invertible!);
(ii) apply the same sequence of row operations to In to get A−1
We can make the process easier by writing the matrices A and In side-by-side in a single augmented n × 2n
matrix (A | In ), then just apply Gauss-Jordan elimination to convert the left-hand-side to In , and let the
right-hand-side “follow along”.

1 0 1
Example. Let A =  1 1 −1
0 1 0

1 0 1
 1 1 −1
0 1 0

. We have



1 0 1
1 0 0
1 0 0
r2 →r2 −r1
−−−−→  0 1 −2 −1 1 0 
0 1 0  −−
0 0 1
0 0 1
0 1 0


1 0 1
1
0 0
r3 →r3 −r2
−−
−−−−→  0 1 −2 −1 1 0 
1 −1 1
0 0 2


1 0 1
1
0 0
r3 → 21 r3
−−−−−→  0 1 −2 −1 1 0 
1
− 21 21
0 0 1
2


1
r1 →r1 −r3
− 21
1 0 0 12
2
r2 →r2 +2r3 
0 1 0 0 0
−−
−−−−−→
1 
1
1
1
0 0 1 2 −2
2
We have now reduced the left-hand-side to its reduced row echelon form, which is In , confirming that A is
invertible. The remaining right-hand-side must be A−1 , so

 1
1
− 21
2
2
1 
A−1 =  0 0
1
1
1
2 −2
2

1 6 4

Example. Let B =
2 4 −1
−1 2 5

1 6 4
 2 4 −1
−1 2 5

. This time:


r2 →r2 −2r1
1 0 0
1 0
1 6
4
r3 →r3 +r1 
0 1 0  −−
0 −8 −9 −2 1
−−−−→
0 8
9
0 0 1
1 0

1 0
1 6
4
r3 →r3 +r2 
0 −8 −9 −2 1
−−
−−−−→
0 0
0 −1 1

0
0 
1

0
0 
1
We could continue with the elimination (exercise: try it!) but the zero row which has appeared in the
bottom of the left-hand-side is clearly not going away again. This means the reduced row echelon form of
B is not In , so B is not invertible!
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