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Elementary Matrix — Definition, Formula & Examples

An elementary matrix is a square matrix you get by performing exactly one row operation on the identity matrix. Multiplying another matrix on the left by it performs that same row operation.

An n×nn \times n matrix EE is elementary if it is obtained from InI_n by a single elementary row operation. There are three types: (1) interchange rows ii and jj (a swap or permutation matrix); (2) multiply row ii by a scalar c≠0c \neq 0; (3) add cc times row jj to row ii, with i≠ji \neq j (a replacement). For any n×pn \times p matrix AA, the product EAEA equals the result of applying that operation to AA. Every elementary matrix is invertible, and its inverse is an elementary matrix of the same type.

Key Formula

E=In+c eiejT,i≠jE = I_n + c\,\mathbf{e}_i\mathbf{e}_j^{T}, \quad i \neq j
Where:
  • EE = the replacement-type elementary matrix that adds cc times row jj to row ii
  • InI_n = the n×nn \times n identity matrix
  • cc = any scalar
  • ei,ej\mathbf{e}_i, \mathbf{e}_j = standard basis column vectors (a 1 in position ii or jj, zeros elsewhere)

Worked Example

Problem: Let A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}. Find the elementary matrix EE for the row operation R2→R2−3R1R_2 \to R_2 - 3R_1, verify that EAEA performs this operation, and find E−1E^{-1}.
Step 1: Apply R2→R2−3R1R_2 \to R_2 - 3R_1 to the identity I2I_2. Row 2 of I2I_2 is (0,1)(0, 1). Subtracting 3(1,0)3(1, 0) gives (−3,1)(-3, 1).
E=(10−31)E = \begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix}
Step 2: Multiply EE on the left of AA. Row 1 is unchanged. Row 2 becomes −3(1,2)+(3,4)=(0,−2)-3(1, 2) + (3, 4) = (0, -2).
EA=(10−31)(1234)=(120−2)EA = \begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & -2 \end{pmatrix}
Step 3: To undo the operation, add 3 times row 1 back to row 2. That reverse operation gives the inverse, which is also an elementary matrix.
E−1=(1031),E−1E=I2E^{-1} = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}, \qquad E^{-1}E = I_2
Answer: E=(10−31)E = \begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix} and EA=(120−2)EA = \begin{pmatrix} 1 & 2 \\ 0 & -2 \end{pmatrix}, which is exactly AA after R2→R2−3R1R_2 \to R_2 - 3R_1. The inverse is E−1=(1031)E^{-1} = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}.

Why It Matters

In a first linear algebra course, writing Gaussian elimination as Ek⋯E2E1A=UE_k \cdots E_2E_1A = U lets you prove that every invertible matrix is a product of elementary matrices and leads directly to the LU factorization. Elementary matrices also explain the determinant rules for row operations. A swap has determinant −1-1, scaling a row by cc has determinant cc, and a replacement has determinant 11.

Common Mistakes

Mistake: Multiplying on the right, computing AEAE instead of EAEA, and expecting a row operation.
Correction: Right-multiplication by EE performs the corresponding column operation on AA. To change rows, put EE on the left.

Related Terms