Row Operations and Elementary Matrices

We show that when we perform elementary row operations on systems of equations represented by

Elementary Matrix Equivalent to Interchanging two Rows of a Matrix

We start with the given system in matrix form Interchange rows (1) and (3) and rewrite the system as Let us now consider the identity matrix

We interchange row (1) and (3) in to obtain the elementary matrix given by
and multiply both sides of system (I) by as follows
Multiply the matrices, using the associativity on the left side, and simplify to

which is the system of equations (II) obtained above by interchanging rows (1) and (3) in the system (I).
Conclusion : Interchanging rows (1) and (3) is equivalent to multiplying (from the left) the two sides of the system by the elementary matrix obtained from the identity matrix .

Elementary Matrix Equivalent to Multiplying a Row by a Constant in a Matrix

Elementary Matrix Equivalent to Adding a Multiple of a Row to Another Row in a Matrix

Inverse of the Elementary Matrices

One of the advantages in using elementary matrices is that their inverse can be obtained without heavy calculations.

Inverse of Elementary Matrix E₁
In the above examples, We obtained the elementary matrix \( E_1 \) by interchanging rows (1) and (3), the inverse of \( E_1 \) is obtained from I₃ by interchanging rows (3) and (1); hence the inverse of \( E_1 \) is given by \[ E_1^{-1} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \]
We now check that
\( E_1 \cdot E_1^{-1} = E_1^{-1} \cdot E_1^ = I_3 \)
\[ E_1 \cdot E_1^{-1} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\] There no need to check that \( E_1^{-1} \cdot E_1^ = I_3 \) beacuse \( E_1 = E_1^{-1} \).

Inverse of Elementary Matrix E₂
In the above examples, We obtained the elementary matrix \( E_2 \) by multiplying row (3) by 2, the inverse of \( E_2 \) is obtained from I₃ by dividing row (3) by 2; hence the inverse of \( E_2 \) is given by \[ E_2^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1/2 \end{bmatrix} \]
We now check that
\( E_2 \cdot E_2^{-1} = E_2^{-1} \cdot E_2^ = I_3 \)
\[ E_2 \cdot E_2^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1/2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]
\[ E_2^{-1} \cdot E_2^ = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1/2 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]

Inverse of Elementary Matrix E₃
In the above examples, We obtained the elementary matrix \( E_3 \) by multiply row (1) by - 2 add it to row (2). The inverse of \( E_3 \) is obtained from I₃ by multiplying row (1) by - 2 and subtracting it from row (2); hence the inverse of \( E_3 \) is given by \[ E_3^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]

We now check that
\( E_3 \cdot E_3^{-1} = E_3^{-1} \cdot E_3^ = I_3 \)
\[ E_3 \cdot E_3^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]
\[ E_3^{-1} \cdot E_3^ = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]

Questions on Elementary Matrices

• Part 1
  Which of the following is an elementary? Explain. \[ A = \begin{bmatrix} 0 & 1 \\ 1 & 0\\ \end{bmatrix} ,\quad B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{bmatrix} ,\quad C = \begin{bmatrix} 1 & 0\\ 4 & 1 \end{bmatrix} \\ ,\quad D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -5 & 0 & 1 \\ \end{bmatrix} ,\quad E = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \\ \end{bmatrix} ,\quad F = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 7 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \]
  
• Part 2
  What is the elementary matrix of the systems of the form \[ A X = B \] for following row operations?
  A) A is 2 by 2 matrix, add 3 times row(1) to row(2)?
  B) A is 3 by 3 matrix, multiply row(3) by - 6.
  C) A is 5 by 5 matrix, multiply row(2) by 10 and add it to row 3.
  
  
• Part 3
  Find the inverse to each elementary matrix found in part 2.

Solutions

• Part 1
  A : interchange rows (1) and (2)
  B: interchange rows (2) and (3)
  C: add 4 times row (1) to row (2)
  D: add - 5 times row (1) to row (3)
  E: is not an elementary matrix
  F: add 7 times row(1) to row (3)
  
• Part 2
  A) \( A = \begin{bmatrix} 1 & 0 \\ 3 & 1\\ \end{bmatrix} \)
  
  B) \( A = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -6 \end{bmatrix} \)
  
  C) \( A = \begin{bmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 &0 & 0 \\ 0 & 10 & 1 &0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} \)
  
• Part 3
  The inverse of the elementary matrices in part 2.
  
  A) Mulriply row(1) by - 3 times and add it to row(3)
  \( A^{-1} = \begin{bmatrix} 1 & 0 \\ -3 & 1\\ \end{bmatrix} \)
  
  B) Divide row (3) by - 6
  \( A^{-1} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -1/6 \end{bmatrix} \)
  
  C) multiply row(2) by - 10 and add it to row(3).
  \( A^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 &0 & 0 \\ 0 & -10 & 1 &0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} \)

More References and links

• Matrices with Examples and Questions with Solutions
• row echelon form calculator
• Row Reduce Augmented Matrices - Calculator
• Linear Algebra
• Add, Subtract and Scalar Multiply Matrices
• Inverse Matrix Questions with Solutions
• Multiplication and Power of Matrices
• Matrix (mathematics)