4.6 Identity Matrix – user's Blog!

4.6 Identity Matrix

1. Identity matrix is a square matrix, usually denoted by the letter I and is also known as unit matrix.

2. All the diagonal elements (from top left to bottom right) of an identity matrix are 1 and the rest of the elements are 0.

For example,

(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) and (\begin{matrix} 1 & 00 \\ \begin{array}{l} 0 \\ 0 \end{array} & \begin{array}{l} 10 \\ 01 \end{array} \end{matrix}) are identity matrices .

3. If I is the identity matrix of order n × n and A is a matrix of the same order, then IA = A and AI = A.

Example 1:

Determine whether each of the following is an identity matrix of

(\begin{matrix} - 2 & 4 \\ 3 & 7 \end{matrix}) .

(a) (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) (b) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})

Solution:

\begin{array}{l} (a) (\begin{matrix} - 2 & 4 \\ 3 & 7 \end{matrix}) (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) \\ = (\begin{matrix} - 2 \times 1 + 4 \times 0 & - 2 \times 0 + 4 \times 1 \\ 3 \times 1 + 7 \times 0 & 3 \times 0 + 7 \times 1 \end{matrix}) \\ = (\begin{matrix} - 2 & 4 \\ 3 & 7 \end{matrix}) \\ Therefore, (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) is an identity matrix . \\ (b) (\begin{matrix} - 2 & 4 \\ 3 & 7 \end{matrix}) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) \\ = (\begin{matrix} - 2 \times 0 + 4 \times 1 & - 2 \times 1 + 4 \times 0 \\ 3 \times 0 + 7 \times 1 & 3 \times 1 + 7 \times 0 \end{matrix}) \\ = (\begin{matrix} 4 & - 2 \\ 7 & 3 \end{matrix}) \\ \neq (\begin{matrix} - 2 & 4 \\ 3 & 7 \end{matrix}) \\ Therefore, (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) is not an identity matrix . \end{array}

Example 2:

Find the product of the following pairs of matrices and determine whether the given matrix is an identity matrix.

\begin{array}{l} (a) (\begin{matrix} - 3 & 2 \\ 5 & 7 \end{matrix}) (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) and (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix} - 3 & 2 \\ 5 & 7 \end{matrix}) \\ (b) (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}) (\begin{matrix} 1 & 8 \\ 5 & 3 \end{matrix}) and (\begin{matrix} 1 & 8 \\ 5 & 3 \end{matrix}) (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}) \end{array}

Solution:

\begin{array}{l} (a) (\begin{matrix} - 3 & 2 \\ 5 & 7 \end{matrix}) (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) \\ = (\begin{matrix} - 3 \times 1 + 2 \times 0 & - 3 \times 0 + 2 \times 1 \\ 5 \times 1 + 7 \times 0 & 5 \times 0 + 7 \times 1 \end{matrix}) = (\begin{matrix} - 3 & 2 \\ 5 & 7 \end{matrix}) \\ (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix} - 3 & 2 \\ 5 & 7 \end{matrix}) \\ = (\begin{matrix} 1 \times - 3 + 0 \times 5 & 1 \times 2 + 0 \times 7 \\ 0 \times - 3 + 1 \times 5 & 0 \times 2 + 1 \times 7 \end{matrix}) = (\begin{matrix} - 3 & 2 \\ 5 & 7 \end{matrix}) \\ ∴ (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) is an identity matrix for (\begin{matrix} - 3 & 2 \\ 5 & 7 \end{matrix}) . \\ (b) (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}) (\begin{matrix} 1 & 8 \\ 5 & 3 \end{matrix}) \\ = (\begin{matrix} 0 \times 1 + 0 \times 5 & 0 \times 8 + 0 \times 3 \\ 1 \times 1 + 1 \times 5 & 1 \times 8 + 1 \times 3 \end{matrix}) = (\begin{matrix} 0 & 0 \\ 6 & 11 \end{matrix}) \\ (\begin{matrix} 1 & 8 \\ 5 & 3 \end{matrix}) (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}) \\ = (\begin{matrix} 1 \times 0 + 8 \times 1 & 1 \times 0 + 8 \times 1 \\ 5 \times 0 + 3 \times 1 & 5 \times 0 + 3 \times 1 \end{matrix}) = (\begin{matrix} 8 & 8 \\ 3 & 3 \end{matrix}) \\ ∴ (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}) is NOT an identity matrix for (\begin{matrix} 1 & 8 \\ 5 & 3 \end{matrix}) . \end{array}