 # Identity Matrix

  
Identity matrices are presented along with their properties including examples and exercises and their detailed solutions.

## Definition of the Identity Matrix

An identity matrix is a square matrix with all diagonal entries equal to $1$ and all other entries equal to zero.
These are examples of identity matrices of dimensions $1 \times 1$,    $2 \times 2$,    $3 \times 3$,    $4 \times 4$ ...
$I_1 = \begin{bmatrix} 1 \\ \end{bmatrix}$ , $I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ , $I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ , $I_4 = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$ , ...
One of the most important properties of the identity matrices is that the product of a square matrix $A$ of dimension $n \times n$ with the identity matrix $I_n$ is equal to $A$. $A I_n = I_n A = A$ The identity matrix is used to define the inverse of a matrix . Matrices $A$ and $B$, of dimensions $n \times n$, are inverse of each other, if $A B = B A = I_n$

## Properties of the Identity Matrices

In what follows, $A$ is a matrix of dimension $n \times n$.
Some of the most important properties of the identity matrices are given below.

1.    The product of an identity matrix $I_n$ by a square matrix $A$ is equal to $A$. $A I_n = I_n A = A$
2.    The product of an identity matrix $I_n$ by itself is equal to itself. $I_n I_n .....I_n = I_n$
3.    The product of a square matrix $A$ by its inverse $A^{-1}$ is equal to the identity matrix $I_n$. $A A^{-1} = A^{-1} A = I_n$
4.    The inverse of the identity matrix $I_n$ is equal to $I_n$. $I_n^{-1} = I_n$
5.    The transpose of the identity matrix $I_n$ is equal to $I_n$. $I_n^T = I_n$
6.    The identity matrix is an orthogonal matrix. (Its columns and rows are orthonormal).
7.    The determinant of an identity matrix is equal to $1$. $Det(I_n) = 1$

## Examples with Solutions

Example 1
Find $x$, $y$ , $z$ and $w$ such that   $\begin{bmatrix} x-2 & y+1 \\ 2z-1 & 2w-2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 6 \\ 9 & 10 \end{bmatrix}$.

Solution
The product on the left of the matrix equation is of the form $A I_2$. According to property $1$, the product of a matrix with an identity matrix is equal to the matrix itself written as $A I_2 = A$. Hence we can write
$\begin{bmatrix} x-2 & y+1 \\ 2z-1 & 2w-2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} x-2 & y+1 \\ 2z-1 & 2w-2 \end{bmatrix}$

Which also gives
$\begin{bmatrix} x-2 & y+1 \\ 2z-1 & 2w-2 \end{bmatrix} = \begin{bmatrix} 2 & 6 \\ 9 & 10 \end{bmatrix}$
Matrices of the same dimensions are equal if their corresponding entries are equal, hence the equations
$x - 2 = 2$
$y + 1 = 6$
$2 z-1 = 9$
$2 w-2 = 10$
Solve the above to obtain:
$x = 4$ , $y = 5$ , $z = 5$ , $w = 6$

Example 2
Simplify the expressions $A^{-1} (A + I_n) - I_n$ where $A$ is a matrix of dimension $n \times n$.

Solution
Use distribution to rewrite the given expression as:
$A^{-1} (A + I_n) - I_n = A^{-1} A + A^{-1} I_n - I_n$
Simplify the right side using the properties 3 and 1 above: $A^{-1} A = I_n$ and $A^{-1} I_n = A^{-1}$
$A^{-1} (A + I_n) - I_n = I_n + A^{-1} - I_n$
Simplify the right side
$A^{-1} (A + I_n) - I_n = A^{-1}$

Example 3
Find matrix $B$ and its inverse $B^{-1}$ given that $A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & x \\ y & 5 \end{bmatrix}$ and $A B = I_2$.

Solution
Substitute $A$, $B$ and the identity matrix $I_2$ in the equation $A B = I_2$ to obtain
$\begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix} \begin{bmatrix} 3 & x \\ y & 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Multiply the matrices on the left
$\begin{bmatrix} 15+2y&5x+10\\ 21+3y&7x+15 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Two matrices of the same dimension are equal if the corresponding entries are equal, hence the equations
$15+2y = 1$ , $21+3y = 0$
$5x+10 = 0$ , $7x+15 = 1$

Solve the above equations to find
$y = - 7$ and $x = - 2$

It can also be shown numerically that
$B A = I_2$
According to property 3 above, matrix $B = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$ and matrix $A$ are inverse of each other and therefore $B^{-1} = A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}$

## Questions (with solutions given below)

In what follows, $A$ , $B$ and $C$ are square matrices of dimension $n \times n$.
• Part 1
Matrices $A$ and $B$ are square matrices such that $A B = I_n$. Find the product $B^{-1} A^{-1}$
• Part 2
Find $a$ and $b$ such that $\begin{bmatrix} 2 & 3 \\ a \cdot b & a+b \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 10 & 7 \end{bmatrix}$.
• Part 3
Simplify the expressions $- A (A^{-1} + I_n) + I_n$.
• Part 4
Given that $B A = C$, write the expression $(B - A^{-1})(B^{-1} + A)$ in terms of $C$ and $C^{-1}$.

### Solutions to the Above Questions

• Part 1
Take the inverse of both sides of the given matrix equation $A B = I_n$
$(A B)^{-1} = I_n^{-1}$
Use the property of inverse of product of matrices: $(A B)^{-1} = B^{-1}A^{-1}$ and the property 4 of identity matrix $I_n^{-1} = I_n$ to rewrite the above equation
$B^{-1}A^{-1} = I_n$

• Part 2
The product of a matrix by the unit matrix is equal to the matrix itself. Hence
$\begin{bmatrix} 2 & 3 \\ a \cdot b & a+b \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ a \cdot b & a+b \end{bmatrix}$
which gives the equation
$\begin{bmatrix} 2 & 3 \\ a \cdot b & a+b \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 10 & 7 \end{bmatrix}$
The equality of the matrices above gives the system of equations
$a b = 10$ and $a + b = 7$
Use the equation $a + b = 7$ to find $b$ in terms of $a$
$b = 7 - a$
Substitute $b$ by $7 - a$ in the equation $a b = 10$ to obtain the quadratic equation
$a (7-a) = 10$
Solve the above equation to find the solutions
$a = 2$ , $a = 5$
Use equation $b = 7 - a$ to find $b$.
$a = 2$ gives $b = 5 b$
$a = 5$ gives $b = 2$
Hence two pairs of solutions:
$a = 2$ and $b = 5$
$a = 7$ and $b = 2$

• Part 3
Using distribution, the given expression may be rewritten as:
$- A (A^{-1} + I_n) + I_n = - A A^{-1} - A I_n + I_n$
Simplify using the properties 3 and 1 above: $A A^{-1} = I_n$ and $A I_n = A$
$= - I_n - A + I_n$
Simplify
$= - A$

• Part 4
Using distribution, the given expression may be rewritten as:
$(B - A^{-1})(B^{-1} + A) = B B^{-1} + B A - A^{-1}B^{-1} - A^{-1} A$
Simplify using the property 3 above: $B B^{-1} = I_n$ and $A^{-1} A = I_n$
$(B - A^{-1})(B^{-1} + A) = I_n + B A - A^{-1}B^{-1} - I_n$
Simplify the right side
$(B - A^{-1})(B^{-1} + A) = B A - A^{-1}B^{-1}$
Use the fact that $B A = C$ and also the property of the inverse of a product of matrices $(B A)^{-1} = A^{-1}B^{-1}$ to rewrite the given expression as
$(B - A^{-1})(B^{-1} + A) = C - C^{-1}$