# 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 � 1 ,    2 � 2 ,    3 � 3 ,    4 � 4 ... One of the most important properties of the identity matrices is that the product of a square matrix A of dimension n � n with the identity matrix In is equal to A .

A In = In A = A

The identity matrix is used to define the inverse of a matrix . Matrices A and B , of dimensions n � n , are inverse of each other, if

A B = B A = In

## Properties of the Identity Matrices

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

1.    The product of an identity matrix In by a square matrix A is equal to A .

A In = In A = A

2.    The product of an identity matrix In by itself is equal to itself.

In In ... In = In

3.    The product of a square matrix A by its inverse A-1 is equal to the identity matrix In.

A A-1 = A-1 A = In

4.    The inverse of the identity matrix In is equal to In.

In-1 = In

5.    The transpose of the identity matrix In is equal to In.

InT = In

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 (In) = 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
$$(B - A^{-1})(B^{-1} + A) = C - C^{-1}$$