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.
The product of an identity matrix In by a square matrix A is equal to A .
A In = In A = A
The product of an identity matrix In by itself is equal to itself.
In In ... In = In
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
The inverse of the identity matrix In is equal to In.
In-1 = In
The transpose of the identity matrix In is equal to In.
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
\( (B - A^{-1})(B^{-1} + A) = C - C^{-1} \)
