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.
- The product of an identity matrix \( I_n \) by a square matrix \( A\) is equal to \( A \). \[ A I_n = I_n A = A \]
- The product of an identity matrix \( I_n \) by itself is equal to itself. \[ I_n I_n .....I_n = I_n \]
- The product of a square matrix \( A \) by its inverse \( A^{-1} \) is equal to the identity matrix \( I_n \). \[ A A^{-1} = A^{-1} A = I_n \]
- The inverse of the identity matrix \( I_n \) is equal to \( I_n \). \[ I_n^{-1} = I_n \]
- The transpose of the identity matrix \( I_n \) is equal to \( I_n \). \[ I_n^T = I_n \]
- The identity matrix is an orthogonal matrix. (Its columns and rows are orthonormal).
- 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
\( (B - A^{-1})(B^{-1} + A) = C - C^{-1} \)
More References and links
-
Linear Algebra - Questions with Solutions
-
Matrices with Examples and Questions with Solutions.
-
Determinant of a Square Matrix.
-
Inverse Matrix Questions with Solutions.
-
Transpose of a Matrix.
-
orthogonal matrix
-
orthonormal
- Elementary Linear Algebra - 7 th Edition - Howard Anton and Chris Rorres
- Introduction to Linear Algebra - Fifth Edition (2016) - Gilbert Strang
- Linear Algebra Done Right - third edition, 2015 - Sheldon Axler
- Linear Algebra with Applications - 2012 - Gareth Williams