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 ...
Column of Matrix
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
    \( (B - A^{-1})(B^{-1} + A) = C - C^{-1} \)



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