I think this phrase is not precise: The product of 'any matrix' and the appropriate identity matrix is always the original matrix, regardless of the order in which the multiplication was performed! In other words, A*I=I*A=A I think this only work when the matrix A is square matrix. For example, we have a 3x2 matrix. To make the statement A*I=A to be true, the identity matrix need to be 2x2 matrix. But to make the statement I*A=A to be true, the identity matrix in this case need to be a 3x3 matrix. You can see these two matrices are not equal.

You are correct. In your 3rd paragraph, you show a right identity matrix such that AI=A and a left identity matrix (IA=A). The right and left identity matrices are equal only if A is square matrix. Nice catch

How do you find the identity matrix for a matrix that is not n by n?

If you have a matrix that is 2x3, the identity matrix will be 3x3 and look like this: 1 0 0 0 1 0 0 0 1 If you have a 3x2, the identity will be 2x2 and look like this 1 0 0 1 You can try these out by your own example. The important part is that the multiplication is defined.

What is the use of identity matrix? IA = A .... and why do we do that?

It's useful when you're solving for messy calculations. For example, If you have A + NA + MA, you could just simply do A(I + N + M) which makes the calculation a lot easier. Also, the product of A and A^-1 is I so that's kinda important. Hope this helps! If you have any questions or need help, please ask! :)

is matrices just the plural version of a matrix?

Yup. same as vertex has the plural vertices.

How do you find the multiplicative inverse of a given matrix? I guess you could create a system of equation with 4 equations and variables and solve (for a 2*2 matrix). But that seems too lengthy. Is there an easier way?

Make two columns, with your given matrix on the left and an identity matrix of the same size on the right. Perform row operations on the matrices. The rule is, whatever operation you do to the left matrix, you must simultaneously do to the right matrix. e.g. if you multiply the top row of your matrix by 5, you must multiply the top row of the identity matrix by 5. Do row operations until you have an identity matrix on the left. The matrix on the right will be the inverse of your original matrix. This is still rather computationally intensive, but that's just a fact of life. Matrix inversion is a computationally difficult task, and there is not always a way to simplify something further.

can we have I matrix with dimensions 1*1? if yes , how we can write it ?

Yes. It would be written [1] https://en.wikipedia.org/wiki/Identity_matrix

Main content

Intro to identity matrices

Google Classroom

Learn what an identity matrix is and about its role in matrix multiplication.

What you should be familiar with before taking this lesson

A matrix is a rectangular arrangement of numbers into rows and columns.

The dimensions of a matrix tell the number of rows and columns of the matrix in that order. Since matrix

A

has

2

rows and

3

columns, it is called a

2 \times 3

matrix.

If this is new to you, we recommend that you check out our intro to matrices.

In matrix multiplication, each entry in the product matrix is the dot product of a row in the first matrix and a column in the second matrix.

If this is new to you, we recommend that you check out our matrix multiplication article.

Definition of identity matrix

The

n \times n

identity matrix, denoted

I_{n}

, is a matrix with

n

rows and

n

columns. The entries on the diagonal from the upper left to the bottom right are all

1

's, and all other entries are

0

For example:

I_{2} = [\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}] I_{3} = [\begin{array}{rr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] I_{4} = [\begin{array}{rr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]

The identity matrix plays a similar role in operations with matrices as the number

1

plays in operations with real numbers. Let's take a look.

Investigation: Multiplying by the identity matrix

Try a few multiplication problems involving the appropriate identity matrix.

I_{2} = [\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}]

and

A = [\begin{array}{rr} 2 & 3 \\ 5 & 1 \end{array}]

I_{2} \cdot A =

I_{3} = [\begin{array}{rr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]

and

A = [\begin{array}{rr} 1 & 5 & 4 \\ 3 & 2 & 2 \\ 4 & 1 & 3 \end{array}]

A \cdot I_{3} =

Let's label the rows of matrix

A

and the columns of matrix

I_{3}

so that it is easier to express

A \cdot I_{3}

We can find

A \cdot I_{3}

as follows:

$\begin{aligned} A \cdot I_{3} & = [\begin{array}{rr} \vec{a_{1}} \cdot \vec{i_{1}} & \vec{a_{1}} \cdot \vec{i_{2}} & \vec{a_{1}} \cdot \vec{i_{3}} \\ \vec{a_{2}} \cdot \vec{i_{1}} & \vec{a_{2}} \cdot \vec{i_{2}} & \vec{a_{2}} \cdot \vec{i_{3}} \\ \vec{a_{3}} \cdot \vec{i_{1}} & \vec{a_{3}} \cdot \vec{i_{2}} & \vec{a_{3}} \cdot \vec{i_{3}} \end{array}] \\ = [\begin{array}{rr} 1 \cdot 1 + 5 \cdot 0 + 4 \cdot 0 & 1 \cdot 0 + 5 \cdot 1 + 4 \cdot 0 & 1 \cdot 0 + 5 \cdot 0 + 4 \cdot 1 \\ 3 \cdot 1 + 2 \cdot 0 + 2 \cdot 0 & 3 \cdot 0 + 2 \cdot 1 + 2 \cdot 0 & 3 \cdot 0 + 2 \cdot 0 + 2 \cdot 1 \\ 4 \cdot 1 + 1 \cdot 0 + 3 \cdot 0 & 4 \cdot 0 + 1 \cdot 1 + 3 \cdot 0 & 4 \cdot 0 + 1 \cdot 0 + 3 \cdot 1 \end{array}] \\ = [\begin{array}{rr} 1 & 5 & 4 \\ 3 & 2 & 2 \\ 4 & 1 & 3 \end{array}] \end{aligned}$ ‍

The conclusion

The product of any square matrix and the appropriate identity matrix is always the original matrix, regardless of the order in which the multiplication was performed! In other words,

A \cdot I = I \cdot A = A

Connections to the real numbers

Multiplicative Identities

The identity matrix

I

plays a similar role to what the number

1

plays in the real number system.

The number $1$ ‍	The identity matrix $I$ ‍
The product of $1$ ‍ and any number $a$ ‍ is $a$ ‍. $(a \cdot 1 = 1 \cdot a = a)$ ‍	The product of a square matrix $A$ ‍ and the appropriate identity matrix $I$ ‍ is $A$ ‍. $(A \cdot I = I \cdot A = A)$ ‍

Multiplicative Inverses

Two real numbers whose product is the multiplicative identity are called multiplicative inverses. For example, the numbers

\frac{1}{3}

and

3

are multiplicative inverses because

\frac{1}{3} \cdot 3 = 1

and

3 \cdot \frac{1}{3} = 1

In fact, all nonzero real numbers have multiplicative inverses. But does this connection hold with matrix operations?

Consider matrices

A

and

B

$A = [\begin{array}{rr} 2 & 3 \\ 3 & 4 \end{array}]$ ‍ ‍ $B = [\begin{array}{rr} - 4 & 3 \\ 3 & - 2 \end{array}]$ ‍

We can multiply to see that

A B = I_{2}

and

B A = I_{2}

Let's label the rows of matrix

A

and the columns of matrix

B

so that it is easier to express

A B

We can find

A B

as follows:

$\begin{aligned} A B & = [\begin{array}{rr} \vec{a_{1}} \cdot \vec{b_{1}} & \vec{a_{1}} \cdot \vec{b_{2}} \\ \vec{a_{2}} \cdot \vec{b_{1}} & \vec{a_{2}} \cdot \vec{b_{2}} \end{array}] \\ = [\begin{array}{rr} 2 \cdot (- 4) + 3 \cdot 3 & 2 \cdot 3 + 3 \cdot (- 2) \\ 3 \cdot (- 4) + 4 \cdot 3 & 3 \cdot 3 + 4 \cdot (- 2) \end{array}] \\ = [\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}] \end{aligned}$ ‍

To find

B A

, be sure to switch the order of the matrices. Because of this, we must re-label the rows and columns of the matrices.

We can find

B A

as follows:

$\begin{aligned} B A & = [\begin{array}{rr} \vec{b_{1}} \cdot \vec{a_{1}} & \vec{b_{1}} \cdot \vec{a_{2}} \\ \vec{b_{2}} \cdot \vec{a_{1}} & \vec{b_{2}} \cdot \vec{a_{2}} \end{array}] \\ = [\begin{array}{rr} (- 4) \cdot 2 + 3 \cdot 3 & - 4 \cdot 3 + 3 \cdot 4 \\ 3 \cdot 2 + (- 2) \cdot 3 & 3 \cdot 3 + (- 2) \cdot 4 \end{array}] \\ = [\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}] \end{aligned}$ ‍

This means that

A

and

B

are multiplicative inverses.

However, as we will see, not all matrices have multiplicative inverses. This is one place where the properties of real numbers differ from the properties of matrices!

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maximumSomething
Posted 8 years ago. Direct link to maximumSomething's post “How do you find the multi...”
How do you find the multiplicative inverse of a matrix?
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(33 votes)
Answer
- patentfox
  Posted a year ago. Direct link to patentfox's post “Stay tuned”
  Stay tuned
  Comment on patentfox's post “Stay tuned”
  (9 votes)
K
Posted 7 years ago. Direct link to K's post “I think this phrase is no...”
I think this phrase is not precise:
The product of 'any matrix' and the appropriate identity matrix is always the original matrix, regardless of the order in which the multiplication was performed! In other words,
A*I=I*A=A

I think this only work when the matrix A is square matrix.

For example, we have a 3x2 matrix. To make the statement A*I=A to be true, the identity matrix need to be 2x2 matrix. But to make the statement I*A=A to be true, the identity matrix in this case need to be a 3x3 matrix. You can see these two matrices are not equal.
Button navigates to signup pageComment on K's post “I think this phrase is no...”
(22 votes)
Answer
- CCDM
  Posted 4 years ago. Direct link to CCDM's post “You are correct. In your...”
  You are correct. In your 3rd paragraph, you show a right identity matrix such that AI=A and a left identity matrix (IA=A). The right and left identity matrices are equal only if A is square matrix. Nice catch
  Button navigates to signup page
  (15 votes)
StanleyBookOwl
Posted 7 years ago. Direct link to StanleyBookOwl's post “How do you find the ident...”
How do you find the identity matrix for a matrix that is not n by n?
Button navigates to signup pageButton navigates to signup page
(18 votes)
Answer
- willy nrobma
  Posted 7 years ago. Direct link to willy nrobma's post “If you have a matrix that...”
  If you have a matrix that is 2x3, the identity matrix will be 3x3 and look like this:
  1 0 0
  0 1 0
  0 0 1
  If you have a 3x2, the identity will be 2x2 and look like this
  1 0
  0 1
  You can try these out by your own example. The important part is that the multiplication is defined.
  Comment on willy nrobma's post “If you have a matrix that...”
  (16 votes)
meltjl
Posted 6 years ago. Direct link to meltjl's post “What is the use of identi...”
What is the use of identity matrix? IA = A .... and why do we do that?
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(4 votes)
Answer
- David Lee
  Posted 6 years ago. Direct link to David Lee's post “It's useful when you're s...”
  It's useful when you're solving for messy calculations. For example, If you have A + NA + MA, you could just simply do A(I + N + M) which makes the calculation a lot easier. Also, the product of A and A^-1 is I so that's kinda important.
  Hope this helps! If you have any questions or need help, please ask! :)
  Button navigates to signup page
  (10 votes)
Ryden Omine
Posted 3 years ago. Direct link to Ryden Omine's post “is matrices just the plur...”
is matrices just the plural version of a matrix?
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(4 votes)
Answer
- loumast17
  Posted 3 years ago. Direct link to loumast17's post “Yup. same as vertex has ...”
  Yup. same as vertex has the plural vertices.
  Button navigates to signup page
  (4 votes)
Suresh123.Ani
Posted 3 years ago. Direct link to Suresh123.Ani's post “How do you find the multi...”
How do you find the multiplicative inverse of a given matrix?

I guess you could create a system of equation with 4 equations and variables and solve (for a 2*2 matrix). But that seems too lengthy.
Is there an easier way?
Button navigates to signup pageComment on Suresh123.Ani's post “How do you find the multi...”
(4 votes)
Answer
- kubleeka
  Posted 3 years ago. Direct link to kubleeka's post “Make two columns, with yo...”
  Make two columns, with your given matrix on the left and an identity matrix of the same size on the right.
  
  Perform row operations on the matrices. The rule is, whatever operation you do to the left matrix, you must simultaneously do to the right matrix. e.g. if you multiply the top row of your matrix by 5, you must multiply the top row of the identity matrix by 5.
  
  Do row operations until you have an identity matrix on the left. The matrix on the right will be the inverse of your original matrix.
  
  This is still rather computationally intensive, but that's just a fact of life. Matrix inversion is a computationally difficult task, and there is not always a way to simplify something further.
  Button navigates to signup page
  (4 votes)
Andrew See
Posted 8 years ago. Direct link to Andrew See's post “Can anyone give an exampl...”
Can anyone give an example of a matrix without a multiplicative inverse?
I understand the idea but would like to see one and think about it.
Button navigates to signup pageComment on Andrew See's post “Can anyone give an exampl...”
(4 votes)
Answer
- Luis F. Vélez
  Posted 7 years ago. Direct link to Luis F. Vélez's post “Those whose determinant i...”
  Those whose determinant is equal to zero for example.
  Comment on Luis F. Vélez's post “Those whose determinant i...”
  (1 vote)
mariamaboutouk
Posted 7 years ago. Direct link to mariamaboutouk's post “can we have I matrix with...”
can we have I matrix with dimensions 1*1?
if yes , how we can write it ?
Button navigates to signup pageComment on mariamaboutouk's post “can we have I matrix with...”
(1 vote)
Answer
- Ky
  Posted 7 years ago. Direct link to Ky's post “Yes. It would be written ...”
  Yes. It would be written [1]
  
  https://en.wikipedia.org/wiki/Identity_matrix
  Button navigates to signup page
  (6 votes)
GEORGE WILLIAM
Posted 5 years ago. Direct link to GEORGE WILLIAM's post “What is the difference be...”
What is the difference between left and right identity matrices?
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(2 votes)
Answer
- CCDM
  Posted 4 years ago. Direct link to CCDM's post “If one has a matrix A whi...”
  If one has a matrix A which is mxn, then the left identity matrix is mxm such that IA=A and the right identity matrix is nxn such that AI=A.
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  (1 vote)
Garret Cervantez
Posted 7 years ago. Direct link to Garret Cervantez's post “Could you have multiplica...”
Could you have multiplicative inverses on two matrices that have different dimensions i.e a 2x3 matrix multiplied a 3x2 matrix that results in the identity matrix?
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(2 votes)
Answer
- kubleeka
  Posted 7 years ago. Direct link to kubleeka's post “Certainly. Invertible mat...”
  Certainly. Invertible matrices are an important part of linear algebra. Just know that matrices are only properly invertible if they are square matrices. So there is no pair of 2x3 and 3x2 matrices that multiply to an identity matrix in both directions.
  Comment on kubleeka's post “Certainly. Invertible mat...”
  (1 vote)

Precalculus

Course: Precalculus > Unit 7

What you should be familiar with before taking this lesson

Definition of identity matrix

Investigation: Multiplying by the identity matrix

The conclusion

Connections to the real numbers

Multiplicative Identities

Multiplicative Inverses

Want to join the conversation?