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Identity Matrix is a type of Square matrix that has one’s along the diagonal and zeros in all the other entries. Identity Matrix is often known as Unit Matrix and Elementary Matrix. In this article, we will discuss more about Identity Matrix, its properties, examples, and important questions.
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Key Terms: Identity Matrix, Square Matrix, Order, Diagonal, Inverse, matrix
Identity Matrix
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The identity matrix is an n × n square matrix with the diagonal containing the ones and all the other entries being zeros. It is represented as simply I or In, where n represents the size of the square matrix.
For Example:
\(I_1 = 1 \\ I_2 = \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix} \\ I_3 = \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} \)
The Identity matrices are also known as diagonal matrices as the elements are present in the diagonals in these matrices.
If we multiplying any matrix by the identity matrix the result will be the provided matrix. The elements of the provided matrix will remain unaffected. For Example:
Let, an Identity Matrix is donated by In × n, where ‘n × n’ shows the order of the matrix.
And A = Any square matrix of order n × n
So,
A × In × n = A
This implies that on multiplying any square matrix with an identity matrix, the elements of the given matrix remain unchanged.
In other terms, an identity matrix is one in which all of the principal diagonals of a square matrix are 1’s and the rest are all 0’s. The examples of 2 × 2 and 3 × 3 Identity Matrices are shown below:
2 × 2 Identity Matrix:
3 × 3 Identity Matrix:
The video below explains this:
Matrices Detailed Video Explanation:
Read More: Invertible Matrices
Properties of Identity Matrix
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1. An identity matrix is always a square matrix:
The Identity matrix is a square matrix as it has the same number of rows and columns. For any whole number n, the identity matrix will be of order n × n.
2. As long as the following rules are followed, an identity matrix can multiply any matrix with any order (dimensions):
- If the identity matrix is the term in the multiplication, it must have dimensions with the same number of columns as the matrix it is multiplying.
Multiplying an identity matrix to a non-unit matrix
- The identity matrix must have the same number of rows as the matrix being multiplied if it is the second term in the multiplication.
Multiplying a non-unit matrix to an identity matrix
- If an identity matrix multiplies a square matrix of the same dimensions, the result will be a square matrix.
Multiplying Square matrix with an Identity Matrix
3. If an identity matrix multiplies another matrix and the multiplication can be solved, the result is the non-unit matrix involved. In terms of mathematics:
\(I_m \cdot A_{m \times n} = A_{m \times n}\) and \(A_{m \times n} \cdot I_m = A_{m \times n}\)
Multiplying an identity matrix to another matrix
4. If a matrix is multiplied by its inverse, the result will be an identity matrix with the same order as the matrices.
\(\begin{align*} AA^{-1}&= \begin{bmatrix} \frac{1}{3} & \frac{1}{3} \\ 0 & \frac{2}{3} \end{bmatrix} \begin{bmatrix} 3 & -\frac{3}{2} \\ 0 & \frac{3}{2} \end{bmatrix} \\ &= \begin{bmatrix} \frac{1}{3} \cdot 3 + \frac{1}{3} \cdot 0 & \frac{1}{3} \cdot (-\frac{3}{2}) + \frac{1}{3} \cdot \frac{3}{2}\\ 0 \cdot 3 + \frac{2}{3} \cdot 0 & 0 \cdot (-\frac{3}{2}) + \frac{2}{3} \cdot \frac{3}{2} \end{bmatrix} \\ &= \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix} \end{align*}\)
Multiplying a matrix by its inverse
Read More: Scalar Matrix
Things to Remember
- The identity matrix is an n × n square matrix with the diagonal containing the ones and all the other entries being zeros.
- It is represented as simply I or In, where n represents the size of the square matrix.
- An identity matrix is always a square matrix.
- When two inverse matrices are multiplied together, the result is an identity matrix.
- The identity matrix multiplied by any matrix ‘A' results in the matrix ‘A' itself.
Also Read:
Sample Questions
Ques: If \(A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}\), then, A+A′ = I. Find ∝. (3 Marks)
Ans:
\(A+A^{\prime} = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} + \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}\)
\(= \begin{bmatrix} 2\cos \alpha & 0 \\ 0 & 2\cos \alpha \end{bmatrix}\)
\(A+A^{\prime} = I\) (Given)
\(\begin{bmatrix} 2\cos \alpha & 0 \\ 0 & 2\cos \alpha \end{bmatrix} = \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}\)
\(2 \cos \alpha = 1\)
\(\cos \alpha = \frac{1}{2}\)
\(\cos \alpha = \cos \frac{\pi}{3}\)
\(\alpha = \frac{\pi}{3}\)
Ques: If \(A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\) and A2 = I. Find relation given by A2 = I. (2 Marks)
Ans:
\(A^2 = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\)
\(= \begin{bmatrix} \alpha^2 + \beta\gamma & \alpha\beta - \alpha\beta \\ \alpha\gamma - \alpha\gamma & \beta\gamma + \alpha^2 \end{bmatrix}\)
\(= \begin{bmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \beta\gamma + \alpha^2 \end{bmatrix}\)
Then according to the question:
\(\begin{bmatrix} \alpha^2 + \beta\gamma & 0 \\ 0 & \beta\gamma + \alpha^2 \end{bmatrix} = \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}\)
\(\Rightarrow \alpha^2 + \beta\gamma = 1 \\ \Rightarrow \alpha^2 + \beta\gamma -1 = 0\)
Ques: If a matrix A and an identity matrix I are given as: \(A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \quad I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), then, find K. So that A2 = KA – 2I (3 Marks)
Ans: A2 = A.A
\(= \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\)
\(= \begin{bmatrix} 9-8 & -6+4\\ 12-8 & -8+4 \end{bmatrix}\)
\(= \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}\)
A2 = KA – 2I
\(\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = K \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
\(\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3K & -2K \\ 4K & -2K \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}\)
\(\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 3K & -2K \\ 4K & -2K \end{bmatrix}\)
∴ K = 1
Ques: Obtain the inverse of the following matrix using elementary operations (5 Marks)
\(A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}\)
Ans: We know that: A = AI
\(\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot A\)
\(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot A \quad [R_1 \Leftrightarrow R_2]\)
\(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{bmatrix} \cdot A \quad [R_3 \rightarrow R_3 - 3R_1]\)
\(\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{bmatrix} = \begin{bmatrix} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -3 & 1 \end{bmatrix} \cdot A \quad [R_1 \rightarrow R_1 - 2R_2]\)
\(\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 5 & -3 & 1 \end{bmatrix} \cdot A \quad [R_3 \rightarrow R_3 + 5R_2]\)
\(\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 & 0 \\ 1 & 0 & 0 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{bmatrix} \cdot A \quad [R_3 \rightarrow \frac{1}{2}R_3]\)
\(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \\ 1 & 0 & 0 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{bmatrix} \cdot A \quad [R_1 \rightarrow R_1+R_3]\)
\(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{bmatrix} \cdot A \quad [R_2 \rightarrow R_2-2R_3]\)
Hence,
\(A^{-1} = \begin{bmatrix} \frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2} \end{bmatrix}\)
Ques: If, \(A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\) Show that (aI+bA)n = anI + nan-1bA. Where I is the identity matrix of order 2 and n ε N (5 Marks)
Ans: When n = 1
(aI+bA)n = anI + nan-1bA
aI + bA = aI + bA
L.H.S = R.H.S
When n = k
(aI + bA)K = AKI + KaK-1bA… (i)
Result is true for n = k
When n = k + 1
(aI + bA)k+1 = (aI + bA). (aI + bA)k
= (aI + bA). (akI + kak-1ba) [From (i)]
= aI (akI + kak-1ba) + bA (akI + kak-1 bA)
= ak+1I + kakba + akba + kak-1 b2A2
\(\because I I = I \\ IA = A =AI\)
= ak+1 + (k+1) akbA (A2 = 0)
Hence result is true for n = k+1
Whenever it is true for n = k.
Hence Proved.
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