Properties of Matrices and Inverse:Definition, Properties and eg-Embibe
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• Written By Preethu

Properties of Matrices and Inverse: Definitions, Properties, Examples

Properties of Matrices and Inverse: Matrices are one of the most powerful tools in Mathematics and greatly simplify our work. The evolution of matrices results from an attempt to find compact and easy methods to solve systems of linear equations. Matrices are not only used to represent the coefficients in a system of linear equations, but their utility far outweighs that.

The concept of an inverse of a matrix is a multidimensional generalisation of the concept of number reciprocal. The product of any number and its reciprocal is $$1$$. Similarly, the product of a square matrix and its inverse is the identity matrix. Matrices and inverse matrices have many properties when the arithmetic operations are performed.

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Properties of Matrices

The arithmetic operations like addition, subtraction, and multiplication can be performed on matrices.

If $$A = \left[ {{a_{ij}}} \right]$$ and $$B = \left[ {{b_{ij}}} \right]$$ are two matrices of the order $$m \times n$$. Then, the sum of the two matrices $$A$$ and $$B$$ is defined as a matrix $$C = {\left[ {{c_{ij}}} \right]_{m \times n}}$$, where $${c_{ij}} = {a_{ij}} + {b_{ij}}$$, for all possible values of $$i$$ and $$j$$.

1. Commutative Law

If $$A = \left[ {{a_{ij}}} \right],\;B = \left[ {{b_{ij}}} \right]$$ are matrices of the same order, say $$m \times n$$, then $$A + B = B + A$$. Now $$A + B = \left[ {{a_{ij}}} \right] + \;\left[ {{b_{ij}}} \right]$$
$$= \left[ {{a_{ij}} + {b_{ij}}} \right]$$
$$= \left[ {{b_{ij}} + {a_{ij}}} \right]$$ (addition of numbers is commutative)
$$= \left( {\left[ {{b_{ij}}\left] + \right[{a_{ij}}} \right]} \right) = B + A$$

2. Associative Law

For any three matrices $$A = \left[ {{a_{ij}}\left] {,\;B = } \right[{b_{ij}}\left] {,\;C = } \right[{c_{ij}}} \right]$$ of the same order, say $$m \times n$$, then $$\left( {A + B} \right) + C = A + (B + C)$$.
Now $$\left( {A + B} \right) + \;C = \left( {\left[ {{a_{ij}}} \right] + \;\left[ {{b_{ij}}} \right]} \right) + \;\left[ {{c_{ij}}} \right]$$
$$= \left[ {{a_{ij}} + {b_{ij}}} \right] + \;\left[ {{c_{ij}}} \right]$$
$$= \left[ {\left( {{a_{ij}} + {b_{ij}}} \right) + \;{c_{ij}}} \right]$$
$$= \left[ {{a_{ij}} + \left( {{b_{ij}} + {c_{ij}}} \right)} \right]$$
$$= \left[ {{a_{ij}}} \right] + \;\left[ {\left( {{b_{ij}} + {c_{ij}}} \right)} \right]$$
$$= \left[ {{a_{ij}}} \right] + \left( {\left[ {{b_{ij}}} \right] + \left[ {{c_{ij}}} \right]} \right)$$
$$\therefore \,\left( {A + B} \right) + \;C = A + \left( {B + C} \right)$$

Let $$A = {\left[ {a_{ij}} \right]}$$ be an $${m \times n}$$ matrix and $$O$$ be an $$m \times n$$ zero matrix, then $$A + O = O + A$$. In other words, $$O$$ is the additive identity for matrix addition.

Let $$A = {\left[ {{a_{ij}}} \right]_{m \times n}}$$ be any matrix. Then there is another matrix, $$– A = {\left[ { – {a_{ij}}} \right]_{m \times n}}$$, such that $$A + \left( {-A} \right) = \left( {-A} \right) + A = 0$$. So $$-A$$ is the additive inverse of $$A$$ or negative of $$A$$.

Properties of Matrix Multiplication

Following are the Properties of Matrix Multiplication

Properties of Scalar Multiplication of a Matrix

If $$A = \left[ {{a_{ij}}} \right],\;B = \left[ {{b_{ij}}} \right]$$ are matrices of the same order, say $$m \times n$$, and $$k$$ and $$l$$ are scalars, then
(i) $$k\left( {A\; + B} \right) = kA + kB$$
$$k\left( {A + B} \right) = k\left( {\left[ {{a_{ij}}} \right] + \;\left[ {{b_{ij}}} \right]} \right)$$
$$= k\left[ {{a_{ij}} + {b_{ij}}} \right]$$
$$= \left[ {k\left( {{a_{ij}} + {b_{ij}}} \right)} \right]$$
$$= \left[ {\left( {k{a_{ij}}} \right) + \;\left( {k{b_{ij}}} \right)} \right]$$
$$= \left[ {k{a_{ij}}} \right] + \;\left[ {k{b_{ij}}} \right]$$
$$= k\left[ {{a_{ij}}\left] { + k} \right[{b_{ij}}} \right] = kA + kB$$

(ii) $$\left( {k + l} \right)A = kA + lA$$
$$\left( {\;k + l} \right)A = \left( {k + l} \right)\left[ {{a_{ij}}} \right]$$
$$= \left[ {\left( {k + l} \right){a_{ij}}} \right] + \;\left[ {k{a_{ij}}} \right] + \;\left[ {l{a_{ij}}} \right]$$
$$= k\left[ {{a_{ij}}\left] { + l} \right[{a_{ij}}} \right] = kA + lA$$

(iii) For any square matrix $$A$$,
$${A^r}{A^s} = {A^{r + s}}$$
And
$${\left( {{A^r}} \right)^s} = {A^{rs}}$$
These are called the power laws.

Properties of Multiplication of Matrices

The product of two matrices $$A$$ and $$B$$ can be calculated if and only if the number of columns in $$A$$ and the number of rows in $$B$$ are equal.

Let $$A = \left[ {{a_{ij}}} \right]$$ be an $$m \times n$$ matrix and $$B = \left[ {{b_{jk}}} \right]$$ be an $$n \times p$$ matrix. Then the product of the matrices $$A$$ and $$B$$ is the matrix $$C$$ of order $$m \times p$$. To get the $$(i,\,k)^{th}$$ element $$C_{ik}$$of the matrix $$C$$, we take the $$i^{th}$$ row of $$A$$ and $$k^{th}$$ column of $$B$$, multiply them, elementwise, and take the sum of all these products.

1. Associative Law
For any three matrices $$A,\,B$$ and $$C$$, we have
$$\left( {AB} \right)\;C = A\;\left( {BC} \right)$$

2. Distributive Law
For three matrices $$A,\,B$$ and $$C$$,
(i) $$A\left( {B + C} \right) = AB + AC$$
(ii) $$\left( {A + B} \right)C = AC + BC$$

3. Existence of Multiplicative Identity

For every square matrix $$A$$, there exists an identity matrix of the same order such that
$$IA = AI = A$$

Properties of Transpose Matrix

If $$A = [a_{ij}]$$ be an $$m \times n$$ matrix, then the matrix obtained by interchanging the rows and columns of $$A$$ is called the transpose of $$A$$. Transpose of the matrix $$A$$ is denoted by $$A’$$ or $$(A^T)$$.
The properties of the transpose of matrices are:
For any matrices $$A$$ and $$B$$ of the same order, we have

(i) The transpose of a transpose of a matrix is the matrix itself.
$$(A’) = A$$

(ii) If a scalar quantity $$k$$ is multiplied by a matrix $$M$$, and taken the transpose of it, it is equal to the scalar multiplied by the transpose of the matrix $$M’$$.
$$\left( {kA} \right)’ = kA’$$, where $$k$$ is any constant

(iii) The sum of transposes of matrices is equal to the transpose of the sum of two matrices.
$$\left( {A + B} \right)’ = A’ + B’$$

(iv) The transpose of the product of two matrices is equal to the product of the transpose of the two matrices in reverse order.
$$(AB)’ = B’A’$$

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Inverse Matrix

If $$A$$ and $$B$$ are two square matrices such that $$AB = BA = I$$, then $$B$$ is the inverse matrix of $$B$$, and it is also denoted by $$A^{-1}$$ and $$A$$ is the inverse of $$B$$. If the inverse of a square matrix exists, then it is unique. Any matrix that has an inverse is said to be invertible. If there is no inverse for a matrix, it is called singular.

Properties of Inverse Matrices

A non-singular matrix is a square matrix whose determinant is non-zero. The non-singular matrix property is satisfied to find the inverse of a matrix.

Consider two non-singular matrices $$A$$ and $$B$$, then the properties of inverse matrices are defined as below:

1. If $$A$$ is non-singular, then $$A^{-1}$$ is non-singular. Also,
$${({A^{ – 1}})^{ – 1}} = A$$
2. If $$A$$ and $$B$$ are non-singular matrices, then $$AB$$ is non-singular matrix. Also,
$${\left( {AB} \right)^{ – 1}} = {B^{ – 1}}\;{A^{ – 1}}$$
Proof: Let $$A$$ and $$B$$ be invertible matrices and let $$C = AB$$.
So, $$C^{-1} = (AB)^{-1}$$
Consider $$C = AB$$.
Multiply both sides on the left by $$A^{-1}$$.
$${A^{ – 1}}C\; = \;{A^{ – 1}}AB\; = \;B$$
So, $${B^{ – 1}}{A^{ – 1}}$$ is the matrix you need to multiply $$C$$ by to get the identity.
Thus, by the definition of inverse $${B^{ – 1}}{A^{ – 1}}\; = {C^{ – 1}} = {\left( {AB} \right)^{ – 1}}$$.
Similarly, $${\left( {ABC} \right)^{ – 1}} = {C^{ – 1}}{B^{ – 1}}{A^{ – 1}}$$
In general, $${\left( {{A_1}\;{A_2} \ldots .{A_n}} \right)^{ – 1}} = A_n^{ – 1}A_{n – 1}^{ – 1} \ldots \ldots A_2^{ – 1}A_1^{ – 1}$$
3. If $$A$$ is non-singular then $${\left( {{A^T}} \right)^{ – 1}} = {\left( {{A^{ – 1}}} \right)^T}{\left( {{A^T}} \right)^{ – 1}} = {\left( {{A^{ – 1}}} \right)^T}$$
4. If $$A$$ and $$B$$ are two non-singular matrices with $$AB = I$$ then $$A$$ and $$B$$ are inverses of each other. So,
$$AB = I$$ and $$BA = I$$
5. If $$k$$ is any non-zero scalar, then $$kA$$  is invertible, then
$${\left( {kA} \right)^{ – 1}} = \left( {\frac{1}{k}} \right){A^{ – 1}}$$
6. If $$A$$ is a square matrix and for $$n = 0,\,2,\,……,A^n$$ is invertible, then
$${\left( {{A^{ – 1}}} \right)^n} = {A^{ – n}}$$

Solved Examples – Properties of Matrices and Inverse

Below are a few solved examples that can help in getting a better idea.

Q.1. The sum of two matrices $$X$$ and $$y$$ is given as $$X + Y = \left[ {\begin{array}{*{20}{c}} 5&2 \\ 0&9 \end{array}} \right]$$. Then, find the value of the sum of matrices in the reverse order, $$Y + X$$.

Solution: By the properties of matrix addition, we know that if $$A = \left[ {{a_{ij}}} \right],\;\,B = \left[ {{b_{ij}}} \right]$$ are matrices of the same order, say $$m \times n$$, then $$A + B = B + A$$.
So, here $$X + Y = Y + X$$
Hence, $$Y + X = \left[ {\begin{array}{*{20}{c}} 5&2 \\ 0&9 \end{array}} \right]$$

Q.2. Find the values of $$x$$ and $$y$$ from the equation $$2\left[ {\begin{array}{*{20}{c}} x&5 \\ 7&{y – 3} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 3&{ – 4} \\ 1&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 7&6 \\ {15}&{14} \end{array}} \right]$$
Solution: We have
$$2\left[ {\begin{array}{*{20}{c}} x&5 \\ 7&{y – 3} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 3&{ – 4} \\ 1&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 7&6 \\ {15}&{14} \end{array}} \right]$$
$$\Rightarrow \left[ {\begin{array}{*{20}{c}} {2x}&{10} \\ {14}&{2y – 6} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 3&{ – 4} \\ 1&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 7&6 \\ {15}&{14} \end{array}} \right]$$
$$\Rightarrow \left[ {\begin{array}{*{20}{c}} {2x + 3}&{10 – 4} \\ {14 + 1}&{2y – 6 + 2} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 7&6 \\ {15}&{14} \end{array}} \right]$$
$$\Rightarrow \left[ {\begin{array}{*{20}{c}} {2x + 3}&6 \\ {15}&{2y – 4} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 7&6 \\ {15}&{14} \end{array}} \right]$$
$$\Rightarrow 2x + 3 = 7$$
$$\Rightarrow 2x = 4$$
$$\therefore x = 2$$
Also, $$2y – 4 = 14$$
$$\Rightarrow 2y = 18$$
$$\therefore y = 9$$
Hence, the value of $$x$$ and $$y$$ is $$2$$ and $$9$$ respectively.

Q.3. $$A,\,B$$ and $$C$$ are three matrices. $$\left( {A + B} \right)C = \left[ {\begin{array}{*{20}{c}} {10} \\ {20} \\ {28} \end{array}} \right].$$ Find the value of $$AC + BC$$
Solution: By the distributive property, the matrix multiplication for three matrices $$A,\,B$$ and $$C$$ is given by
$$A\;\left( {B + C} \right) = AB + AC$$ and $$\left( {A + B} \right)\;C = AC + BC$$
whenever both sides of the equality are defined.
Hence, $$AC + BC = \left( {A + B} \right)C = \left[ {\begin{array}{*{20}{c}} {10} \\ {20} \\ {28} \end{array}} \right]$$.

Q.4. $$A = \left[ {\begin{array}{*{20}{c}} 1&1&{ – 1} \\ 2&0&3 \\ 3&{ – 1}&2 \end{array}} \right]$$, $$B = \left[ {\begin{array}{*{20}{c}} 1&3 \\ 0&2 \\ { – 1}&4 \end{array}} \right]$$ and $$C = \left[ {\begin{array}{*{20}{c}} 1&2&3&{ – 4} \\ 2&0&{ – 2}&1 \end{array}} \right]$$. $$A\left( {BC} \right) = \left[ {\begin{array}{*{20}{c}} 4&4&4&{ – 7} \\ {35}&{ – 2}&{ – 39}&{22} \\ {31}&2&{ – 27}&{11} \end{array}} \right]$$. Find the value of $$(AB)C$$.
Solution: We know that for any three matrices $$A,\,B$$ and $$C$$, we have
$$(AB) C=A (BC)$$ whenever both sides of the equality sign are defined.
Hence, $$\left( {AB} \right)C = \left[ {\begin{array}{*{20}{c}} 4&4&4&{ – 7} \\ {35}&{ – 2}&{ – 39}&{22} \\ {31}&2&{ – 27}&{11} \end{array}} \right]$$

Q.5. $$A = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 2&{ – 1} \end{array}} \right]$$ and $${A^{ – 1}} = \left[ {\begin{array}{*{20}{c}} {\frac{1}{5}}&{\frac{2}{5}} \\ {\frac{2}{5}}&{\frac{{ – 1}}{5}} \end{array}} \right]$$. Find $$(A^{-1})^{-1}$$
Solution: If $$A$$ is non-singular, then so is $$A^{-1}$$ and $$(A^{-1})^{-1} = A$$
Hence, $${({A^{ – 1}})^{ – 1}} = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 2&{ – 1} \end{array}} \right]$$

Q.6. A matrix is given by $$A = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 4&3 \end{array}} \right]$$. Find the $$AI$$.
Solution: For every square matrix $$A$$, there exists an identity matrix of the same order such that
$$IA = AI = A$$
Hence, for the given matrix, $$AI = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 4&3 \end{array}} \right]$$.

Q.7. Find $$A^6$$, where $$A = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 1&1 \end{array}} \right]$$
Solution: By the power rule, we know $${A^6}\; = {A^2}{A^4}\; = \;{A^2}\;\;{\left( {{A^2}} \right)^2}$$
Now $${A^2} = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 2&1 \end{array}} \right]$$
So, $${A^2}{\left( {{A^2}} \right)^2} = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 2&1 \end{array}} \right]{\left[ {\begin{array}{*{20}{c}} 1&0 \\ 2&1 \end{array}} \right]^2}$$
$$= \left[ {\begin{array}{*{20}{c}} 1&0 \\ 2&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0 \\ 3&1 \end{array}} \right]$$
$$= \left[ {\begin{array}{*{20}{c}} 1&0 \\ 5&1 \end{array}} \right]$$

Q.8. Given matrix $$A = \left[ {\begin{array}{*{20}{c}} 5&2 \\ { – 7}&{ – 3} \end{array}} \right].$$ Find the product of $$A A^{-1}$$.
Solution: We know that $$AA^{-1} = I$$.
Hence, the product of the given matrix and its inverse is $$A{A^{ – 1}} = I = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]$$

Summaryof Properties of Matrices and Inverse

Matrices can be operated like numbers in unique ways of addition, subtraction and multiplication. The conditions involved in each of these operations vary. In the way similar to finding the reciprocal of a number, we can find the inverse of a matrix, provided the matrix is invertible. Matrices and inverse matrices are involved in various operations and display several properties. Properties like commutativity, associativity, and distributivity are defined for matrices. Matrix notation and its operations are used in electronic spreadsheet programmes for personal computers, which are then used in various areas of Business and Science, such as budgeting, sales projection, cost estimation, and analysing results.

FAQsof Properties of Matrices and Inverse

Students might be having many questions regarding the Properties of Matrices and Inverse. Here are a few commonly asked questions and answers.

Q.1. What is a non-singular matrix?
Ans: A non-singular matrix is a square matrix whose determinant is non-zero. The non-singular matrix property is satisfied to find the inverse of a matrix.

Q.2. What is the inverse of a matrix?
Ans: If $$A$$ and $$B$$ are two square matrices such that $$AB = BA = I$$, then $$B$$ is the inverse matrix of$$A$$ and is denoted by $$A^{-1}$$, and $$A$$ is the inverse of $$B$$. If the inverse of a square matrix exists, then it is unique. The matrix is said to be invertible.

Q.3. What is the condition of the inverse matrix?
Ans: To get the inverse of a matrix, the matrix must be a square matrix with the same number of rows and columns, and the determinant of the matrix must be non-zero. In simpler words, the matrix must be square and singular.

Q.4. Which are the properties of the transpose of a matrix?
Ans: The properties of the transpose of matrices are:
For any matrices $$A$$ and $$B$$ of suitable orders, we have
(i) $$(A’)’ = A$$
(ii) $$(kA)’ = kA’$$ (where $$k$$ is any constant)
(iii) $$(A + B)’ = A’ + B’$$
(iv) $$(AB)’ = B’ A$$

Q.5. How are matrices used in real life?
Ans: Matrix representations can show many physical operations such as magnification, rotation, and reflection through a plane. This is also used in Genetics, Economics, Sociology, Modern Psychology, and Industrial Management.

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