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1 Million Means: 1 Million in Rupees, Lakhs and Crores

June 5, 2024**Properties of Matrix:** Matrix properties are useful in many procedures that require two or more matrices. Using properties of matrix, all the algebraic operations such as multiplication, reduction, and combination, including inverse multiplication, as well as operations involving many types of matrices, can be done with widespread efficiency. Additive, multiplicative identity, and inverse matrices are included in this study of matrices’ features.

Now let’s learn more about the features of matrix addition, scalar multiplication of matrices, matrix multiplication, transpose matrix, and inverse matrix through examples and frequently asked questions. Keep on reading this article so that you can know more about the properties of matrix and their corresponding numerical counterparts.

A matrix is a rectangular array or table arranged in rows and columns of numbers or variables. Arthur Cayley is the father of matrices. A matrix is denoted by a capital letter (e.g\(A,B,X,…..\)etc). The elements of a matrix are represented by lower case letters with a double subscript (e.g., \({a_{ij}},{b_{ij}},{x_{ij}},….\) etc).

In general, a matrix is denoted by \(A = {\left[ {{a_{ij}}} \right]_{m \times n}}\), where \({a_{ij}}\) belongs to \({i^{{\mathop{\rm th}\nolimits} }}\) row and \({j^{{\rm{th}}}}\) column.

A general representation of matrix with an order \(m×n\) is given by \({[A]_{m \times n}}\), where

- \(m-\) Number of rows in matrix \(A\).
- \(n-\) Number of columns in matrix \(A\).

The three basic yet main operations on matrices are addition, subtraction, and multiplication.

If \(A = {\left[ {{a_{ij}}} \right]_{m \times n}}\) and \(B = {\left[ {{b_{ij}}} \right]_{m \times n}}\) are matrices of the same order, then addition \(A+B\) is a matrix, obtained by adding the corresponding elements of two matrices. i.e.

\({\left[ {{a_{ij}} + {b_{ij}}} \right]_{m \times n}} = A + B\)

Example:

\(\left[ {\begin{array}{*{20}{c}}
{{a_1}}&{{b_1}}\\
{{c_1}}&{{d_1}}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{{a_2}}&{{b_2}}\\
{{c_2}}&{{d_2}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{a_1} + {a_2}}&{{b_1} + {b_2}}\\
{{c_1} + {c_2}}&{{d_1} + {d_2}}
\end{array}} \right]\)

Let two matrices A and B are of the same order, then the subtraction \(A – B = A + ( – B)\) is obtained by subtracting the corresponding elements.

\(A – B = {\left[ {{a_{ij}} – {b_{ij}}} \right]_{m \times n}}\)

Example:

\(\left[ {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}\\ {{c_1}}&{{d_1}} \end{array}} \right] – \left[ {\begin{array}{*{20}{c}} {{a_2}}&{{b_2}}\\ {{c_2}}&{{d_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{a_1} – {a_2}}&{{b_1} – {b_2}}\\ {{c_1} – {c_2}}&{{d_1} – {d_2}} \end{array}} \right]\)

If \(A\) is a matrix of order \(m \times n\) and \(B\) is a matrix of order \(n \times p\), then the order of the product matrix is \(m \times p\).

- Check that the number of columns in the \({{\rm{1}}^{{\rm{st}}}}\) matrix has the same number of rows in the \({{\rm{2}}^{{\rm{nd}}}}\) matrix for the compatibility of matrices for multiplication.
- Multiply the elements of each row of the first matrix by the elements of each column in the second matrix.
- Add the products obtained.
- Put the added products in the respective columns.

If * *\(A = {\left[ {{a_{ij}}} \right]_{m \times n}}\) is a matrix and \(k\) any number, then the matrix which is obtained by multiplying all the elements of \(A\) by the scalar \(k\) is called the scalar multiplication of \(A\) by \(k\).

Then \(k{A_{m \times n}} = {A_{m \times n}}k = {\left[ {k{a_{ij}}} \right]_{m \times n}}\)

A transpose of a matrix is obtained by changing rows to columns and columns to rows. Transpose of the matrix is denoted by \({A^T}\) or \(A’\).

Example:

\(A = \left[ {\begin{array}{*{20}{c}}
{{a_1}}&{{b_1}}\\
{{c_1}}&{{d_1}}
\end{array}} \right],\) then \({A^T} = \left[ {\begin{array}{*{20}{c}}
{{a_1}}&{{c_1}}\\
{{b_1}}&{{d_1}}
\end{array}} \right]\)

The inverse of a matrix \(A\), denoted by \({A^{ – 1}}\) is given by \({A^{ – 1}} = \frac{1}{{\det A}}adj(A),\), where (adj(A)\) is the adjugate of the matrix \(A\) and det \(A\)is the value of the determinant of the matrix \(A\).

Note that \(A \times {A^{ – 1}} = {A^{ – 1}} \times A = I,\) where \(I\) is the identity matrix of the same order as the matrix \(A\).

- In a square matrix, the number of rows and columns are equal.
- A square matrix is a diagonal matrix in which non-diagonal elements are zeroes.
- A square matrix is a scalar matrix in which the non-diagonal elements are zeroes, and diagonal elements are the same.
- A matrix in which all non-diagonal elements are zeroes and diagonal elements are \(1s\) is said to be an identity matrix.
- A matrix is a symmetric matrix if \(A = {A^T}\)
- A matrix is skew-symmetric if \(A = – {A^T}\)
- A matrix is an orthogonal matrix; if \(A \times {A^T} = I\)
- A matrix is said to be a singular matrix, if \(|A| = 0\)
- A matrix is said to be idempotent, if \({A^2} = A\).
- A matrix is said to be involutory, if \({A^2} = I\)

The addition of two matrices follows the commutative law. For two matrices, \(A\) and \(B\), of the same order, we have \(A+B=B+A\).

Now, if there is another matrix \(C\), such that \(A+B=A+C\), then \(B+A=C+A\), and in this case, we can say that \(B=C\).

The addition of matrices follows the commutative law. This means the result of the addition of matrices is the same in whatever manner we group the matrices. For three matrices \(A, B\) and \(C\) of the same order, we have,

\((A+B)+C=A+(B+C)\)

A matrix added to the given matrix, which gives the same matrix as the sum, is an identity matrix for addition.

\(A+O=O+A=A\),

where \(O\) is zero matrix which is the additive identity of the matrix,

- A matrix that is added to the given matrix to get the sum zero is called additive inverse.

\(A+(-A)=0=(-A)+A\),

where \((-A)\) is obtained by changing the sign of every element of \(A\), which is the additive inverse of the matrix. - For two matrices \(A\) and \(B\) of the same order, if we have \(A+B=0=B+A\), then \(B\) is called the additive inverse of \(A\) and also \(A\) is called the additive inverse of \(B\).

Let A, B are two matrices having the same order and \(\lambda \) and \(\mu \) are any two scalars then;

- \(\lambda (A + B) = \lambda A + \lambda B\)
- \((\lambda + \mu )A = \lambda A + \mu A\)
- \(\lambda (\mu A) = (\lambda \mu A) = \mu (\lambda A)\)
- \(( – \lambda A) = – (\lambda A) = \lambda ( – A)\)

Multiplication of two matrices is possible only when the number of columns of the first matrix is equal to the number of rows of the second matrix. If we have a matrix \(A\) of the order \(m×n\) and another matrix \(B\) whose order is \(n×p\), then multiplication of matrix \(A\) with \(B\) is possible. We write it as \(A×B\) or \(AB\). In this case, the order of the product is \(m×p\).

Some properties of the multiplication of two matrices \(A\) and \(B\), then

In general, matrix multiplication is not commutative.

\(AB≠BA\)

However, in some specific cases, two matrices \(A\) and \(B\) follow commutative law. They are

- If \(A=I\), then \(AB=BA\)
- If \(A=B\), then \(AB=BA\)
- If \(A = {B^n}\), then \(AB=BA\)
- If \(A=\)polynomial \((B)\), then \(AB=BA\)
- If \(B\) is invertible and \(A = {B^{ – n}}\), then \(AB=BA\)
- If \(B\) is invertible and \(A=\) polynomial \(\left( {B,{B^{ – 1}}} \right)\). then \(AB=BA\)

Multiplication of matrices is associative.

\((AB)C= A(BC)\)

Multiplication of matrices is distributive over matrix addition.

\(A(B+C)=AB+AC\) and \((A+B)C=AC+BC\)

If \(A\) is an \(m×n\) matrix, then \({I_{m \times m}}{A_{m \times n}} = A = {A_{m \times n}}{I_{n \times n}}\) . Then, \(I\) is said to be a multiplicative identity.

The multiplicative inverse of a matrix \(A\) is a matrix denoted by \({A^{ – 1}}\), such that, \(A \times {A^{ – 1}} = {A^{ – 1}} \times A = I,\), where \(I\) is the identity matrix of the same order as the matrix \(A\).

If \(AB=0\), it is not necessarily that either \(A=0\) or \(B=0\). Both \(A\) and \(B\) can be non-zero matrices.

In general, the cancellation Law does not hold for matrix multiplication. For any three matrices \(A, B\) and \(C\), if \(AB=AC\), then it do not automatically imply \(B=C\). However, \(B=C\), only when \(A\) is invertible.

For any matrix \(A\), we have \(A×O=O×A=O\), where \(O\) is a zero matrix.

Let \(A\) and \(B\) are two matrices, and \(k\) be any real number, then

- \({\left( {{A^T}} \right)^T} = A\)
- \({(A + B)^T} = {A^T} + {B^T}\)
- \({(A \times B)^T} = {B^T} \times {A^T}\)
- \({(kA)^T} = k{A^T}\)

Let \(A\) and \(B\) are two non-singular matrices, then

- \({\left( {{A^{ – 1}}} \right)^{ – 1}} = A\)
- \({(A \times B)^{ – 1}} = {B^{ – 1}} \times {A^{ – 1}}\)
- \({(A \times B \times C)^{ – 1}} = {C^{ – 1}} \times {B^{ – 1}} \times {A^{ – 1}}\)
- \(AB = {I_n}\) where, \(A\) and \(B\) are inverse of each other.
- \({\left( {{A^T}} \right)^{ – 1}} = {\left( {{A^{ – 1}}} \right)^T}\)
- \({\left( {{A^n}} \right)^{ – 1}} = {A^{ – n}}\)where \(A\) is a square matrix and \(n>0\)

- A matrix is said to be singular, whose determinant equal to zero.

\(\det \,\det \,A = 0\) - Determinant of an identity matrix \(\left( {{I_{n \times n}}} \right)\) of any order is \(1\).

\(\left| {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right| = 1\) - The determinants of multiplication or product of two matrices equal to the product of their individual determinants. Let \(A\) and \(B\) are two matrices:

\(\det \,\det \,\left({AB} \right) = \det \,\det \,A \times \det \,\det \,B\) - The determinant of a matrix with power \(n\) is given by

\(\det \,\det \,{A^n} = {\left({\det \,\det \,A} \right)^n}\) - The determinant of the inverse matrix is equal to the reciprocal of the determinant of the matrix.

\(\det \,\det \,\left({ {A^ { – 1}}} \right) = \frac{1}{ {\det \,\det \,A}}\) - Properties of eigenvalues: Let \({\lambda _1},{\lambda _2},{\lambda _3} \ldots \ldots ,{\lambda _n}\) are the Eigenvalues of a square matrix (A), then the determinant of the matrix is equal to the product of the Eigenvalues.

\(\det A = {\lambda _1} \times {\lambda _2} \times {\lambda _3} \ldots \ldots \times {\lambda _n}\)

** Q.1. If \(A = \left[ {\begin{array}{*{20}{c}}
3&2\\
4&0
\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}
3&0\\
3&2
\end{array}} \right].\) Find \(AB\) and \(BA\), check whether they are equal or not?** Given: \(A = \left[ {\begin{array}{*{20}{c}}
3&2\\
4&0
\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}
3&0\\
3&2
\end{array}} \right]\)

Ans:

\( \Rightarrow AB = \left[ {\begin{array}{*{20}{c}} 3&2\\ 4&0 \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}} 3&0\\ 3&2 \end{array}} \right]\)

\( \Rightarrow AB = \left[ {\begin{array}{*{20}{c}} {\left( {3 \times 3} \right) + \left( {2 \times 3} \right)}&{\left( {3 \times 0} \right) + \left( {2 \times 2} \right)}\\ {\left( {4 \times 3} \right) + \left( {0 \times 3} \right)}&{\left( {4 \times 0} \right) + \left( {0 \times 2} \right)} \end{array}} \right]\)

\( \Rightarrow AB = \left[ {\begin{array}{*{20}{c}} {9 + 6}&{0 + 4}\\ {12 + 0}&{0 + 0} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {15}&4\\ {12}&0 \end{array}} \right]\)…(i)

\(BA = \left[ {\begin{array}{*{20}{c}} 3&0\\ 3&2 \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}} 3&2\\ 4&0 \end{array}} \right]\)

\( \Rightarrow BA = \left[ {\begin{array}{*{20}{c}} {\left( {3 \times 3} \right) + \left( {0 \times 4} \right)}&{\left( {3 \times 2} \right) + \left( {0 \times 0} \right)}\\ {\left( {3 \times 3} \right) + \left( {0 \times 4} \right)}&{\left( {3 \times 2} \right) + \left( {2 \times 0} \right)} \end{array}} \right]\)

\( \Rightarrow BA = \left[ {\begin{array}{*{20}{c}} {9 + 0}&{6 + 0}\\ {9 + 8}&{6 + 0} \end{array}} \right]\)

\( \Rightarrow BA = \left[ {\begin{array}{*{20}{c}} 9&6\\ {17}&6 \end{array}} \right]\)….(ii)

From (i) (ii), \(AB≠BA\).

Hence, \(AB\) and \(BA\) are not equal.

** Q.2. Prove that \(A = \left[ {\begin{array}{*{20}{c}}
5&2\\
7&3
\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}
3&{ – 2}\\
{ – 7}&5
\end{array}} \right]\) are inverse to each other under multiplication.** Given, \(A = \left[ {\begin{array}{*{20}{c}}
5&2\\
7&3
\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}
3&{ – 2}\\
{ – 7}&5
\end{array}} \right]\)

Ans:

We know that two matrices are inverse of each other if their product equals the identity matrix.

So, \(AB = \left[ {\begin{array}{*{20}{c}} 5&2\\ 7&3 \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}} 3&{ – 2}\\ { – 7}&5 \end{array}} \right]\)

\( \Rightarrow AB = \left[ {\begin{array}{*{20}{c}} {15 + \left( { – 14} \right)}&{ – 10 + 10}\\ {21 + \left( { – 21} \right)}&{ – 14 + 15} \end{array}} \right]\)

\( \Rightarrow AB = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

\(\therefore AB = I\)

Therefore, the given matrices \(A\) and \(B\) are inverse to each other.

** Q.3. Check the matrices **\(A = \left[ {\begin{array}{*{20}{c}}
2&3\\
2&7
\end{array}} \right]\)

Two matrices are said to be commutative under addition if their sum is the same however the order they added.

\(A + B = \left[ {\begin{array}{*{20}{c}} 2&3\\ 2&7 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 2&{ – 1}\\ { – 1}&1 \end{array}} \right]\)

\( \Rightarrow A + B = \left[ {\begin{array}{*{20}{c}} 4&2\\ 1&8 \end{array}} \right]\)…..(i)

\(B + A = \left[ {\begin{array}{*{20}{c}} 2&{ – 1}\\ { – 1}&1 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 2&3\\ 2&7 \end{array}} \right]\)

\( \Rightarrow B + A = \left[ {\begin{array}{*{20}{c}} {2 + 2}&{\left( { – 1} \right) + 3}\\ {\left( { – 1} \right) + 2}&{1 + 7} \end{array}} \right]\)

\( \Rightarrow B + A = \left[ {\begin{array}{*{20}{c}} 4&2\\ 1&8 \end{array}} \right]\)…..(ii)

From (i),(ii) \(A+B=B+A\)

Hence, the addition of two matrices is commutative.

** Q.4. If \(A = \left[ {\begin{array}{*{20}{c}}
1&2\\
3&4
\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}
4&3\\
2&1
\end{array}} \right],\) then find the value of \(3A+3B\) by using the properties of the matrix.** Given: \(A = \left[ {\begin{array}{*{20}{c}}
1&2\\
3&4
\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}
4&3\\
2&1
\end{array}} \right]\)

Ans:

By the scalar multiplicative property of the matrix, we know that \(\lambda (A + B) = \lambda A + \lambda B\)

\( \Rightarrow 3\left( {\left[ {\begin{array}{*{20}{c}} 1&2\\ 3&4 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 4&3\\ 2&1 \end{array}} \right]} \right)\)

\( \Rightarrow 3\left[ {\begin{array}{*{20}{c}} {1 + 4}&{2 + 3}\\ {3 + 2}&{1 + 4} \end{array}} \right]\)

\( \Rightarrow 3\left[ {\begin{array}{*{20}{c}} 5&5\\ 5&5 \end{array}} \right]\)

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} {3 \times 5}&{3 \times 5}\\ {3 \times 5}&{3 \times 5} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {15}&{15}\\ {15}&{15} \end{array}} \right]\)

Hence, the value of \(3A + 3B\) is \(\left[ {\begin{array}{*{20}{c}} {15}&{15}\\ {15}&{15} \end{array}} \right].\)

** Q.5. Find the value of the \(k\), such that matrix \(A = \left[ {\begin{array}{*{20}{c}}
2&k\\
3&6
\end{array}} \right]\) is a singular matrix.** Given \(A = \left[ {\begin{array}{*{20}{c}}
2&k\\
3&6
\end{array}} \right]\)

Ans:

We know that a matrix is said to be a singular matrix if the determinant of the matrix is zero.

\(\det A = |A| = 0\)

The determinant of the matrix \(A = \left[ {\begin{array}{*{20}{c}} a&c\\ b&d \end{array}} \right]\) Is given by \(ad-bc\).

So, the determinant of the matrix \(A = \left[ {\begin{array}{*{20}{c}} 2&k\\ 3&6 \end{array}} \right] = 0\)

\( \Rightarrow (2 \times 6) – (3 \times k) = 0\)

\( \Rightarrow 12 – 3k = 0\)

\( \Rightarrow 3k = 12\)

\( \Rightarrow k = \frac{{12}}{3}\)

\(\therefore k = 4\)

Hence, the value of \(k\) is \(4\).

Matrix is a concept in linear algebra that is particularly useful. It is a certain arrangement of things, particularly numerals. A matrix is a row-and-column mathematical structure. The aij element in a matrix, such as M, refers to the i-th row and j-th column element. The matrices are shown as square or rectangular brackets or in boxes. The rows and columns of the matrix indicate the horizontal and vertical lines. Entries or elements are the integers in the matrices. Like m-by-n matrices, the matrix size is set in m rows and n columns. The numbers are represented in lower-case letters, whereas the matrices are donated in upper-case letters.

In this article, we discussed the definition of the matrix, the transpose and the inverse of the matrix. We studied the properties related to a matrix such as addition, subtraction and multiplication: cumulative, associative, identity and inverse laws. We also discussed the properties of the transpose of a matrix, properties inverse matrix and scalar multiplication of the matrix, and the solved examples.

** Q.1. What are the properties of scalar multiplication?** Let \(A, B\) are two matrices having the same order and ,\(\lambda \) and \(\mu \) are any two scalars then the properties of scalar multiplication are given below:

Ans:

i. \(\lambda (A + B) = \lambda A + \lambda B\)

ii. \((\lambda + \mu )A = \lambda A + \mu A\)

iii. \(\lambda (\mu A) = (\lambda \mu A) = \mu (\lambda A)\)

iv. \(( – \lambda A) = – (\lambda A) = \lambda ( – A)\)

** Q.2. What are the properties of the square matrix?** The properties of the square matrix are given below:

Ans:

i. In a square matrix, the number of columns and number of rows is equal.

ii. Determinant of the matrix is calculated only for the square matrices.

iii. The inverse operation of the matrix is obtained only for square matrices.

iv. The trace of the square matrix is the sum of diagonal elements of the square matrix.

** Q.3 What is a matrix? Write the properties of determinants?** A matrix is a rectangular array or table arranged in rows and columns of numbers or variables. The properties of the determinants are given below:

Ans:

i. For a singular matrix \(A\), then \(|A| = 0\)

ii. \(\left| {{I_n}} \right| = 1\)

iii. \(|A| = \left| {{A^T}} \right|\)

iv. \(|AB| = |A| \times |B|\)

v. \(\left| {{A^{ – 1}}} \right| = \frac{1}{{|A|}}\)

** Q.4. Explain Properties of Matrix in addition.** The properties of the matrix addition are given below:

Ans:

1. \(A + B = B + A\)

ii. \((A + B) + C = A + (B + C)\)

iii. \(A + O = O + A = A\)

** Q.5. Can multiplication of matrix is commutative?** In general, the multiplication of the matrix is not commutative under multiplication. For two matrices \(A\) and \(B, AB≠BA\).

Ans:

However, in few specific cases, the multiplication of two matrices is commutative.