Non Singular Matrix: Definition, Formula, Properties & Solved Examples

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Jasmine Grover

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Non-Singular Matrix, also known as a regular matrix, is the most frequent form of a square matrix that comprises real numbers or complex numbers. Non Singular matrix can be defined as a square matrix whose determinant is a non-zero value and the non-singular matrix property is to be satisfied to find the inverse of a matrix. Its fundamental property is that there is always another matrix for any invertible matrix that, when multiplied by the first, provides the identity matrix with the same dimensions. Various types of non-singular matrices are frequently utilised in Matrices-to-Geometry applications. Cryptography is one of the most important uses of the inverse of a non-singular square matrix. 

Key Terms: Singular Matrix, Non-Singular Matrix, Determinant, Inverse Matrix, Square Matrix, Non-Zero Determinant, Rows, Columns, Cryptography, Geometry, Matrix, Regular matrix


Non-Singular Matrix

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A non-singular matrix is a square matrix with a non-zero determinant. Because it has a determinant value, the non-singular matrix is invertible, and its inverse may be obtained. If the matrix is square A =  \(\begin{bmatrix}a & b \\[0.3em]c & d \\[0.3em] \end{bmatrix}\). The determinant of this matrix A has a non-zero value, indicating a non-singular matrix. |A| =|ad - bc| ≠ 0.

Non-Singular Matrix

Non-Singular Matrix

Read More: Singular Matrix

  • Non-Singular Matrix Properties

The following are amongst the most important characteristics of a non-singular matrix.

  1. The determinant of a non-singular matrix is a non-zero value.
  2. The non-singular matrix is invertible since its determinant can be computed.
  3. The non-singular matrix is square because determinants can only be calculated for non-singular matrices.
  4. The multiplication of two non-singular matrices is a non-singular matrix.
  5. If A is a non-singular matrix and k is a constant, then kA is also a non-singular matrix.

Discover about the Chapter video:

Determinants Detailed Video Explanation:

Read More: Matrix Multiplication


Mathematical Properties of Non-Singular Matrix

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Some mathematical properties of non-singular matrix are: 

  • If the given matrix A is non-singular then so is A-1 and (A-1)-1 = A
  • If the given matrices A and B non-singular then so is AB and (AB)-1 = B-1 A-1
  • If the given matrix A is non-singular then so is (AT)-1 = (A-1)T
  • If the given matrices A and B non-singular and AB = In, then A and B are inverses of each other
  • If the given matrix A is non-singular then it will have only one unique inverse
  • If the given matrices A and B are non-singular and they both have inverse, then AB will also have one inverse, (AB)-1 = B-1 A-1

Condition for Non-Singular Matrix

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|A| ≠ 0 

For A2x2 =  \(\begin{bmatrix}a & b \\[0.3em]c & d \\[0.3em] \end{bmatrix}\)

|ad x bc| ≠ 0

How To Find A Non-Singular Matrix?

The determinant can be used to determine if it is a single or non-singular matrix. A non-singular matrix's determinant is a non-zero number. The determinant of a matrix can be defined using row or column operations or by utilising the co-factor of the matrix components to obtain the determinant.

Read More: Applications of Determinants and Matrices


Rules For Row and Column Operations of A Determinant 

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Given below are the rules that are helps to perform the row and column operations on determinants: 

  • If the rows and columns are interchanged, the value of the determinant remains unchanged.
  • If any two rows or two columns are interchanged, then the sign of the determinant changes. 
  • In case, any two rows or columns of a matrix are equal, then the value of the determinant will be zero.
  • In case every element of a particular row/column is multiplied by a constant, then the value of the determinant will also get multiplied by the constant.
  • The determinant can be expressed as a sum of determinants if the elements of a row or a column are expressed as a sum of elements.
  • The value of the determinant remains unchanged if the elements of a row or column are added or subtracted with the corresponding multiples of elements of another row or column. 

In the simplest square matrix of order 1×1 matrix, which has only one number, the determinant becomes the number itself. Elaborated below are the methods for how to calculate the determinants for the second-order and third-order matrices.

Read More: Types of Matrices


Calculating 2D Determinants

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The determinant formula may be used to find the determinant of any 2d square matrix or a square matrix of order 2x2:

\(C =\begin{bmatrix}a & b \\[0.3em]c & d \\[0.3em] \end{bmatrix}\)

Its two-dimensional determinant may be determined as follows:

\(|C| =\begin{bmatrix}a & b \\[0.3em]c & d \\[0.3em] \end{bmatrix}\)

|C| = (a × d) - (b × c)

Read More: Symmetric and Skew Symmetric Matrices

Calculating 3D Determinants

This is the calculation of any 3d square matrix or a square matrix of order 3x3.

\(C = \begin{pmatrix} a1 & b1 & c1 \\ a2 & b2 & c1 \\ a3 & b3& c3 \end{pmatrix}\)

The determinant is calculated using the first row. The components a1, b1, and c1 are multiplied with their respective co-factors, and the total of the elements' products with their respective co-factors yields the value of the square matrix's determinant. Alternatively, the elements of any row or column of the matrix can be utilised to get the matrix's determinant.

|C| = a1\(\begin{bmatrix}b2 & c2 \\[0.3em]b3 & c3 \\[0.3em] \end{bmatrix}\)  – b1\(\begin{bmatrix}a2 & c2 \\[0.3em]a3 & c3 \\[0.3em] \end{bmatrix}\) + c1\(\begin{bmatrix}a2 & b2 \\[0.3em]a3 & b3 \\[0.3em] \end{bmatrix}\)

|C| = a1((b2 x c3) – (b3 x c2)) − b1((a2 x c3) – (a3 x c2)) + c1((a2 x b3) – (a3 x b2))


Inverse of a Matrix

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The inverse of a matrix seems to be another matrix that produces the multiplicative identity when multiplied with the supplied matrix. The adjoint of a matrix divided by the determinant of a matrix is the basic formula for the inverse of a matrix for order 2 x 2. A-1 = (1 / |A|) x Adj A

Inverse of a Matrix

Inverse of a Matrix

Read More: Inverse Matrix Formula


Applications of Non-Singular Matrix

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In our daily lives, invertible matrices are employed in a variety of applications. They're useful for a variety of jobs, but 3D transformations are where they flourish. Here are some real-world instances of invertible matrices.

  • Invertible matrices can be used to encrypt communication. Encrypting a transmission may be accomplished in a variety of ways, with coding becoming increasingly prevalent in recent years.
  • To decrypt a message, cryptographers, especially those who created the specific encryption scheme, employ invertible matrices.
  • In computer animation, invertible matrices are utilised to generate the three-dimensional visuals you see on the screen.

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Things to Remember

  • A non-singular matrix is a square matrix with a non-zero determinant. 
  • Since it has a determinant value, the non-singular matrix is invertible, and its inverse may be obtained.
  • Only a determinant of zero may make a matrix singular. A non-singular matrix is defined as one with a non-zero determinant.
  • If the matrix has an inverse, the identity matrix is obtained by multiplying the matrix by its inverse.
  • Most importantly, if a person can identify an inverse for a matrix, it is undoubtedly non-singular.
  • A person must ensure that the matrix fits all the invertible matrix theorem's requirements. This check is required to determine if the matrix is single or non-singular.
  • The only unique value representation of a matrix is its determinant. Any row or column of the given matrix contains the matrix's determinant. The total of the product of the components and their co-factors inside a given row or column is the determinant of a matrix.

Read More: Operations on Matrices


Sample Questions

Ques. Calculate the determinant of the matrix \(\begin{bmatrix}1 & -4 \\[0.3em]3 & 5 \\[0.3em] \end{bmatrix}\) And find its singularity. (3 Marks)

Ans. We know that 

A = \(\begin{bmatrix}1 & -4 \\[0.3em]3 & 5 \\[0.3em] \end{bmatrix}\)

For a general matrix A = \(\begin{bmatrix}a & b \\[0.3em]c & d \\[0.3em] \end{bmatrix}\), the determinant is |A| =|ad - bc|.

|A| = 1 x [(5) - (3) x (-4)] 

= 5 - (-12) = 5 + 12 = 17

|A| = 17, and is a non-singular matrix.

Ques. For a matrix \(\begin{bmatrix}4 & -1 & 0\\[0.3em]2 & 3 & 5\\[0.3em]-1 & 7 & 2\\[0.3em] \end{bmatrix}\) find the singularity of the matrix. (3 Marks)

Ans. Here, we are given that

A = \(\begin{bmatrix}4 & -1 & 0\\[0.3em]2 & 3 & 5\\[0.3em]-1 & 7 & 2\\[0.3em] \end{bmatrix}\)

Using co-factors of the 1st row, calculate the determinant

|A| = 4 \(\begin{bmatrix}3 & 5 \\[0.3em]7 & 2 \\[0.3em] \end{bmatrix}\) – (-1)\(\begin{bmatrix}2 & 5 \\[0.3em]-1 & 2 \\[0.3em] \end{bmatrix}\) + 0 \(\begin{bmatrix}2 & 3 \\[0.3em]-1 & 7 \\[0.3em] \end{bmatrix}\)

|A| = 4 x [3 x (2) – 5 x (7)] +1 x [2 x (2) - (-1) x (5)] + 0 x [2 x (7) - ((-1) x (3)]

|A| = 4 x (6 – 35) + 1 x (4 + 5) + 0 x (14 + 3)

|A| = 4 x (-29) + 1 x (9) + 0 x (17)

|A| = -116 + 19 + 0 = -97

|A| is -97, and is a non-singular matrix.

Ques. If A = \(\begin{pmatrix} -3 & 1 \\ 5 & 0 \end{pmatrix}\) and B = \(\begin{pmatrix} 0 & 1/5 \\ 1 & 3/5 \end{pmatrix}\) ,Then show that A is an invertible matrix and B is its inverse. (5 Marks)

Ans. Given that 

A = \(\begin{pmatrix} -3 & 1 \\ 5 & 0 \end{pmatrix}\) and B = \(\begin{pmatrix} 0 & 1/5 \\ 1 & 3/5 \end{pmatrix}\), Now, finding the determinant of A,

|A| = \(\begin{pmatrix} -3 & 1 \\ 5 & 0 \end{pmatrix}\)

= – 3(0) – 1(5)

= 0 – 5

= – 5 ≠ 0

Thus, A is an invertible matrix.

We know that if A is invertible and B is its inverse, AB = BA = I, where I is an identity matrix.

AB = \(\begin{pmatrix} -3 & 1 \\ 5 & 0 \end{pmatrix} \times \begin{pmatrix} 0 & 1/5 \\ 1 & 3/5 \end{pmatrix}\)

\(\begin{pmatrix} -3.0 + 1.1 & (-3).(\frac{1}{5}) + 1.(\frac{3}{5}) \\ 5.0 + 0.1 & 5.(\frac{1}{5}) + 0.(\frac{3}{5}) \end{pmatrix}\)

\(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)

BA = \(\begin{pmatrix} 0 & 1/5 \\ 1 & 3/5 \end{pmatrix} \times \begin{pmatrix} -3 & 1 \\ 5 & 0 \end{pmatrix} \)

\(\begin{pmatrix} 0. (-3) + (\frac{1}{5}).5 & 0.1 + .(\frac{1}{5}). 0 \\ 1. (-3) + (\frac{3}{5}).5 & 1.1 + .(\frac{3}{5}). 0 \end{pmatrix}\)

\(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)

AB = BA = 1

A is invertible, and B is its inverse.

Ques. What are the different sorts of Matrices? (5 Marks)

Ans. Matrices come in a variety of shapes and sizes. The following are a few of them:

  • A row matrix with only one row.
  • A column matrix with only one column.
  • A diagonal matrix is termed an identity or unit Matrix if all major diagonal elements are 1.
  • A square matrix has the same rows and columns as a rectangular matrix.
  • A scalar Matrix is one in which all of the components in a diagonal Matrix are equal.
  • Matrices with the same number of entries are equal Matrices are called equal matrices.

Ques. Express the matrix A as the sum of the symmetric and skew-symmetric matrix, where A = \(\begin{bmatrix}2 & 4 &-6 \\[0.3em]7 & 3 & 5 \\[0.3em]1 & -2 & 4 \\[0.3em] \end{bmatrix} \) (5 Marks)

Ans. The given matrix is 

A = \(\begin{bmatrix}2 & 4 &-6 \\[0.3em]7 & 3 & 5 \\[0.3em]1 & -2 & 4 \\[0.3em] \end{bmatrix} \) then A’ = \(\begin{bmatrix}2 & 7 & 1 \\[0.3em]4 & 3 & -2 \\[0.3em]-6 & 5 & 4 \\[0.3em] \end{bmatrix} \)

Hence

Symmetric matrix = (A + A’) / 2 = (1/2) x \(\begin{bmatrix}2 & 4 &-6 \\[0.3em]7 & 3 & 5  \\[0.3em]1 & -2 & 4  \\[0.3em] \end{bmatrix} \) + \(\begin{bmatrix}2 & 7 & 1 \\[0.3em]4 & 3 & -2 \\[0.3em]-6 & 5 & 4 \\[0.3em] \end{bmatrix} \)

= (1/2) \(\begin{bmatrix}4 & 11 &-5 \\[0.3em]11 & 6 & 3  \\[0.3em]-5 & 3 & 8  \\[0.3em] \end{bmatrix} \)

\(\begin{bmatrix}2 & 11/2 &-5/2 \\[0.3em]11/2 & 3 & 3/2  \\[0.3em]-5/2 & 3/2 & 4  \\[0.3em] \end{bmatrix} \)

Skew-Symmetric Matrix = (A – A’) / 2 = (1/2) x \(\begin{bmatrix}2 & 4 &-6 \\[0.3em]7 & 3 & 5 \\[0.3em]1 & -2 & 4 \\[0.3em] \end{bmatrix}\) – \(\begin{bmatrix}2 & 7 & 1 \\[0.3em]4 & 3 & -2 \\[0.3em]-6 & 5 & 4 \\[0.3em] \end{bmatrix} \)

= (1/2) \(\begin{bmatrix}0 & -3 &-7 \\[0.3em]3 & 0 & 7  \\[0.3em]7 & -7 & 0  \\[0.3em] \end{bmatrix} \)

\(\begin{bmatrix}0 & -3/2 &-7/2 \\[0.3em]3/2 & 0 & 7/2  \\[0.3em]7/2 & -7/2 & 0  \\[0.3em] \end{bmatrix} \)

Therefore, 

Symmetric matrix + Skew - Symmetric matrix

= ((A + A’) / 2) + ((A - A’) / 2)

\(\begin{bmatrix}2 & 11/2 &-5/2 \\[0.3em]11/2 & 3 & 3/2  \\[0.3em]-5/2 & 3/2 & 4  \\[0.3em] \end{bmatrix} \) + \(\begin{bmatrix}0 & -3/2 &-7/2 \\[0.3em]3/2 & 0 & 7/2  \\[0.3em]7/2 & -7/2 & 0  \\[0.3em] \end{bmatrix} \)

\(\begin{bmatrix}2 & 4 &-6 \\[0.3em]7 & 3 & 5 \\[0.3em]1 & -2 & 4 \\[0.3em] \end{bmatrix} \) = A

Ques. If A = \(\begin{bmatrix}1 & 3 &2 \\[0.3em]2 & 0 & -1 \\[0.3em]1 & 2 & 3 \\[0.3em] \end{bmatrix} \), then show that A satisfies the equation A3 – 4A2 – 3A + 11I = 0 (5 Marks)

Ans. A2 = A x A

\(\begin{bmatrix}1 & 3 &2 \\[0.3em]2 & 0 & -1 \\[0.3em]1 & 2 & 3 \\[0.3em] \end{bmatrix} \times \begin{bmatrix}1 & 3 &2 \\[0.3em]2 & 0 & -1 \\[0.3em]1 & 2 & 3 \\[0.3em] \end{bmatrix} \)

\(\begin{bmatrix}1+6+2 & 3+0+4 &2-3+6 \\[0.3em]2+0-1 & 6+0-2 & 4+0--3 \\[0.3em]1+4+3 & 3+0+6 & 2-2+9 \\[0.3em] \end{bmatrix}\)

\( \begin{bmatrix}9 & 7 &5 \\[0.3em]1 & 4 & 1 \\[0.3em]8 & 9 & 9 \\[0.3em] \end{bmatrix} \)

And

A3 = A2 x A

\( \begin{bmatrix}9 & 7 &5 \\[0.3em]1 & 4 & 1 \\[0.3em]8 & 9 & 9 \\[0.3em] \end{bmatrix} \) x \(\begin{bmatrix}1 & 3 &2 \\[0.3em]2 & 0 & -1 \\[0.3em]1 & 2 & 3 \\[0.3em] \end{bmatrix} \)

\(\begin{bmatrix}9+14+5 & 27+0+10 & 18-7+15 \\[0.3em]1+8+1 & 3+0+2 & 2-4+3 \\[0.3em]8+18+9 & 24+0+18 & 16-9+27 \\[0.3em] \end{bmatrix}\)

\( \begin{bmatrix}28 & 37 &26 \\[0.3em]10 & 5 & 1 \\[0.3em]35 & 42 & 34 \\[0.3em] \end{bmatrix} \)

Now, A3 – 4A2 – 3A + 11I

\( \begin{bmatrix}28 & 37 &26 \\[0.3em]10 & 5 & 1 \\[0.3em]35 & 42 & 34 \\[0.3em] \end{bmatrix} \) – 4\( \begin{bmatrix}9 & 7 &5 \\[0.3em]1 & 4 & 1 \\[0.3em]8 & 9 & 9 \\[0.3em] \end{bmatrix} \) – 3\(\begin{bmatrix}1 & 3 &2 \\[0.3em]2 & 0 & -1 \\[0.3em]1 & 2 & 3 \\[0.3em] \end{bmatrix} \) + 11 \(\begin{bmatrix}1 & 0 &0 \\[0.3em]0 & 1 &0 \\[0.3em]0 & 0 & 1 \\[0.3em] \end{bmatrix} \)

\(\begin{bmatrix}28-36-3+11 & 37-28-9+0 & 26-20-6+0 \\[0.3em]10-4-6+0 & 5-16+0+11 & 1-4+3+0 \\[0.3em]35-32-3+0 & 42-36-6+0 & 34-36-9+11 \\[0.3em] \end{bmatrix}\)

\(\begin{bmatrix}0& 0 &0 \\[0.3em]0 & 0 &0 \\[0.3em]0 & 0 & 0 \\[0.3em] \end{bmatrix} \)

Ques. Let \(A = \begin{bmatrix}2 & 3 \\[0.3em]-1 & 2 \\[0.3em] \end{bmatrix} \) then show that A2 – 4A + 7I = 0. Also calculate A5. (5 Marks)

Ans. Given that, 

\(A = \begin{bmatrix}2 & 3 \\[0.3em]-1 & 2  \\[0.3em] \end{bmatrix} \)

A2 = A x A

\(\begin{bmatrix}2 & 3 \\[0.3em]-1 & 2  \\[0.3em] \end{bmatrix} \times \begin{bmatrix}2 & 3 \\[0.3em]-1 & 2  \\[0.3em] \end{bmatrix}\)

\(\begin{bmatrix}1 & 12 \\[0.3em]-4 & 1  \\[0.3em] \end{bmatrix}\)

- 4A = – 4 x \(\begin{bmatrix}2 & 3 \\[0.3em]-1 & 2  \\[0.3em] \end{bmatrix}\)

\(\begin{bmatrix}-8 & -12 \\[0.3em]4 & -8  \\[0.3em] \end{bmatrix}\)

7I = \(\begin{bmatrix}7 & 0 \\[0.3em]0 & 7  \\[0.3em] \end{bmatrix}\)

A3 – 4A + 7I

A3 = A3 . A

= 4 x (4A - 7I) – 7I

= 16A – 28I – 7A 

= 9A – 28I

A5 = A3 . A3

= (9A – 28I) . (4A - 7I)

= 36A3 – 63A -112A + 196I

= 36(4A – 7I) – 175A + 196I

= - 31A – 56I

= – 31 \(\begin{bmatrix}2 & 3 \\[0.3em]-1 & 2  \\[0.3em] \end{bmatrix}\) – 56 \(\begin{bmatrix}1 & 0 \\[0.3em]0 & 1  \\[0.3em] \end{bmatrix}\)

\(\begin{bmatrix}-118 & -93 \\[0.3em]31 & -118  \\[0.3em] \end{bmatrix}\)

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CBSE CLASS XII Related Questions

  • 1.
    Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]


      • 2.
        Find:

        If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

          • \(0\)
          • \(-2\)
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        • 3.

          At a birthday party, children are being served orange juice in conical cups, as shown in the figure. 


          Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
          On the basis of the above information, answer the following questions :


            • 4.
              Find the vector and cartesian equations of the line passing through the point of intersection of the lines \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j}) \) and \( \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) \) and parallel to the line \( \frac{x - 1}{-2} = \frac{7 - y}{-3} = z \).


                • 5.
                  Find:

                  If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

                    • \(p = -\frac{2}{3}, \, q = \frac{5}{3}\)
                    • \(p = -\frac{8}{3}, \, q = \frac{20}{3}\)
                    • \(p = \frac{20}{3}, \, q = -\frac{8}{3}\)
                    • \(p = 0, \, q = 0\)

                  • 6.

                    Find:
                    Let \(A=[a_{ij}]\) be a \(2\times2\) matrix whose elements are given by \[ a_{ij}=\frac{(2i-j)^2}{3} \] Find the transpose matrix \(A'\).

                      • \(\begin{bmatrix} \frac{1}{3} & 3 \\ 0 & \frac{4}{3} \end{bmatrix}\)
                      • \(\begin{bmatrix} \frac{1}{3} & 0 \\ 3 & \frac{4}{3} \end{bmatrix}\)
                      • \(\begin{bmatrix} \frac{4}{3} & 3 \\ 1 & 0 \end{bmatrix}\)
                      • \(\begin{bmatrix} \frac{4}{3} & 0 1 & \frac{3}{3} \end{bmatrix}\)
                    CBSE CLASS XII Previous Year Papers

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