Singular Matrix | Definition, Properties, Solved Examples

A singular matrix is a square matrix (i.e., a matrix where the number of rows is equal to the number of columns ) whose determinant is zero. This means it can't be inverted. In other words, you can't multiply it by another matrix to get the identity matrix. There is no matrix B such that when you multiply matrix A by B, you get the identity matrix I (i.e., AB = I).

Why is the Inverse of a Singular Matrix Not Defined?

We know that the formula to determine the inverse of a matrix is equal to the adjoint of the matrix divided by the determinant, i.e.,

A^-1 = (adj A) / |A|.

From the definition of a singular matrix, we know that ∣A∣ = 0. Since the determinant of A is zero, the matrix is singular, meaning its inverse does not exist.

Singular Matrix Examples

Various examples of singular matrices are:

• \left[\begin{array}{ccc} 1 & 2 \\ 4 & 8\end{array}\right],
• \left[\begin{array}{ccc} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{array}\right],
• \left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 7 & 8 & 9 \end{array}\right], etc.

Theorem to Generate Singular Matrices

Let A be a matrix of order n × k and B be a matrix of order k × n, where n>k. Then, the product AB, which is an n × n matrix, is always singular.

This means that if you multiply any n × k matrix by any k × n matrix, with n > k, the resulting square matrix will always be singular (i.e., its determinant will be zero).
The intuition is that the rank of the product AB cannot exceed k, which is less than n, so AB cannot be full-rank and thus must be singular.

To generate a singular matrix of size n × n:

• Randomly generate a matrix A of size n × k (with k < n).
• Randomly generate a matrix B of size k × n.
• Compute AB. The resulting matrix is guaranteed to be singular

Example: If n = 3, k = 1.

Solution:

Let A = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}_{(3\times1),} \text{and,} \ B = \begin{bmatrix} 4 & 5 & 6 \end{bmatrix}_{ (1 \times 3)}
Then the product (AB) is :
AB = A \cdot B = \begin{bmatrix}1 \\2 \\3\end{bmatrix}\begin{bmatrix}4 & 5 & 6\end{bmatrix}= \begin{bmatrix}1 \cdot 4 & 1 \cdot 5 & 1 \cdot 6 \\2 \cdot 4 & 2 \cdot 5 & 2 \cdot 6 \\3 \cdot 4 & 3 \cdot 5 & 3 \cdot 6\end{bmatrix}= \begin{bmatrix}4 & 5 & 6 \\8 & 10 & 12 \\12 & 15 & 18\end{bmatrix}
AB is 3×3, but the rows are linearly dependent → Singular

Properties of a Singular Matrix

The following are the properties of the Singular Matrix:

• Every singular matrix must be a square matrix, i.e., a matrix that has an equal number of rows and columns.
• The determinant of a singular matrix is equal to zero.
• As the determinant of a singular matrix is zero, its inverse is not defined.
• A zero matrix of any order is a singular matrix, as its determinant is zero.
• In a singular matrix, some rows and columns are linearly dependent.
• The rank of a singular matrix will be less than the order of the matrix, i.e., Rank (A) < Order of A.
• A matrix that has any two rows or any two columns identical is singular, as the determinant of such a matrix is zero.
• When a row or column's elements in a matrix are all zeros, then the matrix is singular, as its determinant is zero.
• When one row (or column) of a matrix is a scalar multiple of another row (or column), then the matrix is singular. It's with Matrices. Its determinant is zero. 

Singular Vs Non-Singular Matrix

Let us consider that A and B are two square matrices of order "n × n".

• If AB = BA = I, where I is an identity or unit matrix of order n, then B is said to be the inverse matrix of A.
• Matrix A is non-singular (i.e., it has an inverse) if it satisfies this condition.
• If A does not have an inverse (i.e., no matrix B exists such that AB = BA = I), then A is called a singular matrix.

Thus:

• Non-singular matrix: Has an inverse.
• Singular matrix: Does not have an inverse.

Differences between the Singular Matrix and Non-Singular Matrix can be understood using the table given below.

Identifying a Singular Matrix

Follow the conditions given below to determine whether the given matrix is singular or not.

• Determine whether the given matrix is square or not.
• If the given matrix is square, then find the determinant of the matrix.

⇒ If |A|= 0, then the given matrix is singular.

⇒ If |A|≠0, then the given matrix is non-singular.

Formula for Determinant of "2 × 2" Matrix

If A = \left[\begin{array}{cc} a & b\\ c & d \end{array}\right]      is a "2 × 2" matrix, then its determinant is 

|A|= [ad – bc]

Formula for Determinant of "3 × 3" Matrix

If A = \left[\begin{array}{ccc} a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3} \end{array}\right]     is a "3 × 3" matrix, then its determinant is 

|A|= a₁(b₂c₃ – b₃c₂) – a₂(b₁c₃ – b₃c₁) + a₃(b₁c₂ – b₂c₁)

Learn in detail:

• Detereminant of 2 × 2 Matrix
• Detereminant of 3 × 3 Matrix
• Detereminant of 4 × 4 Matrix

Solved Questions on Singular Matrix

Question 1: Find the value of k if the matrix given below is singular A = \left[\begin{array}{cc} k & -4\\ 5 & 2 \end{array}\right]

Solution:

Given matrix A = \left[\begin{array}{cc} k & -4\\ 5 & 2 \end{array}\right]

We know that the determinant of a singular matrix is zero, i.e., det A = 0
⇒ (2×k) – (–4 × 5) = 0
⇒ 2k + 20 = 0
⇒ 2k = -20
⇒ k = –20/2 = –10

Hence, the value of k if the given matrix is a singular matrix is –10.

Question 2: Determine the inverse of the matrix given below P = \left[\begin{array}{cc} -3 & 4\\ 6 & -8 \end{array}\right]

Solution:

Given matrix  P = \left[\begin{array}{cc} -3 & 4\\ 6 & -8 \end{array}\right]
P^-1 = Adj P / |P|

Now, let us find the determinant of the matrix P.
|P| = (–3 × –8) – (6 × 4)
|P| = 24 – 24 = 0

Since, the determinant of matrix P = 0, it is a singular matrix, and its inverse matrix doesn't exist.

Question 3: Determine whether the given matrix is singular or not A = \left[\begin{array}{ccc} 1 & 0 & -3\\ 0 & 5 & 2\\ -1 & 4 & 0 \end{array}\right]

Solution:

Given matrix A = \left[\begin{array}{ccc} 1 & 0 & -3\\ 0 & 5 & 2\\ -1 & 4 & 0 \end{array}\right]

To determine whether the given matrix is singular or not, we have to find its determinant.

det A = 1[(5 × 0) – (4 × 2)] – 0[(0 × 0) – (2 × –1)] + (-3) [(0 × 4) – (–1 × 5)]
⇒ |A| = (1 × -8) – 0 + (–3 × 5) 
⇒ |A| = –8 – 15 = –23 ≠ 0

Since the determinant of the given matrix is not equal to zero, it is a non-singular matrix.

Question 4: Find the value of b if the matrix given below is singular B = \left[\begin{array}{cc} 9 & b\\ 6 & -2 \end{array}\right]

Solution:

Given matrix B = \left[\begin{array}{cc} 9 & b\\ 6 & -2 \end{array}\right]

We know that the determinant of a singular matrix is zero, i.e., det B = 0

⇒ (9 × –2) – (6 × b) = 0
⇒ –18 – 6b = 0
⇒ –6b = 18
⇒ b = 18/–6 = –3

Hence, the value of b if the given matrix is a singular matrix is –3.

Question 5: Find x if A=\begin{pmatrix}x + 1 & x^2 & 1 \\0 & 1 & 4 \\1 & x & x+3\end{pmatrix} value of x is the singular matrix?

Solution:

If A is singular matrix then it means its determinant is 0 i.e., det A = 0,

(x+1) (0-1) -x(x+3-4) +2(1-0) =0
-x -1 -x² + x + 2=0
-x² +1 =0
x² =1
x = ±1

Answer: x=1 or -1 for A to be singular.

Question 6: Determine whether the following system has a unique solution or not: 2x+y+2z =3, x + z=5, 4x + y +4z =7.

Solution:

If we write the given system in the matrix form then the corresponding matrix equation is AX = B, then the coefficient matrix is,

A = \begin{pmatrix}2 & 1 & 2 \\1 & 0 & 1 \\4 & 1 & 4\end{pmatrix}

If the determinant of A is NOT zero or if A is non-singular matrix, then only the system has a unique solution. Otherwise it may have infinite or no solutions.

det A = 2(0 - 1) -1(4 - 4) + 2(1 - 0)a
= -2 +0 +2
= 0

Answer: The above given system does NOT have a unique solution because the determinant is ZERO. So, it either has an infinite number of solutions or it has no solution.

Question 7: Given the matrix A, for what value of x is the matrix singular?
A(x) = \begin{pmatrix}x^2 & 2x \\4 & 2x-4 \\\end{pmatrix}

Solution:

Calculate the determinant of A(x):
det(A(x)) = x²(2x-4) -2x(4) = 2x³ - 4x² - 8x

Factor the determinant:
det(A(x)) = 2x( x² -2x - 4) = 2x( x- 4) ( x + 1)

The matrix is singular when the determinant is 0, so the singular values of x are x = 0, and x = 4, and x = -1.

Question 8: Consider the matrix B and determine the values of x for which the matrix is singular.
B(x) = \begin{pmatrix}x^2-4 & x+1 \\2x-2 & x^2-4 \\\end{pmatrix}

Solution:

Determinant of B(x):
det(B(x)) = (x² - 4)(x² -4) - ( x + 1)(2x - 2)
= (x² - 4)² - 2x(x +1 -2) = (x² - 4)² - 2x( x - 1)
=(x⁴ - 8x² +16) -2x(x - 1) = x⁴ - 8x² +16 - 2x² + 2x
= x(x + 1).

The matrix is singular when the determinant is zero, which occurs for x =0, x = -1, x = 2.

Question 9: Given a matrix C(x), find the values of x for which the matrix is singular.
C(x) = \begin{pmatrix}x & 1 & 1 \\2x & x & 1 \\3x^2 & 3x & 1\end{pmatrix}

Solution:

determinant of C(x):

det(C(x)) = x(x - 3x) -1(2x - 3x²) +1(2x(3x) - x(3x²)
= x( -2x) -2x + 3x² +1(6x² - 3x³)
= -2x² -2x + 3x² + 6x² - 3x³
= -3x³ +7x² -2x = x(-3x² + 7x -2) = x[-3x(x - 2) +1 (x - 2)]
= x(-3x + 1)(x - 2) , x= 0, x = 1/3 , x = 2

So the matrix C(x) is singular when x =0, x= 1/3, x =2.

Question 10: Consider the system of linear equations:

x + ky + z = 3
2x + 4y + 3z =7
x + 2y +(k + 1) = 5

Determine the values of 𝑘 for which the matrix is singular. Then, discuss the solution of the system for those values of k.

Solution:

1. From the Matrix and Augmented Matrix

the coefficient matrix A and the augmented matrix [A | b] are: A = \begin{pmatrix}1 & k & 1 \\2 & 4 & 3 \\1 & 2 & k+1\end{pmatrix}, [A \mid B]\begin{bmatrix}1 & k & 1 & | & 3 \\2 & 4 & 3 & | & 7\\1 & 2 & k+1 & | & 5\end{bmatrix}

2. Calculate the Determinant of A
compute the determinant of A:
det(A) = 1(4(k+1) - 6) - k(2(k + 1) -3) + 1(4 - 4)
= (4k - 2) -k(2k -1 ) + 0
= 4k -2 -2k² + k
= -2k² + 5k -2

A matrix is singular when it's determinant is zero:
-2k² + 5k -2 = 0
-2k² + 4k + k -2 =0
-2k(k - 2) + 1( k-2 ) =0
(-2k + 1) (k - 2) =0

So, -2k + 1 =0, k - 2 =0
k = 1/2 or k = 2.

• For k = 1/2 , substitute k = 1/2 into the augmented matrix and perform row reduction to check if the system is consistent. If there is no inconsistency, the system has infinitely many solutions. For now, the system has no solution for k = 1/2, because of contradiction in third row.
• for k = 2: substitute k =2 into the augmented matrix and perform row reduction. Again check if the system is consistent. There is no contradiction, and the system is consistent. which means the system when k = 2 has infinitely many solutions.

Practice Questions on Singular Matrix

Question 1: Determine whether the following system has a unique solution, infinitely many solutions, or no solution:

2x + 3y - z = 1
4x + 6y -2z = 2
-2x -3y + z = -1

Question 2: Find the value of k for which the matrix is singular and determine if the system of equations has a unique solution, infinitely many solutions, or no solution:

x + y + k = k
2x + 2y + 2z = 2k
3x + 3y + 3z = 3k

Question 3: For which values of k does the following system of equations have a unique solution?

x + 2y + kz =1
2x + 4y + 3z =4
3x + 6y + 4z = 7

Question 4: Determine if the following matrix is singular or non-singular \begin{pmatrix}1 & 0 & 2 \\0 & 0 & 0 \\3 & 4 & 5\end{pmatrix}

Question 5: Find the value of a for which the matrix is singular, and hence does not have an inverse: B = \begin{pmatrix}a & 2 & 3 \\4 & 5 & 6 \\7 & 8 & 9\end{pmatrix}

Question 6: Determine for which values of b the following matrix is invertible: \begin{pmatrix}1 & b & 1 \\0 & 1 & 1 \\0 & 0 & 1\end{pmatrix}

Question 7: Given the Matrix C(x), find the values of x for which the matrix is singular \begin{pmatrix}x^2 & x + 1 & 2x \\x^2 - x & 2x & 1 \\2x - 1 & x^2 + x & 3\end{pmatrix}

Question 8: Given the symmetric matrix, find the values of x for which the matrix A is singular A = \begin{pmatrix}x^2 + 2 & x & x - 1 \\x & x^2 + 1 & x + 2 \\x - 1 & x + 2 & x^2\end{pmatrix}

Answer Key

1. The system has infinitely many solutions.
2. The system has infinitely many solutions for all values of k.
3. The system has no unique solution for any value of k.
4. As the determinant is 0, it is a Singular Matrix.
5. a = 1.
6. The matrix is invertible for all values of b.
7. x = 0, ±1
8. x = ±1

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