Non-Singular Matrix

A non-singular matrix is a square matrix that has a nonzero determinant. This means that the matrix is invertible, and its inverse exists. Mathematically, a square matrix \( A \) of order \( n \) is non-singular if:

\[ \det(A) \neq 0 \]

Non-singular matrices play a crucial role in linear algebra, particularly in solving systems of linear equations, transformations, and determinant-based operations.

Properties of a Non-Singular Matrix

1. Determinant is Non-Zero

The defining property of a non-singular matrix is that its determinant is not zero:

\[ \det(A) \neq 0 \]

Example: Consider the matrix:

\[ A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \]

The determinant is calculated as:

\[ \det(A) = (2 \times 4) – (3 \times 1) = 8 – 3 = 5 \neq 0 \]

Since the determinant is nonzero, \( A \) is a non-singular matrix.

2. Inverse Exists

For a non-singular matrix \( A \), there exists an inverse matrix \( A^{-1} \) such that:

\[ A A^{-1} = A^{-1} A = I \]

where \( I \) is the identity matrix.

Example: The inverse of the matrix:

\[ B = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix} \]

can be found using the formula:

\[ B^{-1} = \frac{1}{\det(B)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]

where \( \det(B) = (4 \times 6) – (7 \times 2) = 24 – 14 = 10 \).

Thus,

\[ B^{-1} = \frac{1}{10} \begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix} \]

3. Rank is Equal to Order

A non-singular \( n \times n \) matrix has full rank, meaning:

\[ \text{rank}(A) = n \]

Example: The matrix:

\[ C = \begin{bmatrix} 5 & 1 \\ 2 & 3 \end{bmatrix} \]

has two linearly independent rows, so its rank is 2, which is equal to its order \( n = 2 \). Hence, it is non-singular.

4. Unique Solution to \( Ax = b \)

If \( A \) is a non-singular matrix, then the system of linear equations:

\[ Ax = b \]

has a unique solution given by:

\[ x = A^{-1} b \]

Example: Consider the system:

\[ \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 5 \end{bmatrix} \]

The coefficient matrix is non-singular because:

\[ \det(A) = (3 \times 4) – (2 \times 1) = 12 – 2 = 10 \neq 0 \]

Thus, the system has a unique solution.

5. Eigenvalues are Non-Zero

A non-singular matrix does not have zero as an eigenvalue. If \( A \) is non-singular, then:

\[ \lambda_i \neq 0 \quad \forall i \]

Example: The matrix:

\[ D = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} \]

has eigenvalues found by solving:

\[ \det(D – \lambda I) = 0 \]

which results in:

\[ (2-\lambda)(3-\lambda) – (1 \times 1) = 0 \]

\[ 6 – 2\lambda – 3\lambda + \lambda^2 – 1 = 0 \]

\[ \lambda^2 – 5\lambda + 5 = 0 \]

The solutions are nonzero, confirming that \( D \) is non-singular.

6. The Product of Two Non-Singular Matrices is Non-Singular

If \( A \) and \( B \) are non-singular matrices of the same order, then their product is also non-singular:

\[ \det(AB) = \det(A) \times \det(B) \neq 0 \]

Example: Consider:

\[ A = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 4 & 2 \\ 1 & 5 \end{bmatrix} \]

Since:

\[ \det(A) = (2 \times 3) – (1 \times 1) = 6 – 1 = 5 \neq 0 \]

\[ \det(B) = (4 \times 5) – (2 \times 1) = 20 – 2 = 18 \neq 0 \]

their product is also non-singular.