Inverse of Matrix - Matrices

Matrices are defined as a rectangular array of numbers or functions. Since it is rectangular array, it is 2-dimensional. The two dimensions here are the number of rows(m) and the number of columns(n) respectively. Let us look at some examples.

\(A = \begin{bmatrix} 7 & 14 & 91 \\ -2 & 1 & 3 \end{bmatrix}\)

\(B = \begin{bmatrix} -2 & -6 & 3 \\ 2 & 7 & -3 \\ -9 & 9 & 1\\ 8 & 5 & 7 \end{bmatrix}\)

In the above examples, we have matrices A and B. The general notation of a matrix is:

A = [aij]m × n where 1 ≤ I ≤ m, 1 ≤ j ≤ n and i, j ∈Z.

If the matrix has m rows and n columns, it is said to be a matrix of the order m×n. We call this an m by n matrix. So, A is a 2×3 matrix and B is a 4×3 matrix. Also, if a matrix is of m×n order, it will have mn elements.  The basic mathematical operations like addition, subtraction, multiplication and division can be performed on matrices. Apart from them, there are some elementary operations, also called transformations, which can be performed on matrices.

Rules for Transformation of Matrix:

According to transformations,

  • Any two rows (or columns) of a matrix can be exchanged,
  • The elements of any row (or column) of a matrix can be multiplied by a non-zero number
  • The elements of any row (or column) can be added with the corresponding elements of another row (or column) which is multiplied by a non-zero number.

Let us consider three matrices X, A and B such that X = AB. To determine the inverse of a matrix using elementary transformation, we convert the given matrix into an identity matrix.

If the inverse of matrix A, A-1 exists then to determine A-1 using elementary row operations

  • Write A = IA, where I is the identity matrix of the same order as A
  • Apply a sequence of row operations till we get identity matrix on the LHS and use the same elementary operations on the RHS to get I = BA. The matrix B on the RHS is the inverse of matrix A.
  • To find the inverse of A using column operations, write A = IA and apply column operations sequentially till I = AB is obtained, where B is the inverse matrix of A.

To understand this concept better let us take a look at the following example.

Example: Find the inverse of matrix A given below:

\(A = \begin{bmatrix} 1 & 3\\ 6 & -1 \end{bmatrix}\)

Solution: Let  A = IA

Or \(\begin{bmatrix} 1 & 6\\ 3 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}A\)

Applying R2→ R2 – 6R1, we have

\(\begin{bmatrix} 1 & 3\\ 0 & -19 \end{bmatrix} = \begin{bmatrix} 1 & 0\\ -6 & 1 \end{bmatrix}A\)

Applying R2→ (-1/19)R2 , we have

\(\begin{bmatrix} 1 & 3\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0\\ \frac{6}{19} & -\frac{1}{19} \end{bmatrix}A\)

Applying R1→ R1 – 3R2, we have

\(\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{19} & \frac{3}{19}\\ \frac{6}{19} & -\frac{1}{19} \end{bmatrix}A\)

Thus, the inverse of matrix A is given by:

\(A^{-1} = \begin{bmatrix} \frac{1}{19} & \frac{3}{19}\\ \frac{6}{19} & -\frac{1}{19} \end{bmatrix}\)<

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Practise This Question

Let F(α)=cosαsinα0sinαcosα0001, where αϵR.
Then [F(α)]1 is equal to