Inverse Matrices

If \(A=\left[ \begin{matrix}4 & -7  \\-3 & 2  \\\end{matrix} \right]\), we have

M11 = Minor of A11 = 2, M12 = Minor of A12 = —3,

M21 = Minor of A21 = —7, M22 = Minor of A22 = 4

COFACTORS: Let A = [aij] be a square matrix of order n. Then the cofactor Cij of aij in A is equal to (—1) i+j. times the determinant of the sub-matrix of order (n-1) obtained by leaving ith row and jth column of A.

If follows from this definition that Cij =Cofactor of aij in A.

= (-1) i+j Mij, where Mij is minor of aij in A.

Thus, Cij = Mij if i+j is even and, Cij = – Mij if i+j is odd.

\(A=\left[ \begin{matrix}1 & 2 & 3  \\-3 & 2 & -1  \\2 & -4 & 3  \\\end{matrix} \right]\). If , then we have

C11 = (-1)1+1 M11= M11 = \(\left| \begin{matrix}2 & -1  \\-4 & 3  \\\end{matrix} \right|\) = 2,

C12 = (-) 1+2M12 = – M12 = \(\left| \begin{matrix}-3 & -1  \\2 & 3  \\\end{matrix} \right|\) = 7,

C13 = (-1)1+3 M13 = M13 = \(\left| \begin{matrix}-3 & 2  \\2 & -4  \\\end{matrix} \right|\) = 8,

C23 = (-1)2+3 M23 = – M23 = \(-\left| \begin{matrix}1 & 2  \\2 & -4  \\\end{matrix} \right|\) = 8 etc.

ADJOINT OF A SQUARE MATRIX:

DEFINITION: Let A = [aij] be a square matrix of order n and let Cij be cofactor of aij in A. Then the transpose of the matrix of cofactors of elements of A is called the adjoint of A and is denoted by adj A.

Thus, adj A = [Cij] T.

(adj A) ij = Cij = Cofactor of aji in A.

Find the adjoint of matrix \(A=\left[ \begin{matrix}p & q  \\r & s  \\\end{matrix} \right]\).

SOLUTION: We have,

Cofactor of A11 = s,

Cofactor of A12 = —r,

Cofactor of A21 = -q

And, Cofactor of A22 = p.

\(\therefore adj\,A=\left[ \begin{matrix}s & -r  \\-q & p  \\\end{matrix} \right]\,\,=\,\,\left[ \begin{matrix}s & -q  \\-r & p  \\\end{matrix} \right]\).

It is evident from this example that the adjoint of a square matrix of order 2 can be easily obtained by interchanging the diagonal elements and changing signs of off-diagonal elements.

If \(A=\left[ \begin{matrix}-2 & 3  \\-5 & 4  \\\end{matrix} \right]\), then by the above rule, we have,

\(adj\,A=\left[ \begin{matrix}4 & -3  \\5 & -2  \\\end{matrix} \right]\).

Find the adjoint of matrix \(A=\left[ \begin{matrix}1 & 1 & 1  \\2 & 1 & -3  \\- & 1 & 2  \\\end{matrix} \right]\)

SOLUTION: Let Cij be cofactor of aji in A. Then cofactors of elements of A are given by

\({{C}_{11}}=\left| \begin{matrix}1 & -3  \\2 & 3  \\\end{matrix} \right|=9\), \({{C}_{12}}=-\left| \begin{matrix}2 & -3  \\-1 & 3  \\\end{matrix} \right|=-3\).

\({{C}_{13}}=\left| \begin{matrix}2 & 1  \\-1 & 2  \\\end{matrix} \right|=5\), \({{C}_{21}}=-\left| \begin{matrix}1 & 1  \\2 & 3  \\\end{matrix} \right|=-1\).

\({{C}_{22}}=\left| \begin{matrix}1 & 1  \\-1 & 3  \\\end{matrix} \right|=4\), \({{C}_{23}}=-\left| \begin{matrix}1 & 1  \\-1 & 2  \\\end{matrix} \right|=-3\).

\({{C}_{31}}=\left| \begin{matrix}1 & 1  \\1 & -3  \\\end{matrix} \right|=-4\), \({{C}_{32}}=-\left| \begin{matrix}1 & 1  \\2 & -3  \\\end{matrix} \right|=5\).

\({{C}_{33}}=\left| \begin{matrix}1 & 1  \\2 & 1  \\\end{matrix} \right|=-1\).

∴ \(adj\,A={{\left[ \begin{matrix}9 & -3 & 5  \\-1 & 4 & -3  \\-4 & 5 & -1  \\\end{matrix} \right]}^{T}}=\left[ \begin{matrix}9 & -1 & -4  \\-3 & 4 & 5  \\5 & -3 & -1  \\\end{matrix} \right]\).

PROPERTIES OF ADJOINT: The following are some properties of adjoint of a square matrix which are states as theorems.

Let A be a square matrix of order n. Then A (adj A) = |A|In = (adj A) A. Let A be a non-singular square matrix of order n. Then |adj A| = |A|n-1.

If A and B are non-singular square matrices of the same order, then adj AB = (adj B) (adj A).

If A is an invertible square matrix, then adj AT = (adj A) T.

If A is a non-singular square matrix, then adj ( adj A) = |A|n-2 A.

If A is a symmetric matrix, then adj A is also a symmetric matrix.

INVERSE OF A MATRIX

DEFINITION A square matrix of order n is invertible if there exists a square matrix B of the same order such that AB = I„ = BA.

In such a case, we say that the inverse of A is B and we write, A-1= B.

PROPERTIES OF INVERSE: The inverse of a matrix has the following properties

Every invertible matrix possesses a unique inverse.

A square matrix is invertible iff it is non-singular.

The inverse of A is given by \({{A}^{-1}}=\frac{1}{|A|}.adjA\).

AB = AC ⇒ B = C

BA = CA ⇒ B = C

Let \(A=\left[ \begin{matrix}1 & 2  \\3 & 6  \\\end{matrix} \right],\,B=\left[ \begin{matrix}-2 & 0  \\0 & 0  \\\end{matrix} \right],\,C=\left[ \begin{matrix}0 & 0  \\-1 & 0  \\\end{matrix} \right]\). Then, \(AB=\left[ \begin{matrix}1 & 2  \\3 & 6  \\\end{matrix} \right]\left[ \begin{matrix}-2 & 0  \\0 & 0  \\\end{matrix} \right]=\left[ \begin{matrix}-2 & 0  \\-6 & 0  \\\end{matrix} \right]\).

And,

\(AC = \left[ \begin{matrix}1 & 2  \\3 & 6  \\\end{matrix} \right]\left[ \begin{matrix}0 & 0  \\-1 & 0  \\\end{matrix} \right]=\left[ \begin{matrix}-2 & 0  \\-6 & 0  \\\end{matrix} \right]\).

Clearly, AB = BC but B ≠ C.

(Reversal law) if A and B are invertible matrices of the same order, then AB is invertible and (AB)-1 = B-1 A-1

If A is an invertible square matrix, then AT is also invertible and (AT)-1 = (A-1) T.

If the non-singular matrix A is symmetric, then A-1 is also symmetric.

If A is a non-singular matrix, then |A¯¹| = |A|¯¹ i.e,. |A¯¹| = \(\frac{1}{\left| A \right|}\).