# 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.

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| = $$\frac{1}{\left| A \right|}$$.