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| = \(\frac{1}{\left| A \right|}\).