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Adjoint and Inverse of a Matrix

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definition

Defination:  The adjoint of a square matrix A = `[a_(ij)]_(n × n)` is defined as the transpose of the matrix `[A_(ij)]_(n × n)`, where `A_(ij)` is the cofactor of the element `a_(ij)`. Adjoint of the matrix A is denoted by adj A.
Let A = `[(a_11,a_12,a_13),(a_21,a_22,a_23),(a_31,a_32,a_33)]`

Then adj A = Transpose of `[(A_11,A_12,A_13),(A_21,A_22,A_23),(A_31,A_32,A_33)] = [(A_11,A_21,A_31),(A_12,A_22,A_32),(A_13,A_23,A_33)]`

theorem

If A be any given square matrix of order n, then
A(adj A) = (adj A) A = |A| I , 
where I is the identity matrix of order n
Verification 
Let A =`[(a_11,a_12,a_13),(a_21,a_22,a_23),(a_31,a_32,a_33)]` , then adj A =` [(A_11,A_21,A_31),(A_12,A_22,A_32),(A_13,A_23,A_33)]`
Since sum of product of elements of a row (or a column) with corresponding cofactors is equal to |A| and otherwise zero, we have
A (adj A) =  `|(|A|,0,0),(0,|A|,0),(0,0,|A|)| = |A| [(1,0,0),(0,1,0),(0,0,1)]`
Similarly, we can show  (adj A) A = A I
Hence A (adj A) = (adj A) A = A I

notes

Definition :  A square matrix A is said to be singular if A = 0.
For example, the determinant of matrix A = `[(1,2),(4,8)]` is zero
Hence A is a singular matrix.

Definition :  A square matrix A is said to be non-singular if A ≠ 0
Let A = `[(1,4),(3,2)]`. Then |A| = `|(1,4)(3,2)|` = 4 - 6 = 2 ≠ 0
Hence A is a nonsingular matrix.
We state the following theorems without proof.

Theorem : If A and B are nonsingular matrices of the same order, then AB and BA are also nonsingular matrices of the same order.
Theorem: The determinant of the product of matrices is equal to product of their respective determinants, that is, |AB| = |A| |B| , where A and B are square matrices of the same order
Remark:  We know that (adj A) A = |A| I = `[(|A|,0,0),(0,|A|,0),(0,0,|A|)]` , |A| ≠ 0
Writing determinants of matrices on both sides, we have
`|(adj A)A| = |(|A|,0,0),(0,|A|,0),(0,0,|A|)|`

i.e,. `|(adjA)| |A| = |A|^3 |(1,0,0),(0,1,0),(0,0,1)|`

i.e. `|(adj A)| |A| = |A|3 (1) i.e. |(adj A)| = |A|2`

i.e. `|(adj A)| = |A|^2`

In general, if A is a square matrix of order n, then |adj(A)| = |A|^(n – 1).

Theorem:  A square matrix A is invertible if and only if A is nonsingular matrix.

Proof:  Let A be invertible matrix of order n and I be the identity matrix of order n. Then, there exists a square matrix B of order n such that 
AB = BA = I
Now   AB = I.  So |AB| = I or |A| |B| = 1  (since |I| =1, |AB|=|A||B|)
This gives |A| ≠ 0.
Hence A is nonsingular.
Conversely, let A be nonsingular. Then A ≠ 0
Now A (adj A) = (adj A) A = |A| I                  (Theorem 1)
or `A(1/|A| adj A) = (1/|A| adj A) A = I`
or `AB = BA = I` ,where  `B = 1/|A| adj A`
Thus, A is invertible and `A^-1 = 1/ |A|adj A`

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