Adjoint Of A matrix & Inverse Of A Matrix?

Adjoint of a Matrix:

Let  A = \( [a_{ij}]_{n×n}\) denote a square matrix. The adjoint of  A is the transpose of matrix obtained by replacing each element of A by its cofactor in |A|. It is denoted as adj A = \( [b_{ij}]_{n×n} b_{ij} \) = \( A_{ji}\)  where  \( A_{ji}\) represents the cofactor of the \( (j,i)^{th} \)  element of the matrix A. In simple termsthe transpose of the cofactor matrix of  is known as the adjoint of A or adj A.

Let A = \(\left[\begin{matrix}
a_{11} & a_{12} & a_{13}\cr
a_{21} & a_{22} & a_{23} \cr
a_{31} & a_{32} & a_{33}\cr
\end{matrix}\right]  \)

Then adj A = \(\left[\begin{matrix}
A_{11} & A_{12} & A_{13}\cr
A_{21} & A_{22} & A_{23} \cr
A_{31} & A_{32} & A_{33}\cr
\end{matrix}\right]^T  \)

Where \(\left[
\begin{matrix}
A_{11} & A_{12} & A_{13}\cr
A_{21} & A_{22} & A_{23} \cr
A_{31} & A_{32} & A_{33}\cr
\end{matrix}
\right]^T  \)
 represent the cofactor matrix.

\(~~~~~~~~~~~~~~~\) adj A = \(\left[\begin{matrix}
A_{11} & A_{12} & A_{13}\cr
A_{21} & A_{22} & A_{23} \cr
A_{31} & A_{32} & A_{33}\cr
\end{matrix}\right]  \)

Adjoint - matrix & Inverse Of A Matrix

Properties of Adjoint

1. For a square matrix A of order n the following property holds true

A ( adj A ) = (adj A) A = |A| In , where In  is an identity matrix of order n .

2. adj kA = \( k^{n-1} \) ( adj A )

3. adj (AB) = ( adj B) (adj A )

For a better understanding let us take an example.

Example:Find adj A for  A = \(\left[\begin{matrix}
4 & -3 & 5\cr
1 & 0 & 3 \cr
-1 & 5 & 2 \cr
\end{matrix}\right]\)

Solution: The cofactors of A  are

\( C_{11} \)\(\left[\begin{matrix}
0 & 3 \cr
5  & 2 \cr
\end{matrix}
\right]^T  \)
= (0 – 15) = -15

Similarly we can calculate the other cofactors which come out to be

C12 = – 5

C13 = 5

C21 = 31

C22 = -13

C23 = -17

C31 = -9\( C_{31} \) = – 9

C32 = -7

C33 = 3

Since the transpose of the cofactor matrix of A is adj A

⇒ adj A = \(\left[\begin{matrix}
-15 & -5 & 5\cr
31 & 13 & -17 \cr
-9 & -7 & 3 \cr
\end{matrix}\right]^T\)

⇒ adj A = \(\left[\begin{matrix}
-15 & 31 & -9\cr
-5 & 13 & -7 \cr
5 & -17 & 3 \cr
\end{matrix}\right]\)

Let us see what happens if the adjoint is multiplied by matrix .

A × adj A = \(\left[\begin{matrix}
4 & -3 & 5\cr
1 & 0 & 3 \cr
-1 & 5 & 2 \cr
\end{matrix}\right]\)

\(\left[\begin{matrix}
-15 & 31 & -9\cr
-5 & 13 & -7 \cr
5 & -17 & 3 \cr
\end{matrix}\right]\)

We get A × adj A = \(\left[\begin{matrix}
-20 & 0 & 0\cr
0 & -20 & 0 \cr
0 & 0 & -20 \cr
\end{matrix}\right]\)

⇒ A × adj A = (-20) \(\left[\begin{matrix}
1 & 0 & 0\cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
\end{matrix}\right]\)

Also det A = -20

So this justifies the property of the determinant which states that A(adj A) = (adj A)A = |A| In

Inverse of a matrix:

We have learnt that a square matrix A is called as singular if |A| = 0 and it is called as non-singular if |A|≠ 0.

Inverse of a square matrix A = \( [a_{ij}]_{n×n} \) (which is denoted as \( A^{-1} \)) is the matrix  B such that

B = \( [b_{ij}]_{n × n} \) such that AB = BA = In

In simpler terms the product of adj A with \( \frac {1}{|A|}\) gives the inverse of a matrix A.

Inverse of a matrix if it exists is unique.

Since AB = In

Then using determinants

|AB| = |I|

\( \Rightarrow |A||B| \) = 1d|I| = 1 )

  1. d)

This indicates that |A| ≠ 0. Therefore A is non-singular.

Now as we know, A(adj A) = ( adj A)A = |A|In (By property of adjoint)

\( \Rightarrow \frac {1}{|A|} (A(adj A)) \)= \(\frac {1}{|A|}(adj A)A \) = I ————— (1)

Since AB = BA = I, from equation 1 we can say that B = \( \frac {1}{|A|} \) adj A..

B represents the inverse of the matrix A and is given as B = \( A^{-1} \) = \( \frac {1}{|A|} adj A \) .

Let us solve the example given above to find out the inverse.

Example:Find \( A^{-1}\) for A = \( \left[\begin{matrix}
4 & 3 & 5\cr
1 & 0 & 3 \cr
-1 & 5 & 2 \cr
\end{matrix}\right]\)

We have already figured out the adjoint of the given matrix which is

adj A = \( \left[\begin{matrix}
-15 & -31 & -9\cr
5 & 13 & 7 \cr
5 & 17 & 3 \cr
\end{matrix}\right]\)

Now for determining the inverse of the given matrix let us find its determinant

|A|= -20

As we know, \( A^{-1} = \frac {1}{|A|} adj A \)

We get, \(A^{-1} = \frac {1}{20} \left[\begin{matrix}
-15 & -31 & -9\cr
5 & 13 & 7 \cr
5 & 17 & 3 \cr
\end{matrix}\right]\)

This gives the required inverse of the given matrix A.

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Practise This Question

 A =[0110]is an involutary matrix.