The adjoint of a matrix (also called the adjugate of a matrix) is defined as the transpose of the cofactor matrix of that particular matrix. For a matrix A, the adjoint is denoted as adj (A) . On the other hand, the inverse of a matrix A is that matrix which when multiplied by the matrix A give an identity matrix. The inverse of a Matrix A is denoted by A-1 .
Table of Contents in Matrices
Download this lesson as PDF:- Adjoint and Inverse of a Matrix PDF
Adjoint of a Matrix
Let the determinant of a square matrix A be ∣ A ∣ \left| A \right| ∣ A ∣
I f A = [ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ] T h e n ∣ A ∣ = ∣ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ∣ If A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right]\;\; Then \;\;\left| A \right|=\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right| I f A = ⎣ ⎢ ⎡ a 1 1 a 2 1 a 3 1 a 1 2 a 2 2 a 3 2 a 1 3 a 2 3 a 3 3 ⎦ ⎥ ⎤ T h e n ∣ A ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ a 1 1 a 2 1 a 3 1 a 1 2 a 2 2 a 3 2 a 1 3 a 2 3 a 3 3 ∣ ∣ ∣ ∣ ∣ ∣ ∣
The matrix formed by the cofactors of the elements in is [ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ] \left[ \begin{matrix} {{A}_{11}} & {{A}_{12}} & {{A}_{13}} \\ {{A}_{21}} & {{A}_{22}} & {{A}_{23}} \\ {{A}_{31}} & {{A}_{32}} & {{A}_{33}} \\ \end{matrix} \right] ⎣ ⎢ ⎡ A 1 1 A 2 1 A 3 1 A 1 2 A 2 2 A 3 2 A 1 3 A 2 3 A 3 3 ⎦ ⎥ ⎤
Where A 11 = ( − 1 ) 1 + 1 ∣ a 22 a 23 a 32 a 33 ∣ = a 22 a 33 − a 23 . a 32 {{A}_{11}}={{\left( -1 \right)}^{1+1}}\left| \begin{matrix} {{a}_{22}} & {{a}_{23}} \\ {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|={{a}_{22}}{{a}_{33}}-{{a}_{23}}.\,{{a}_{32}} A 1 1 = ( − 1 ) 1 + 1 ∣ ∣ ∣ ∣ ∣ a 2 2 a 3 2 a 2 3 a 3 3 ∣ ∣ ∣ ∣ ∣ = a 2 2 a 3 3 − a 2 3 . a 3 2
A 12 = ( − 1 ) 1 + 2 ∣ a 21 a 23 a 31 a 3 ∣ = − a 21 . a 33 + a 23 . a 31 ; A 13 = ( − 1 ) 1 + 3 ∣ a 21 a 22 a 31 a 32 ∣ = a 21 a 32 − a 22 a 31 ; {{A}_{12}}={{\left( -1 \right)}^{1+2}}\left| \begin{matrix} {{a}_{21}} & {{a}_{23}} \\ {{a}_{31}} & a\ 3 \\ \end{matrix} \right|=-{{a}_{21}}.\,{{a}_{33}}+{{a}_{23}}.\,{{a}_{31}};{{A}_{13}}={{\left( -1 \right)}^{1+3}}\left| \begin{matrix} {{a}_{21}} & {{a}_{22}} \\ {{a}_{31}} & {{a}_{32}} \\ \end{matrix} \right|={{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}}; A 1 2 = ( − 1 ) 1 + 2 ∣ ∣ ∣ ∣ ∣ a 2 1 a 3 1 a 2 3 a 3 ∣ ∣ ∣ ∣ ∣ = − a 2 1 . a 3 3 + a 2 3 . a 3 1 ; A 1 3 = ( − 1 ) 1 + 3 ∣ ∣ ∣ ∣ ∣ a 2 1 a 3 1 a 2 2 a 3 2 ∣ ∣ ∣ ∣ ∣ = a 2 1 a 3 2 − a 2 2 a 3 1 ;
A 21 = ( − 1 ) 2 + 1 ∣ a 12 a 13 a 32 a 33 ∣ = − a 12 a 33 + a 13 . a 32 ; A 22 = ( − 1 ) 2 + 2 ∣ a 11 a 13 a 31 a 33 ∣ = a 11 a 33 − a 13 . a 31 ; {{A}_{21}}={{\left( -1 \right)}^{2+1}}\left| \begin{matrix} {{a}_{12}} & {{a}_{13}} \\ {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|=-{{a}_{12}}{{a}_{33}}+{{a}_{13}}.\,{{a}_{32}};{{A}_{22}}={{\left( -1 \right)}^{2+2}}\left| \begin{matrix} {{a}_{11}} & {{a}_{13}} \\ {{a}_{31}} & {{a}_{33}} \\ \end{matrix} \right|={{a}_{11}}{{a}_{33}}-{{a}_{13}}.\,{{a}_{31}}; A 2 1 = ( − 1 ) 2 + 1 ∣ ∣ ∣ ∣ ∣ a 1 2 a 3 2 a 1 3 a 3 3 ∣ ∣ ∣ ∣ ∣ = − a 1 2 a 3 3 + a 1 3 . a 3 2 ; A 2 2 = ( − 1 ) 2 + 2 ∣ ∣ ∣ ∣ ∣ a 1 1 a 3 1 a 1 3 a 3 3 ∣ ∣ ∣ ∣ ∣ = a 1 1 a 3 3 − a 1 3 . a 3 1 ;
A 23 = ( − 1 ) 2 + 3 ∣ a 11 a 12 a 31 a 32 ∣ = − a 11 a 32 + a 12 . a 31 ; A 31 = ( − 1 ) 3 + 1 ∣ a 12 a 13 a 22 a 23 ∣ = a 12 a 23 − a 13 . a 22 ; {{A}_{23}}={{\left( -1 \right)}^{2+3}}\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} \\ {{a}_{31}} & {{a}_{32}} \\ \end{matrix} \right|=-{{a}_{11}}{{a}_{32}}+{{a}_{12}}.\,{{a}_{31}};{{A}_{31}}={{\left( -1 \right)}^{3+1}}\left| \begin{matrix} {{a}_{12}} & {{a}_{13}} \\ {{a}_{22}} & {{a}_{23}} \\ \end{matrix} \right|={{a}_{12}}{{a}_{23}}-{{a}_{13}}.\,{{a}_{22}}; A 2 3 = ( − 1 ) 2 + 3 ∣ ∣ ∣ ∣ ∣ a 1 1 a 3 1 a 1 2 a 3 2 ∣ ∣ ∣ ∣ ∣ = − a 1 1 a 3 2 + a 1 2 . a 3 1 ; A 3 1 = ( − 1 ) 3 + 1 ∣ ∣ ∣ ∣ ∣ a 1 2 a 2 2 a 1 3 a 2 3 ∣ ∣ ∣ ∣ ∣ = a 1 2 a 2 3 − a 1 3 . a 2 2 ;
A 32 = ( − 1 ) 3 + 2 ∣ a 11 a 13 a 21 a 23 ∣ = − a 11 a 23 + a 13 . a 21 ; A 33 = ( − 1 ) 3 + 3 ∣ a 11 a 12 a 21 a 22 ∣ = a 11 a 22 − a 12 . a 21 ; {{A}_{32}}={{\left( -1 \right)}^{3+2}}\left| \begin{matrix} {{a}_{11}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{23}} \\ \end{matrix} \right|=-{{a}_{11}}{{a}_{23}}+{{a}_{13}}.\,{{a}_{21}};{{A}_{33}}={{\left( -1 \right)}^{3+3}}\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} \\ {{a}_{21}} & {{a}_{22}} \\ \end{matrix} \right|={{a}_{11}}{{a}_{22}}-{{a}_{12}}.\,{{a}_{21}}; A 3 2 = ( − 1 ) 3 + 2 ∣ ∣ ∣ ∣ ∣ a 1 1 a 2 1 a 1 3 a 2 3 ∣ ∣ ∣ ∣ ∣ = − a 1 1 a 2 3 + a 1 3 . a 2 1 ; A 3 3 = ( − 1 ) 3 + 3 ∣ ∣ ∣ ∣ ∣ a 1 1 a 2 1 a 1 2 a 2 2 ∣ ∣ ∣ ∣ ∣ = a 1 1 a 2 2 − a 1 2 . a 2 1 ;
Then the transpose of the matrix of co-factors is called the adjoint of the matrix A and is written as
adj A. a d j A = [ A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 ] adj\,A=\left[ \begin{matrix} {{A}_{11}} & {{A}_{21}} & {{A}_{31}} \\ {{A}_{12}} & {{A}_{22}} & {{A}_{32}} \\ {{A}_{13}} & {{A}_{23}} & {{A}_{33}} \\ \end{matrix} \right] a d j A = ⎣ ⎢ ⎡ A 1 1 A 1 2 A 1 3 A 2 1 A 2 2 A 2 3 A 3 1 A 3 2 A 3 3 ⎦ ⎥ ⎤
The product of a matrix A and its adjoint is equal to unit matrix multiplied by the determinant A.
Let A be a square matrix, then (Adjoint A). A = A. (Adjoint A) = | A |. I
Let A = [ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ] a n d a d j A = [ A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 ] A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right] \;\;and \;\;adj \;A\;=\;\left[ \begin{matrix} {{A}_{11}} & {{A}_{21}} & {{A}_{31}} \\ {{A}_{12}} & {{A}_{22}} & {{A}_{32}} \\ {{A}_{13}} & {{A}_{23}} & {{A}_{33}} \\ \end{matrix} \right] A = ⎣ ⎢ ⎡ a 1 1 a 2 1 a 3 1 a 1 2 a 2 2 a 3 2 a 1 3 a 2 3 a 3 3 ⎦ ⎥ ⎤ a n d a d j A = ⎣ ⎢ ⎡ A 1 1 A 1 2 A 1 3 A 2 1 A 2 2 A 2 3 A 3 1 A 3 2 A 3 3 ⎦ ⎥ ⎤
A. (adj. A) = [ 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 ] =\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right]\times \left[ \begin{matrix} {{A}_{11}} & {{A}_{21}} & {{A}_{31}} \\ {{A}_{12}} & {{A}_{22}} & {{A}_{32}} \\ {{A}_{13}} & {{A}_{23}} & {{A}_{33}} \\ \end{matrix} \right] = ⎣ ⎢ ⎡ a 1 1 a 2 1 a 3 1 a 1 2 a 2 2 a 3 2 a 1 3 a 2 3 a 3 3 ⎦ ⎥ ⎤ × ⎣ ⎢ ⎡ A 1 1 A 1 2 A 1 3 A 2 1 A 2 2 A 2 3 A 3 1 A 3 2 A 3 3 ⎦ ⎥ ⎤
= [ a 11 A 11 + a 12 A 12 + a 13 A 13 a 11 A 21 + a 12 A 22 + a 13 A 23 a 11 A 31 + a 12 A 32 + a 13 A 33 a 21 A 11 + a 22 A 12 + a 23 A 13 a 21 A 21 + a 22 A 22 + a 23 A 23 a 21 A 31 + a 22 A 32 + a 23 A 33 a 31 A 11 + a 32 A 12 + a 33 A 13 a 31 A 21 + a 32 A 22 + a 33 A 23 a 31 A 31 + a 32 A 32 + a 33 A 33 ] =\left[ \begin{matrix} {{a}_{11}}{{A}_{11}}+{{a}_{12}}{{A}_{12}}+{{a}_{13}}{{A}_{13}} & {{a}_{11}}{{A}_{21}}+{{a}_{12}}{{A}_{22}}+{{a}_{13}}{{A}_{23}} & {{a}_{11}}{{A}_{31}}+{{a}_{12}}{{A}_{32}}+{{a}_{13}}{{A}_{33}} \\ {{a}_{21}}{{A}_{11}}+{{a}_{22}}{{A}_{12}}+{{a}_{23}}{{A}_{13}} & {{a}_{21}}{{A}_{21}}+{{a}_{22}}{{A}_{22}}+{{a}_{23}}{{A}_{23}} & {{a}_{21}}{{A}_{31}}+{{a}_{22}}{{A}_{32}}+{{a}_{23}}{{A}_{33}} \\ {{a}_{31}}{{A}_{11}}+{{a}_{32}}{{A}_{12}}+{{a}_{33}}{{A}_{13}} & {{a}_{31}}{{A}_{21}}+{{a}_{32}}{{A}_{22}}+{{a}_{33}}{{A}_{23}} & {{a}_{31}}{{A}_{31}}+{{a}_{32}}{{A}_{32}}+{{a}_{33}}{{A}_{33}} \\ \end{matrix} \right] = ⎣ ⎢ ⎡ a 1 1 A 1 1 + a 1 2 A 1 2 + a 1 3 A 1 3 a 2 1 A 1 1 + a 2 2 A 1 2 + a 2 3 A 1 3 a 3 1 A 1 1 + a 3 2 A 1 2 + a 3 3 A 1 3 a 1 1 A 2 1 + a 1 2 A 2 2 + a 1 3 A 2 3 a 2 1 A 2 1 + a 2 2 A 2 2 + a 2 3 A 2 3 a 3 1 A 2 1 + a 3 2 A 2 2 + a 3 3 A 2 3 a 1 1 A 3 1 + a 1 2 A 3 2 + a 1 3 A 3 3 a 2 1 A 3 1 + a 2 2 A 3 2 + a 2 3 A 3 3 a 3 1 A 3 1 + a 3 2 A 3 2 + a 3 3 A 3 3 ⎦ ⎥ ⎤
= [ ∣ A ∣ 0 0 0 ∣ A ∣ 0 0 0 ∣ A ∣ ] = ∣ A ∣ [ 1 0 0 0 1 0 0 0 1 ] = ∣ A ∣ I . \left[ \begin{matrix} \left| A \right| & 0 & 0 \\ 0 & \left| A \right| & 0 \\ 0 & 0 & \left| A \right| \\ \end{matrix} \right]=\left| A \right|\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]=\left| A \right|I. ⎣ ⎢ ⎡ ∣ A ∣ 0 0 0 ∣ A ∣ 0 0 0 ∣ A ∣ ⎦ ⎥ ⎤ = ∣ A ∣ ⎣ ⎢ ⎡ 1 0 0 0 1 0 0 0 1 ⎦ ⎥ ⎤ = ∣ A ∣ I .
How to Find the Adjoint of a Matrix and Its Inverse?
Example Problems on How to Find the Adjoint of a Matrix
Example 1: If A= -A then x + y is equal to
(a) 2 (b) -1 (c) 0 (d) 12
Solution:
(c) A = -A; A is skew-symmetric matrix; diagonal elements of A are zeros
x = 0 , y = 0 ∴x + y = 0
Example 2: If A and B are two skew-symmetric matrices of order n, then,
(a) AB is a skew-symmetric matrix (b) AB is a symmetric matrix
(c) AB is a symmetric matrix if A and B commute (d)None of these
Solution:
(c) We are given A’ = -A and B’ = -B;
Now, (AB)’ = B’A’ = (-B) (-A) = BA = AB, if A and B commute.
Example 3: Let A and B be two matrices such that AB’ + BA’ = O. If A is skew symmetric ,then BA
(a) Symmetric (b) Skew symmetric (c) Invertible (d) None of these
Solution:
(c) we have, (BA)’ = A’B’ = -AB’ [ A is skew symmetric]; = BA’ = B(-A) = -BA BA is skew symmetric.
Example 4: Let A = [ 1 2 3 1 3 4 1 4 3 ] , =\left[ \begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 4 \\ 1 & 4 & 3 \\ \end{matrix} \right], = ⎣ ⎢ ⎡ 1 1 1 2 3 4 3 4 3 ⎦ ⎥ ⎤ ,
then the co-factors of elements of A are given by –
Solution:
Co-factors of the elements of any matrix are obtain by eliminating all the elements of the same row and column and calculating the determinant of the remaining elements.
A 11 = ∣ 3 4 4 3 ∣ = 3 × 3 − 4 × 4 = − 7 {{A}_{11}}=\left| \begin{matrix} 3 & 4 \\ 4 & 3 \\ \end{matrix} \right|=3\times 3-4\times 4=-7 A 1 1 = ∣ ∣ ∣ ∣ ∣ 3 4 4 3 ∣ ∣ ∣ ∣ ∣ = 3 × 3 − 4 × 4 = − 7
A 12 = − ∣ 1 4 1 3 ∣ = 1 , A 13 = ∣ 1 3 1 4 ∣ = 1 ; A 21 = − ∣ 2 3 4 3 ∣ = 6 , A 22 = ∣ 1 3 1 3 ∣ = 0 {{A}_{12}}=-\left| \begin{matrix} 1 & 4 \\ 1 & 3 \\ \end{matrix} \right|=1,{{A}_{13}}=\left| \begin{matrix} 1 & 3 \\ 1 & 4 \\ \end{matrix} \right|=1; {{A}_{21}}=-\left| \begin{matrix} 2 & 3 \\ 4 & 3 \\ \end{matrix} \right|=6,{{A}_{22}}=\left| \begin{matrix} 1 & 3 \\ 1 & 3 \\ \end{matrix} \right|=0 A 1 2 = − ∣ ∣ ∣ ∣ ∣ 1 1 4 3 ∣ ∣ ∣ ∣ ∣ = 1 , A 1 3 = ∣ ∣ ∣ ∣ ∣ 1 1 3 4 ∣ ∣ ∣ ∣ ∣ = 1 ; A 2 1 = − ∣ ∣ ∣ ∣ ∣ 2 4 3 3 ∣ ∣ ∣ ∣ ∣ = 6 , A 2 2 = ∣ ∣ ∣ ∣ ∣ 1 1 3 3 ∣ ∣ ∣ ∣ ∣ = 0
A 23 = − ∣ 1 2 1 4 ∣ = − 2 , A 31 = ∣ 2 3 3 4 ∣ = − 1 ; A 32 = − ∣ 1 3 1 4 ∣ = − 1 , A 33 = ∣ 1 2 1 3 ∣ = 1 {{A}_{23}}=-\left| \begin{matrix} 1 & 2 \\ 1 & 4 \\ \end{matrix} \right|=-2,\,\,\,\,{{A}_{31}}=\left| \begin{matrix} 2 & 3 \\ 3 & 4 \\ \end{matrix} \right|=-1;\,\,\,\,{{A}_{32}}=-\left| \begin{matrix} 1 & 3 \\ 1 & 4 \\ \end{matrix} \right|=-1, \;\;\;{{A}_{33}}=\left| \begin{matrix} 1 & 2 \\ 1 & 3 \\ \end{matrix} \right|=1 A 2 3 = − ∣ ∣ ∣ ∣ ∣ 1 1 2 4 ∣ ∣ ∣ ∣ ∣ = − 2 , A 3 1 = ∣ ∣ ∣ ∣ ∣ 2 3 3 4 ∣ ∣ ∣ ∣ ∣ = − 1 ; A 3 2 = − ∣ ∣ ∣ ∣ ∣ 1 1 3 4 ∣ ∣ ∣ ∣ ∣ = − 1 , A 3 3 = ∣ ∣ ∣ ∣ ∣ 1 1 2 3 ∣ ∣ ∣ ∣ ∣ = 1
∴ A d j A = ∣ − 7 6 − 1 1 0 − 1 1 − 2 1 ∣ \,\,\,Adj\,\,A=\left| \begin{matrix} -7 & 6 & -1 \\ 1 & 0 & -1 \\ 1 & -2 & 1 \\ \end{matrix} \right| A d j A = ∣ ∣ ∣ ∣ ∣ ∣ ∣ − 7 1 1 6 0 − 2 − 1 − 1 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣
Example 5: Which of the following statements are false –
(a) If | A | = 0, then | adj A | = 0;
(b) Adjoint of a diagonal matrix of order 3 × 3 is a diagonal matrix;
(c) Product of two upper triangular matrices is an upper triangular matrix;
(d) adj (AB) = adj (A) adj (B);
Solution:
(d) We have, adj (AB) = adj (B) adj (A) and not adj (AB) = adj (A) adj (B)
Inverse of a Matrix
If A and B are two square matrices of the same order, such that AB = BA = I (I = unit matrix)
Then B is called the inverse of A, i.e. B = A–1 and A is the inverse of B. Condition for a square matrix A to possess an inverse is that the matrix A is non-singular, i.e., | A | ≠ 0. If A is a square matrix and B is its inverse then AB = I. Taking determinant of both sides | AB | = | I | or | A | | B | = I. From this relation it is clear that | A | ≠ 0, i.e. the matrix A is non-singular.
How to find the inverse of a matrix by using the adjoint matrix?
We know that, A . ( A d j A ) = ∣ A ∣ I o r A . ( A d j A ) ∣ A ∣ = I ( P r o v i d e d ∣ A ∣ ≠ 0 ) A.\left( Adj\,A \right)=\left| A \right|I\;\; or \;\;\;\frac{A.\left( Adj\,A \right)}{\left| A \right|}=I\;\; (Provided \left| A \right|\ne 0) A . ( A d j A ) = ∣ A ∣ I o r ∣ A ∣ A . ( A d j A ) = I ( P r o v i d e d ∣ A ∣ = 0 )
And A . A − 1 = I ; A.{{A}^{-1}}=I; A . A − 1 = I ; A − 1 = 1 ∣ A ∣ ( A d j . A ) {{A}^{-1}}=\frac{1}{\left| A \right|}\left( Adj.\,A \right) A − 1 = ∣ A ∣ 1 ( A d j . A )
Properties of Inverse and Adjoint of a Matrix
Property 1: For a square matrix A of order n, A adj(A) = adj(A) A = |A|I, where I is the identitiy matrix of order n.
Property 2: A square matrix A is invertible if and only if A is a non-singular matrix.
Problems on Finding the Inverse of a Matrix
Illustration 1: Let A = [ 1 0 − 1 3 4 5 0 − 6 − 7 ] . =\left[ \begin{matrix} 1 & 0 & -1 \\ 3 & 4 & 5 \\ 0 & -6 & -7 \\ \end{matrix} \right]. = ⎣ ⎢ ⎡ 1 3 0 0 4 − 6 − 1 5 − 7 ⎦ ⎥ ⎤ . What is inverse of A ?
Solution:
By using the formula A-1 = a d j A ∣ A ∣ w e c a n o b t a i n t h e v a l u e o f A − 1 =\frac{adj\,A}{\left| A \right|}\; we\; can\; obtain\; the\; value\; of \;{{A}^{-1}} = ∣ A ∣ a d j A w e c a n o b t a i n t h e v a l u e o f A − 1
We have A 11 = [ 4 5 − 6 − 7 ] = 2 A 12 = − [ 3 5 0 − 7 ] = 21 {{A}_{11}}=\left[ \begin{matrix} 4 & 5 \\ -6 & -7 \\ \end{matrix} \right]=2\,\,\,{{A}_{12}}=-\left[ \begin{matrix} 3 & 5 \\ 0 & -7 \\ \end{matrix} \right]=21 A 1 1 = [ 4 − 6 5 − 7 ] = 2 A 1 2 = − [ 3 0 5 − 7 ] = 2 1
And similarly A 13 = − 18 , A 31 = 4 , A 32 = − 8 , A 33 = 4 , A 21 = + 6 , A 22 = − 7 , A 23 = 6 {{A}_{13}}=-18,{{A}_{31}}=4,{{A}_{32}}=-8,{{A}_{33}}=4,{{A}_{21}}=+6,{{A}_{22}}=-7,{{A}_{23}}=6 A 1 3 = − 1 8 , A 3 1 = 4 , A 3 2 = − 8 , A 3 3 = 4 , A 2 1 = + 6 , A 2 2 = − 7 , A 2 3 = 6
adj A = [ 2 6 4 21 − 7 − 8 − 18 6 4 ] =\left[ \begin{matrix} 2 & 6 & 4 \\ 21 & -7 & -8 \\ -18 & 6 & 4 \\ \end{matrix} \right] = ⎣ ⎢ ⎡ 2 2 1 − 1 8 6 − 7 6 4 − 8 4 ⎦ ⎥ ⎤
Also ∣ A ∣ = ∣ 1 0 − 1 3 4 5 0 − 6 − 7 ∣ = { 4 × ( − 7 ) − ( − 6 ) × 5 − 3 × ( − 6 ) } \left| A \right|=\left| \begin{matrix} 1 & 0 & -1 \\ 3 & 4 & 5 \\ 0 & -6 & -7 \\ \end{matrix} \right|=\left\{ 4\times \left( -7 \right)-\left( -6 \right)\times 5-3\times \left( -6 \right) \right\} ∣ A ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 3 0 0 4 − 6 − 1 5 − 7 ∣ ∣ ∣ ∣ ∣ ∣ ∣ = { 4 × ( − 7 ) − ( − 6 ) × 5 − 3 × ( − 6 ) }
=-28+30+18=20 A − 1 = a d j A ∣ A ∣ = 1 20 [ 2 6 4 21 − 7 − 8 − 18 6 4 ] {{A}^{-1}}=\frac{adj\,A}{\left| A \right|}=\frac{1}{20}\left[ \begin{matrix} 2 & 6 & 4 \\ 21 & -7 & -8 \\ -18 & 6 & 4 \\ \end{matrix} \right] A − 1 = ∣ A ∣ a d j A = 2 0 1 ⎣ ⎢ ⎡ 2 2 1 − 1 8 6 − 7 6 4 − 8 4 ⎦ ⎥ ⎤
Illustration 2: If the product of a matrix A and [ 1 1 2 0 ] i s t h e m a t r i x [ 3 2 1 1 ] , \left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right] \;is\; the\; matrix \;\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \\ \end{matrix} \right], [ 1 2 1 0 ] i s t h e m a t r i x [ 3 1 2 1 ] ,
then A-1 is given by:
( a ) [ 0 − 1 2 − 4 ] ( b ) [ 0 − 1 − 2 − 4 ] ( c ) [ 0 1 2 − 4 ] (a) \left[ \begin{matrix}\;\;\;\; 0 & -1 \\ 2 & -4 \\ \end{matrix} \right]\;\;\;\; (b) \left[ \begin{matrix} 0 & -1 \\ -2 & -4 \\ \end{matrix} \right] \;\;\;\; (c)\left[ \begin{matrix} 0 & 1 \\ 2 & -4 \\ \end{matrix} \right] ( a ) [ 0 2 − 1 − 4 ] ( b ) [ 0 − 2 − 1 − 4 ] ( c ) [ 0 2 1 − 4 ]
(d) None of these
Solution:
(a) We know if AB = C, then B − 1 A − 1 = C − 1 ⇒ A − 1 = B C − 1 {{B}^{-1}}{{A}^{-1}}={{C}^{-1}}\Rightarrow {{A}^{-1}}=B{{C}^{-1}} B − 1 A − 1 = C − 1 ⇒ A − 1 = B C − 1 by using this formula we will get value of A-1 in the above problem.
Here, A [ 1 1 2 0 ] = [ 3 2 1 1 ] ⇒ A − 1 = [ 1 1 2 0 ] [ 3 2 1 1 ] − 1 = [ 1 1 2 0 ] [ 1 − 2 − 1 3 ] = [ 0 1 2 − 4 ] A\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \\ \end{matrix} \right]\Rightarrow {{A}^{-1}}=\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]{{\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \\ \end{matrix} \right]}^{-1}}=\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -2 \\ -1 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 1 \\ 2 & -4 \\ \end{matrix} \right] A [ 1 2 1 0 ] = [ 3 1 2 1 ] ⇒ A − 1 = [ 1 2 1 0 ] [ 3 1 2 1 ] − 1 = [ 1 2 1 0 ] [ 1 − 1 − 2 3 ] = [ 0 2 1 − 4 ]
Illustration 3: Let A = [ 2 1 − 1 0 1 0 1 3 − 1 ] a n d B = [ 1 2 5 2 3 1 − 1 1 1 ] . P r o v e t h a t ( A B ) − 1 = B − 1 A − 1 . A =\left[ \begin{matrix} 2 & 1 & -1 \\ 0 & 1 & 0 \\ 1 & 3 & -1 \\ \end{matrix} \right]\;and\; B =\left[ \begin{matrix} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1 \\ \end{matrix} \right]. \;Prove\; that \;{{\left( AB \right)}^{-1}}={{B}^{-1}}{{A}^{-1}}. A = ⎣ ⎢ ⎡ 2 0 1 1 1 3 − 1 0 − 1 ⎦ ⎥ ⎤ a n d B = ⎣ ⎢ ⎡ 1 2 − 1 2 3 1 5 1 1 ⎦ ⎥ ⎤ . P r o v e t h a t ( A B ) − 1 = B − 1 A − 1 .
Solution:
By obtaining | AB | and adj AB we can obtain ( A B ) − 1 {{\left( AB \right)}^{-1}} ( A B ) − 1 by using the formula ( A B ) − 1 = a d j A B ∣ A B ∣ . {{\left( AB \right)}^{-1}}=\frac{adj\,AB}{\left| AB \right|}. ( A B ) − 1 = ∣ A B ∣ a d j A B . Similarly we can also obtain the values of B-1 and A-1 Then by multiplying B-1 and A-1 we can prove the given problem.
Here, A B = [ 2 1 − 1 0 1 0 1 3 − 1 ] [ 1 2 5 2 3 1 − 1 1 1 ] = [ 2 + 2 + 1 4 + 3 − 1 10 + 1 − 1 0 + 2 + 0 0 + 3 + 0 0 + 1 + 0 1 + 6 + 1 2 + 9 − 1 5 + 3 − 1 ] = [ 5 6 10 2 3 1 8 10 7 ] AB=\left[ \begin{matrix} 2 & 1 & -1 \\ 0 & 1 & 0 \\ 1 & 3 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 2+2+1 & 4+3-1 & 10+1-1 \\ 0+2+0 & 0+3+0 & 0+1+0 \\ 1+6+1 & 2+9-1 & 5+3-1 \\ \end{matrix} \right]=\left[ \begin{matrix} 5 & 6 & 10 \\ 2 & 3 & 1 \\ 8 & 10 & 7 \\ \end{matrix} \right] A B = ⎣ ⎢ ⎡ 2 0 1 1 1 3 − 1 0 − 1 ⎦ ⎥ ⎤ ⎣ ⎢ ⎡ 1 2 − 1 2 3 1 5 1 1 ⎦ ⎥ ⎤ = ⎣ ⎢ ⎡ 2 + 2 + 1 0 + 2 + 0 1 + 6 + 1 4 + 3 − 1 0 + 3 + 0 2 + 9 − 1 1 0 + 1 − 1 0 + 1 + 0 5 + 3 − 1 ⎦ ⎥ ⎤ = ⎣ ⎢ ⎡ 5 2 8 6 3 1 0 1 0 1 7 ⎦ ⎥ ⎤
Now, ∣ A B ∣ = ∣ 5 6 10 2 3 1 8 10 7 ∣ = 5 ( 21 − 10 ) − 6 ( 14 − 8 ) + 10 ( 20 − 24 ) = 55 − 36 − 40 = − 21. \left| AB \right|=\left| \begin{matrix} 5 & 6 & 10 \\ 2 & 3 & 1 \\ 8 & 10 & 7 \\ \end{matrix} \right|=5\left( 21-10 \right)-6\left( 14-8 \right)+10\left( 20-24 \right)=55-36-40=-21. ∣ A B ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ 5 2 8 6 3 1 0 1 0 1 7 ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 5 ( 2 1 − 1 0 ) − 6 ( 1 4 − 8 ) + 1 0 ( 2 0 − 2 4 ) = 5 5 − 3 6 − 4 0 = − 2 1 .
The matrix of cofactors of | AB | is = [ 3 ( 7 ) − 1 ( 10 ) − { 2 ( 7 ) − 8 ( 1 ) } a 2 ( 10 ) − 3 ( 8 ) − { 6 ( 7 ) − 10 ( 10 ) } 5 ( 7 ) − 8 ( 10 ) − { 5 ( 10 ) − 6 ( 8 ) } 6 ( 1 ) − 10 ( 3 ) − { 5 ( 1 ) − 2 ( 10 ) } 5 ( 3 ) − 6 ( 2 ) ] = [ 11 − 6 − 4 58 − 45 − 2 − 24 15 3 ] \left[ \begin{matrix} 3\left( 7 \right)-1\left( 10 \right) & -\left\{ 2\left( 7 \right)-8\left( 1 \right) \right\} &a 2\left( 10 \right)-3\left( 8 \right) \\ -\left\{ 6\left( 7 \right)-10\left( 10 \right) \right\} & 5\left( 7 \right)-8\left( 10 \right) & -\left\{ 5\left( 10 \right)-6\left( 8 \right) \right\} \\ 6\left( 1 \right)-10\left( 3 \right) & -\left\{ 5\left( 1 \right)-2\left( 10 \right) \right\} & 5\left( 3 \right)-6\left( 2 \right) \\ \end{matrix} \right]=\left[ \begin{matrix} 11 & -6 & -4 \\ 58 & -45 & -2 \\ -24 & 15 & 3 \\ \end{matrix} \right] ⎣ ⎢ ⎡ 3 ( 7 ) − 1 ( 1 0 ) − { 6 ( 7 ) − 1 0 ( 1 0 ) } 6 ( 1 ) − 1 0 ( 3 ) − { 2 ( 7 ) − 8 ( 1 ) } 5 ( 7 ) − 8 ( 1 0 ) − { 5 ( 1 ) − 2 ( 1 0 ) } a 2 ( 1 0 ) − 3 ( 8 ) − { 5 ( 1 0 ) − 6 ( 8 ) } 5 ( 3 ) − 6 ( 2 ) ⎦ ⎥ ⎤ = ⎣ ⎢ ⎡ 1 1 5 8 − 2 4 − 6 − 4 5 1 5 − 4 − 2 3 ⎦ ⎥ ⎤
adj AB =[ 11 58 − 24 − 6 − 45 15 − 4 − 2 3 ] S o , ( A B ) − 1 = a d j A B ∣ A B ∣ = − 1 21 [ 11 58 − 24 − 6 − 45 15 − 4 − 2 3 ] \left[ \begin{matrix} 11 & 58 & -24 \\ -6 & -45 & 15 \\ -4 & -2 & 3 \\ \end{matrix} \right] So, \,\,{{\left( AB \right)}^{-1}}=\frac{adj\,AB}{\left| AB \right|}=\frac{-1}{21}\left[ \begin{matrix} 11 & 58 & -24 \\ -6 & -45 & 15 \\ -4 & -2 & 3 \\ \end{matrix} \right] ⎣ ⎢ ⎡ 1 1 − 6 − 4 5 8 − 4 5 − 2 − 2 4 1 5 3 ⎦ ⎥ ⎤ S o , ( A B ) − 1 = ∣ A B ∣ a d j A B = 2 1 − 1 ⎣ ⎢ ⎡ 1 1 − 6 − 4 5 8 − 4 5 − 2 − 2 4 1 5 3 ⎦ ⎥ ⎤
Next, ∣ B ∣ = ∣ 1 2 5 2 3 1 − 1 1 1 ∣ = 1 ( 3 − 1 ) − 2 ( 2 + 1 ) + 5 ( 2 + 3 ) = 21 \left| B \right|=\left| \begin{matrix} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1 \\ \end{matrix} \right|=1\left( 3-1 \right)-2\left( 2+1 \right)+5\left( 2+3 \right)=21 ∣ B ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 2 − 1 2 3 1 5 1 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 1 ( 3 − 1 ) − 2 ( 2 + 1 ) + 5 ( 2 + 3 ) = 2 1
∴ B − 1 a d j B ∣ B ∣ = 1 21 [ 2 3 − 13 − 3 6 9 5 − 3 − 1 ] ; ∣ A ∣ = [ 2 1 − 1 0 1 0 1 3 − 1 ] = 1 ( − 2 + 1 ) = − 1 {{B}^{-1}}\frac{adj\,B}{\left| B \right|}=\frac{1}{21}\left[ \begin{matrix} 2 & 3 & -13 \\ -3 & 6 & 9 \\ 5 & -3 & -1 \\ \end{matrix} \right]; \;\;\left| A \right|=\left[ \begin{matrix} 2 & 1 & -1 \\ 0 & 1 & 0 \\ 1 & 3 & -1 \\ \end{matrix} \right]=1\left( -2+1 \right)=-1 B − 1 ∣ B ∣ a d j B = 2 1 1 ⎣ ⎢ ⎡ 2 − 3 5 3 6 − 3 − 1 3 9 − 1 ⎦ ⎥ ⎤ ; ∣ A ∣ = ⎣ ⎢ ⎡ 2 0 1 1 1 3 − 1 0 − 1 ⎦ ⎥ ⎤ = 1 ( − 2 + 1 ) = − 1
∴ A − 1 = a d j A ∣ A ∣ = 1 − 1 [ − 1 − 2 1 0 − 1 0 − 1 − 5 2 ] \,\,{{A}^{-1}}=\frac{adj\,A}{\left| A \right|}=\frac{1}{-1}\left[ \begin{matrix} -1 & -2 & 1 \\ 0 & -1 & 0 \\ -1 & -5 & 2 \\ \end{matrix} \right] A − 1 = ∣ A ∣ a d j A = − 1 1 ⎣ ⎢ ⎡ − 1 0 − 1 − 2 − 1 − 5 1 0 2 ⎦ ⎥ ⎤
∴ B − 1 A − 1 = − 1 21 [ 2 3 − 13 − 3 6 9 5 − 3 − 1 ] [ − 1 − 2 1 0 − 1 0 − 1 − 5 2 ] {{B}^{-1}}{{A}^{-1}}=-\frac{1}{21}\left[ \begin{matrix} 2 & 3 & -13 \\ -3 & 6 & 9 \\ 5 & -3 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & -2 & 1 \\ 0 & -1 & 0 \\ -1 & -5 & 2 \\ \end{matrix} \right] B − 1 A − 1 = − 2 1 1 ⎣ ⎢ ⎡ 2 − 3 5 3 6 − 3 − 1 3 9 − 1 ⎦ ⎥ ⎤ ⎣ ⎢ ⎡ − 1 0 − 1 − 2 − 1 − 5 1 0 2 ⎦ ⎥ ⎤
= − 1 21 [ 11 58 − 24 − 6 − 45 15 − 4 − 2 3 ] T h u s , ( A B ) − 1 = B − 1 A − 1 =-\frac{1}{21}\left[ \begin{matrix} 11 & 58 & -24 \\ -6 & -45 & 15 \\ -4 & -2 & 3 \\ \end{matrix} \right] \;\;Thus, \;\;\;{{\left( AB \right)}^{-1}}={{B}^{-1}}{{A}^{-1}} = − 2 1 1 ⎣ ⎢ ⎡ 1 1 − 6 − 4 5 8 − 4 5 − 2 − 2 4 1 5 3 ⎦ ⎥ ⎤ T h u s , ( A B ) − 1 = B − 1 A − 1
Illustration 4: If A = [ 0 2 y z x y − z x − y z ] s a t i s f i e s A ’ = A − 1 , =\left[ \begin{matrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \\ \end{matrix} \right] satisfies\; A’={{A}^{-1}}, = ⎣ ⎢ ⎡ 0 x x 2 y y − y z − z z ⎦ ⎥ ⎤ s a t i s f i e s A ’ = A − 1 ,
then
( a ) x = ± 1 / 6 , y = ± 1 / 6 , z = ± 1 / 3 ( b ) x = ± 1 / 2 , y = ± 1 / 6 , z = ± 1 / 3 (a) x=\pm 1/\sqrt{6},y=\pm 1/\sqrt{6},z=\pm 1/\sqrt{3}\;\; \;\;\;\;\;\;\; (b) x=\pm 1/\sqrt{2},y=\pm 1/\sqrt{6},z=\pm 1/\sqrt{3} ( a ) x = ± 1 / 6 , y = ± 1 / 6 , z = ± 1 / 3 ( b ) x = ± 1 / 2 , y = ± 1 / 6 , z = ± 1 / 3
( c ) x = ± 1 / 6 , y = ± 1 / 2 , z = ± 1 / 3 ( d ) x = ± 1 / 2 , y = ± 1 / 3 , z = ± 1 / 2 (c) x=\pm 1/\sqrt{6},y=\pm 1/\sqrt{2},z=\pm 1/\sqrt{3} \;\;\;\;\;\;\;\;\;\;\;\; (d) x=\pm 1/\sqrt{2},y=\pm 1/3,z=\pm 1/\sqrt{2} ( c ) x = ± 1 / 6 , y = ± 1 / 2 , z = ± 1 / 3 ( d ) x = ± 1 / 2 , y = ± 1 / 3 , z = ± 1 / 2
Solution:
(b) Given that A ’ = A − 1 A’={{A}^{-1}} A ’ = A − 1 and we know that A A − 1 = I A{{A}^{-1}}=I A A − 1 = I and therefore A A ’ = I . AA’=I. A A ’ = I . Using the multiplication method we can obtain values of x, y and z.
A ’ = A − 1 ⇔ A A ’ = 1 A’={{A}^{-1}}\Leftrightarrow AA’=1 A ’ = A − 1 ⇔ A A ’ = 1
Now, A A ’ = [ 0 2 y z x y − z x − y z ] [ 0 x x 2 y y − y z − z z ] = [ 4 y 2 + z 2 2 y 2 − z 2 − 2 y 2 + z 2 y 2 − z 2 x 2 + y 2 + z 2 x 2 − y 2 − z 2 − 2 y 2 + z 2 x 2 − y 2 − z 2 x 2 + y 2 + z 2 ] AA’=\left[ \begin{matrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \\ \end{matrix} \right]\left[ \begin{matrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \\ \end{matrix} \right]=\left[ \begin{matrix} 4{{y}^{2}}+{{z}^{2}} & 2{{y}^{2}}-{{z}^{2}} & -2{{y}^{2}}+{{z}^{{}}} \\ 2{{y}^{2}}-{{z}^{2}} & {{x}^{2}}+{{y}^{2}}+{{z}^{2}} & {{x}^{2}}-{{y}^{2}}-{{z}^{2}} \\ -2{{y}^{2}}+{{z}^{2}} & {{x}^{2}}-{{y}^{2}}-{{z}^{2}} & {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \\ \end{matrix} \right] A A ’ = ⎣ ⎢ ⎡ 0 x x 2 y y − y z − z z ⎦ ⎥ ⎤ ⎣ ⎢ ⎡ 0 2 y z x y − z x − y z ⎦ ⎥ ⎤ = ⎣ ⎢ ⎡ 4 y 2 + z 2 2 y 2 − z 2 − 2 y 2 + z 2 2 y 2 − z 2 x 2 + y 2 + z 2 x 2 − y 2 − z 2 − 2 y 2 + z x 2 − y 2 − z 2 x 2 + y 2 + z 2 ⎦ ⎥ ⎤
Thus, A A ’ = I ⇒ 4 y 2 + z 2 = 1 , 2 y 2 − z 2 = 0 , x 2 + y 2 + z 2 = 1 , x 2 − y 2 − z 2 = 0 AA’=I\;\;\;\;\;\;\; \Rightarrow 4{{y}^{2}}+{{z}^{2}}=1,2{{y}^{2}}-{{z}^{2}}=0, \;\;\;\;\;\;\; {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=1,{{x}^{2}}-{{y}^{2}}-{{z}^{2}}=0 A A ’ = I ⇒ 4 y 2 + z 2 = 1 , 2 y 2 − z 2 = 0 , x 2 + y 2 + z 2 = 1 , x 2 − y 2 − z 2 = 0
x = ± 1 / 2 , y = ± 1 / 6 , z = ± 1 / 3 x=\pm 1/\sqrt{2},y=\pm 1/\sqrt{6},z=\pm 1/\sqrt{3} x = ± 1 / 2 , y = ± 1 / 6 , z = ± 1 / 3
Illustration 5: If A = [ 0 1 2 1 2 3 3 x 1 ] a n d A − 1 = [ 1 / 2 − 1 / 2 1 / 2 − 4 3 y 5 / 2 − 3 / 2 1 / 2 ] , =\left[ \begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \\ \end{matrix} \right] \;\;and \;\;{{A}^{-1}}=\left[ \begin{matrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & y \\ 5/2 & -3/2 & 1/2 \\ \end{matrix} \right], = ⎣ ⎢ ⎡ 0 1 3 1 2 x 2 3 1 ⎦ ⎥ ⎤ a n d A − 1 = ⎣ ⎢ ⎡ 1 / 2 − 4 5 / 2 − 1 / 2 3 − 3 / 2 1 / 2 y 1 / 2 ⎦ ⎥ ⎤ ,
then
( a ) x = 1 , y = − 1 ( b ) x = − 1 , y = 1 ( c ) x = 2 , y = − 1 / 2 ( d ) x = 1 / 2 , y = 1 2 (a) x=1,y=-1\;\;\;\; (b) x=-1,y=1\;\;\;\;\; (c)x=2,y=-1/2 \;\;\;\;(d) x=1/2,y=\frac{1}{2} ( a ) x = 1 , y = − 1 ( b ) x = − 1 , y = 1 ( c ) x = 2 , y = − 1 / 2 ( d ) x = 1 / 2 , y = 2 1
Solution:
(a) We know A A − 1 = I , A{{A}^{-1}}=I, A A − 1 = I , hence by solving it we can obtain the values of x and y.
We have
[ 1 0 0 0 1 0 0 0 1 ] = A A − 1 = [ 0 1 2 1 2 3 3 x 1 ] [ 1 / 2 − 1 / 2 1 / 2 − 4 3 y 5 / 2 − 3 / 2 1 / 2 ] = [ 1 0 y + 1 0 1 2 ( y + 1 ) 4 ( 1 − x ) 3 ( x − 1 ) 2 + x y ] \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]=A{{A}^{-1}}=\left[ \begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & y \\ 5/2 & -3/2 & 1/2 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & y+1 \\ 0 & 1 & 2\left( y+1 \right) \\ 4\left( 1-x \right) & 3\left( x-1 \right) & 2+xy \\ \end{matrix} \right] ⎣ ⎢ ⎡ 1 0 0 0 1 0 0 0 1 ⎦ ⎥ ⎤ = A A − 1 = ⎣ ⎢ ⎡ 0 1 3 1 2 x 2 3 1 ⎦ ⎥ ⎤ ⎣ ⎢ ⎡ 1 / 2 − 4 5 / 2 − 1 / 2 3 − 3 / 2 1 / 2 y 1 / 2 ⎦ ⎥ ⎤ = ⎣ ⎢ ⎡ 1 0 4 ( 1 − x ) 0 1 3 ( x − 1 ) y + 1 2 ( y + 1 ) 2 + x y ⎦ ⎥ ⎤
⇒ 1 − x = 0 , x − 1 = 0 ; y + 1 = 0 , y + 1 = 0 , 2 + x y = 1 \Rightarrow \,\,\,1-x=0,x-1=0;y+1=0,y+1=0,2+xy=1 ⇒ 1 − x = 0 , x − 1 = 0 ; y + 1 = 0 , y + 1 = 0 , 2 + x y = 1
∴ x = 1, y = -1