Find Inverse and Adjoint of Matrices with Their Properties Worksheet. It is denoted by adj A. a22;{{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}};A23=(−1)2+3∣∣∣∣∣a11a31a12a32∣∣∣∣∣=−a11a32+a12.a31;A31=(−1)3+1∣∣∣∣∣a12a22a13a23∣∣∣∣∣=a12a23−a13.a22; A32=(−1)3+2∣a11a13a21a23∣=−a11a23+a13. A = A. Log in. [100010001]=AA−1=[0121233x1][1/2−1/21/2−43y5/2−3/21/2]=[10y+1012(y+1)4(1−x)3(x−1)2+xy]\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]⎣⎢⎡100010001⎦⎥⎤=AA−1=⎣⎢⎡01312x231⎦⎥⎤⎣⎢⎡1/2−45/2−1/23−3/21/2y1/2⎦⎥⎤=⎣⎢⎡104(1−x)013(x−1)y+12(y+1)2+xy⎦⎥⎤, ⇒ 1−x=0,x−1=0;y+1=0,y+1=0,2+xy=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+xy=1, Example Problems on How to Find the Adjoint of a Matrix. Find Inverse and Adjoint of Matrices with Their Properties Worksheet. Now, (AB)’ = B’A’ = (-B) (-A) = BA = AB, if A and B commute. Adjoint of a Matrix Let A = [ a i j ] be a square matrix of order n . Free Matrix Adjoint calculator - find Matrix Adjoint step-by-step. In terms of components, Illustration 2: If the product of a matrix A and [1120] is the matrix [3211],\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right] \;is\; the\; matrix \;\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \\ \end{matrix} \right],[1210]isthematrix[3121], (a)[ 0−12−4] (b)[0−1−2−4] (c)[012−4](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)[02−1−4](b)[0−2−1−4](c)[021−4]. Play Matrices â Inverse of a 3x3 Matrix using Adjoint. The term "Hermitian" is used interchangeably as opposed to "Self-Adjoint". Example 1: If A= -A then x + y is equal to, (c) A = -A; A is skew-symmetric matrix; diagonal elements of A are zeros. {{\left( AB \right)}^{-1}}=\frac{adj\,AB}{\left| AB \right|}.(AB)−1=∣AB∣adjAB. Transpose of a Matrix – Properties ( Part 1 ) Play Transpose of a Matrix – Properties ( Part 2 ) Play Transpose of a Matrix – Properties ( Part 3 ) ... Matrices – Inverse of a 2x2 Matrix using Adjoint. Illustration 3: Let A=[21−101013−1] and B=[125231−111]. Example 3: Let A and B be two matrices such that AB’ + BA’ = O. Definition of Adjoint of a Matrix. where Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share ⦠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. We strongly recommend you to refer below as a prerequisite of this. (Adj A)=∣A∣I or A. a21;{{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}};A32=(−1)3+2∣∣∣∣∣a11a21a13a23∣∣∣∣∣=−a11a23+a13.a21;A33=(−1)3+3∣∣∣∣∣a11a21a12a22∣∣∣∣∣=a11a22−a12.a21; Then the transpose of the matrix of co-factors is called the adjoint of the matrix A and is written as, adj A. adj A=[A11A21A31A12A22A32A13A23A33]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]adjA=⎣⎢⎡A11A12A13A21A22A23A31A32A33⎦⎥⎤.