Adjoint and Inverse of a Matrix

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

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Adjoint of a Matrix

Let the determinant of a square matrix A be A\left| A \right|

IfA=[a11a12a13a21a22a23a31a32a33]    Then    A=a11a12a13a21a22a23a31a32a33If 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|

The matrix formed by the cofactors of the elements in is [A11A12A13A21A22A23A31A32A33]\left[ \begin{matrix} {{A}_{11}} & {{A}_{12}} & {{A}_{13}} \\ {{A}_{21}} & {{A}_{22}} & {{A}_{23}} \\ {{A}_{31}} & {{A}_{32}} & {{A}_{33}} \\ \end{matrix} \right]

Where A11=(1)1+1a22a23a32a33=a22a33a23.a32{{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}}

A12=(1)1+2a21a23a31a 3=a21.a33+a23.a31;A13=(1)1+3a21a22a31a32=a21a32a22a31;{{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}};

A21=(1)2+1a12a13a32a33=a12a33+a13.a32;A22=(1)2+2a11a13a31a33=a11a33a13.a31;{{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}};

A23=(1)2+3a11a12a31a32=a11a32+a12.a31;A31=(1)3+1a12a13a22a23=a12a23a13.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}};

A32=(1)3+2a11a13a21a23=a11a23+a13.a21;A33=(1)3+3a11a12a21a22=a11a22a12.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}};

Then the transpose of the matrix of co-factors is called the adjoint of the matrix A and is written as

adj A. adjA=[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]

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=[a11a12a13a21a22a23a31a32a33]    and    adj  A  =  [A11A21A31A12A22A32A13A23A33]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. (adj. A) =[a11a12a13a21a22a23a31a32a33]×[A11A21A31A12A22A32A13A23A33]=\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]

=[a11A11+a12A12+a13A13a11A21+a12A22+a13A23a11A31+a12A32+a13A33a21A11+a22A12+a23A13a21A21+a22A22+a23A23a21A31+a22A32+a23A33a31A11+a32A12+a33A13a31A21+a32A22+a33A23a31A31+a32A32+a33A33]=\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]

= [A000A000A]=A[100010001]=AI.\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.

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 =[123134143],=\left[ \begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 4 \\ 1 & 4 & 3 \\ \end{matrix} \right],

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.

A11=3443=3×34×4=7{{A}_{11}}=\left| \begin{matrix} 3 & 4 \\ 4 & 3 \\ \end{matrix} \right|=3\times 3-4\times 4=-7

A12=1413=1,A13=1314=1;A21=2343=6,A22=1313=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

A23=1214=2,A31=2334=1;A32=1314=1,      A33=1213=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

AdjA=761101121\,\,\,Adj\,\,A=\left| \begin{matrix} -7 & 6 & -1 \\ 1 & 0 & -1 \\ 1 & -2 & 1 \\ \end{matrix} \right|

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.(AdjA)=AI    or      A.(AdjA)A=I    (ProvidedA0)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)

And A.A1=I;A.{{A}^{-1}}=I; A1=1A(Adj.A){{A}^{-1}}=\frac{1}{\left| A \right|}\left( Adj.\,A \right)

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 =[101345067].=\left[ \begin{matrix} 1 & 0 & -1 \\ 3 & 4 & 5 \\ 0 & -6 & -7 \\ \end{matrix} \right]. What is inverse of A ?

Solution:

By using the formula A-1 =adjAA  we  can  obtain  the  value  of  A1=\frac{adj\,A}{\left| A \right|}\; we\; can\; obtain\; the\; value\; of \;{{A}^{-1}}

We have A11=[4567]=2A12=[3507]=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

And similarly A13=18,A31=4,A32=8,A33=4,A21=+6,A22=7,A23=6{{A}_{13}}=-18,{{A}_{31}}=4,{{A}_{32}}=-8,{{A}_{33}}=4,{{A}_{21}}=+6,{{A}_{22}}=-7,{{A}_{23}}=6

adj A =[26421781864]=\left[ \begin{matrix} 2 & 6 & 4 \\ 21 & -7 & -8 \\ -18 & 6 & 4 \\ \end{matrix} \right]

Also A=101345067={4×(7)(6)×53×(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\}

=-28+30+18=20 A1=adjAA=120[26421781864]{{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]

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],

then A-1 is given by:

(a)[        0124]        (b)[0124]        (c)[0124](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]

(d) None of these

Solution:

(a) We know if AB = C, then B1A1=C1A1=BC1{{B}^{-1}}{{A}^{-1}}={{C}^{-1}}\Rightarrow {{A}^{-1}}=B{{C}^{-1}} by using this formula we will get value of A-1 in the above problem.

Here, A[1120]=[3211]A1=[1120][3211]1=[1120][1213]=[0124]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]

Illustration 3: Let A=[211010131]  and  B=[125231111].  Prove  that  (AB)1=B1A1.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}}.

Solution:

By obtaining | AB | and adj AB we can obtain (AB)1{{\left( AB \right)}^{-1}} by using the formula (AB)1=adjABAB.{{\left( AB \right)}^{-1}}=\frac{adj\,AB}{\left| AB \right|}. 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, AB=[211010131][125231111]=[2+2+14+3110+110+2+00+3+00+1+01+6+12+915+31]=[56102318107]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]

Now, AB=56102318107=5(2110)6(148)+10(2024)=553640=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.

The matrix of cofactors of | AB | is = [3(7)1(10){2(7)8(1)}a2(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)]=[11645845224153]\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]

adj AB =[11582464515423]So,(AB)1=adjABAB=121[11582464515423]\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]

Next, B=125231111=1(31)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

B1adjBB=121[2313369531];    A=[211010131]=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

A1=adjAA=11[121010152]\,\,{{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]

B1A1=121[2313369531][121010152]{{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]

=121[11582464515423]    Thus,      (AB)1=B1A1=-\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}}

Illustration 4: If A =[02yzxyzxyz]satisfies  A=A1,=\left[ \begin{matrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \\ \end{matrix} \right] satisfies\; 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}

(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}

Solution:

(b) Given that A=A1A’={{A}^{-1}} and we know that AA1=IA{{A}^{-1}}=I and therefore AA=I.AA’=I. Using the multiplication method we can obtain values of x, y and z.

A=A1AA=1A’={{A}^{-1}}\Leftrightarrow AA’=1

Now, AA=[02yzxyzxyz][0xx2yyyzzz]=[4y2+z22y2z22y2+z2y2z2x2+y2+z2x2y2z22y2+z2x2y2z2x2+y2+z2]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]

Thus, AA=I              4y2+z2=1,2y2z2=0,              x2+y2+z2=1,x2y2z2=0AA’=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

x=±1/2,y=±1/6,z=±1/3x=\pm 1/\sqrt{2},y=\pm 1/\sqrt{6},z=\pm 1/\sqrt{3}

Illustration 5: If A =[0121233x1]    and    A1=[1/21/21/243y5/23/21/2],=\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],

then

(a)x=1,y=1        (b)x=1,y=1          (c)x=2,y=1/2        (d)x=1/2,y=12(a) x=1,y=-1\;\;\;\; (b) x=-1,y=1\;\;\;\;\; (c)x=2,y=-1/2 \;\;\;\;(d) x=1/2,y=\frac{1}{2}

Solution:

(a) We know AA1=I,A{{A}^{-1}}=I, hence by solving it we can obtain the values of x and y.

We have

[100010001]=AA1=[0121233x1][1/21/21/243y5/23/21/2]=[10y+1012(y+1)4(1x)3(x1)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]

1x=0,x1=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

∴ x = 1, y = -1