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

• Introduction to Matrices
• Types of Matrices
• Matrix Operations
• Adjoint and Inverse of a Matrix
• Rank of a Matrix and Special Matrices
• Solving Linear Equations using Matrix

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Adjoint of a Matrix

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

\(If A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right]\;\; Then \;\;\left| A \right|=\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|\)

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

Where \({{A}_{11}}={{\left( -1 \right)}^{1+1}}\left| \begin{matrix} {{a}_{22}} & {{a}_{23}} \\ {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|={{a}_{22}}{{a}_{33}}-{{a}_{23}}.\,{{a}_{32}}\)

\({{A}_{12}}={{\left( -1 \right)}^{1+2}}\left| \begin{matrix} {{a}_{21}} & {{a}_{23}} \\ {{a}_{31}} & a\ 3 \\ \end{matrix} \right|=-{{a}_{21}}.\,{{a}_{33}}+{{a}_{23}}.\,{{a}_{31}};{{A}_{13}}={{\left( -1 \right)}^{1+3}}\left| \begin{matrix} {{a}_{21}} & {{a}_{22}} \\ {{a}_{31}} & {{a}_{32}} \\ \end{matrix} \right|={{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}};\)

\({{A}_{21}}={{\left( -1 \right)}^{2+1}}\left| \begin{matrix} {{a}_{12}} & {{a}_{13}} \\ {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|=-{{a}_{12}}{{a}_{33}}+{{a}_{13}}.\,{{a}_{32}};{{A}_{22}}={{\left( -1 \right)}^{2+2}}\left| \begin{matrix} {{a}_{11}} & {{a}_{13}} \\ {{a}_{31}} & {{a}_{33}} \\ \end{matrix} \right|={{a}_{11}}{{a}_{33}}-{{a}_{13}}.\,{{a}_{31}};\)

\({{A}_{23}}={{\left( -1 \right)}^{2+3}}\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} \\ {{a}_{31}} & {{a}_{32}} \\ \end{matrix} \right|=-{{a}_{11}}{{a}_{32}}+{{a}_{12}}.\,{{a}_{31}};{{A}_{31}}={{\left( -1 \right)}^{3+1}}\left| \begin{matrix} {{a}_{12}} & {{a}_{13}} \\ {{a}_{22}} & {{a}_{23}} \\ \end{matrix} \right|={{a}_{12}}{{a}_{23}}-{{a}_{13}}.\,{{a}_{22}};\)

\({{A}_{32}}={{\left( -1 \right)}^{3+2}}\left| \begin{matrix} {{a}_{11}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{23}} \\ \end{matrix} \right|=-{{a}_{11}}{{a}_{23}}+{{a}_{13}}.\,{{a}_{21}};{{A}_{33}}={{\left( -1 \right)}^{3+3}}\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} \\ {{a}_{21}} & {{a}_{22}} \\ \end{matrix} \right|={{a}_{11}}{{a}_{22}}-{{a}_{12}}.\,{{a}_{21}};\)

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

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

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 \(A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right] \;\;and \;\;adj \;A\;=\;\left[ \begin{matrix} {{A}_{11}} & {{A}_{21}} & {{A}_{31}} \\ {{A}_{12}} & {{A}_{22}} & {{A}_{32}} \\ {{A}_{13}} & {{A}_{23}} & {{A}_{33}} \\ \end{matrix} \right]\)

A. (adj. A) \(=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right]\times \left[ \begin{matrix} {{A}_{11}} & {{A}_{21}} & {{A}_{31}} \\ {{A}_{12}} & {{A}_{22}} & {{A}_{32}} \\ {{A}_{13}} & {{A}_{23}} & {{A}_{33}} \\ \end{matrix} \right]\)

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

= \(\left[ \begin{matrix} \left| A \right| & 0 & 0 \\ 0 & \left| A \right| & 0 \\ 0 & 0 & \left| A \right| \\ \end{matrix} \right]=\left| A \right|\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]=\left| A \right|I.\)

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

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.

\({{A}_{11}}=\left| \begin{matrix} 3 & 4 \\ 4 & 3 \\ \end{matrix} \right|=3\times 3-4\times 4=-7\)

\({{A}_{12}}=-\left| \begin{matrix} 1 & 4 \\ 1 & 3 \\ \end{matrix} \right|=1,{{A}_{13}}=\left| \begin{matrix} 1 & 3 \\ 1 & 4 \\ \end{matrix} \right|=1; {{A}_{21}}=-\left| \begin{matrix} 2 & 3 \\ 4 & 3 \\ \end{matrix} \right|=6,{{A}_{22}}=\left| \begin{matrix} 1 & 3 \\ 1 & 3 \\ \end{matrix} \right|=0\)

\({{A}_{23}}=-\left| \begin{matrix} 1 & 2 \\ 1 & 4 \\ \end{matrix} \right|=-2,\,\,\,\,{{A}_{31}}=\left| \begin{matrix} 2 & 3 \\ 3 & 4 \\ \end{matrix} \right|=-1;\,\,\,\,{{A}_{32}}=-\left| \begin{matrix} 1 & 3 \\ 1 & 4 \\ \end{matrix} \right|=-1, \;\;\;{{A}_{33}}=\left| \begin{matrix} 1 & 2 \\ 1 & 3 \\ \end{matrix} \right|=1\)

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

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, \(A.\left( Adj\,A \right)=\left| A \right|I\;\; or \;\;\;\frac{A.\left( Adj\,A \right)}{\left| A \right|}=I\;\; (Provided \left| A \right|\ne 0)\)

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

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

Solution:

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

We have \({{A}_{11}}=\left[ \begin{matrix} 4 & 5 \\ -6 & -7 \\ \end{matrix} \right]=2\,\,\,{{A}_{12}}=-\left[ \begin{matrix} 3 & 5 \\ 0 & -7 \\ \end{matrix} \right]=21\)

And similarly \({{A}_{13}}=-18,{{A}_{31}}=4,{{A}_{32}}=-8,{{A}_{33}}=4,{{A}_{21}}=+6,{{A}_{22}}=-7,{{A}_{23}}=6\)

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

Also \(\left| A \right|=\left| \begin{matrix} 1 & 0 & -1 \\ 3 & 4 & 5 \\ 0 & -6 & -7 \\ \end{matrix} \right|=\left\{ 4\times \left( -7 \right)-\left( -6 \right)\times 5-3\times \left( -6 \right) \right\}\)

=-28+30+18=20 \({{A}^{-1}}=\frac{adj\,A}{\left| A \right|}=\frac{1}{20}\left[ \begin{matrix} 2 & 6 & 4 \\ 21 & -7 & -8 \\ -18 & 6 & 4 \\ \end{matrix} \right]\)

Illustration 2: If the product of a matrix A and \(\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right] \;is\; the\; matrix \;\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \\ \end{matrix} \right],\)

then A^-1 is given by:

\((a) \left[ \begin{matrix}\;\;\;\; 0 & -1 \\ 2 & -4 \\ \end{matrix} \right]\;\;\;\; (b) \left[ \begin{matrix} 0 & -1 \\ -2 & -4 \\ \end{matrix} \right] \;\;\;\; (c)\left[ \begin{matrix} 0 & 1 \\ 2 & -4 \\ \end{matrix} \right]\)

(d) None of these

Solution:

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

Here, \(A\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \\ \end{matrix} \right]\Rightarrow {{A}^{-1}}=\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]{{\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \\ \end{matrix} \right]}^{-1}}=\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -2 \\ -1 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 1 \\ 2 & -4 \\ \end{matrix} \right]\)

Illustration 3: Let \(A =\left[ \begin{matrix} 2 & 1 & -1 \\ 0 & 1 & 0 \\ 1 & 3 & -1 \\ \end{matrix} \right]\;and\; B =\left[ \begin{matrix} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1 \\ \end{matrix} \right]. \;Prove\; that \;{{\left( AB \right)}^{-1}}={{B}^{-1}}{{A}^{-1}}.\)

Solution:

By obtaining | AB | and adj AB we can obtain \({{\left( AB \right)}^{-1}}\) by using the formula \({{\left( AB \right)}^{-1}}=\frac{adj\,AB}{\left| AB \right|}.\) 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, \(AB=\left[ \begin{matrix} 2 & 1 & -1 \\ 0 & 1 & 0 \\ 1 & 3 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 2+2+1 & 4+3-1 & 10+1-1 \\ 0+2+0 & 0+3+0 & 0+1+0 \\ 1+6+1 & 2+9-1 & 5+3-1 \\ \end{matrix} \right]=\left[ \begin{matrix} 5 & 6 & 10 \\ 2 & 3 & 1 \\ 8 & 10 & 7 \\ \end{matrix} \right]\)

Now, \(\left| AB \right|=\left| \begin{matrix} 5 & 6 & 10 \\ 2 & 3 & 1 \\ 8 & 10 & 7 \\ \end{matrix} \right|=5\left( 21-10 \right)-6\left( 14-8 \right)+10\left( 20-24 \right)=55-36-40=-21.\)

The matrix of cofactors of | AB | is = \(\left[ \begin{matrix} 3\left( 7 \right)-1\left( 10 \right) & -\left\{ 2\left( 7 \right)-8\left( 1 \right) \right\} &a 2\left( 10 \right)-3\left( 8 \right) \\ -\left\{ 6\left( 7 \right)-10\left( 10 \right) \right\} & 5\left( 7 \right)-8\left( 10 \right) & -\left\{ 5\left( 10 \right)-6\left( 8 \right) \right\} \\ 6\left( 1 \right)-10\left( 3 \right) & -\left\{ 5\left( 1 \right)-2\left( 10 \right) \right\} & 5\left( 3 \right)-6\left( 2 \right) \\ \end{matrix} \right]=\left[ \begin{matrix} 11 & -6 & -4 \\ 58 & -45 & -2 \\ -24 & 15 & 3 \\ \end{matrix} \right]\)

adj AB =\(\left[ \begin{matrix} 11 & 58 & -24 \\ -6 & -45 & 15 \\ -4 & -2 & 3 \\ \end{matrix} \right] So, \,\,{{\left( AB \right)}^{-1}}=\frac{adj\,AB}{\left| AB \right|}=\frac{-1}{21}\left[ \begin{matrix} 11 & 58 & -24 \\ -6 & -45 & 15 \\ -4 & -2 & 3 \\ \end{matrix} \right]\)

Next, \(\left| B \right|=\left| \begin{matrix} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1 \\ \end{matrix} \right|=1\left( 3-1 \right)-2\left( 2+1 \right)+5\left( 2+3 \right)=21\)

∴ \({{B}^{-1}}\frac{adj\,B}{\left| B \right|}=\frac{1}{21}\left[ \begin{matrix} 2 & 3 & -13 \\ -3 & 6 & 9 \\ 5 & -3 & -1 \\ \end{matrix} \right]; \;\;\left| A \right|=\left[ \begin{matrix} 2 & 1 & -1 \\ 0 & 1 & 0 \\ 1 & 3 & -1 \\ \end{matrix} \right]=1\left( -2+1 \right)=-1\)

∴ \(\,\,{{A}^{-1}}=\frac{adj\,A}{\left| A \right|}=\frac{1}{-1}\left[ \begin{matrix} -1 & -2 & 1 \\ 0 & -1 & 0 \\ -1 & -5 & 2 \\ \end{matrix} \right]\)

∴ \({{B}^{-1}}{{A}^{-1}}=-\frac{1}{21}\left[ \begin{matrix} 2 & 3 & -13 \\ -3 & 6 & 9 \\ 5 & -3 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & -2 & 1 \\ 0 & -1 & 0 \\ -1 & -5 & 2 \\ \end{matrix} \right]\)

\(=-\frac{1}{21}\left[ \begin{matrix} 11 & 58 & -24 \\ -6 & -45 & 15 \\ -4 & -2 & 3 \\ \end{matrix} \right] \;\;Thus, \;\;\;{{\left( AB \right)}^{-1}}={{B}^{-1}}{{A}^{-1}}\)

Illustration 4: If A \(=\left[ \begin{matrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \\ \end{matrix} \right] satisfies\; A’={{A}^{-1}},\)

then

\((a) x=\pm 1/\sqrt{6},y=\pm 1/\sqrt{6},z=\pm 1/\sqrt{3}\;\; \;\;\;\;\;\;\; (b) x=\pm 1/\sqrt{2},y=\pm 1/\sqrt{6},z=\pm 1/\sqrt{3}\)

\((c) x=\pm 1/\sqrt{6},y=\pm 1/\sqrt{2},z=\pm 1/\sqrt{3} \;\;\;\;\;\;\;\;\;\;\;\; (d) x=\pm 1/\sqrt{2},y=\pm 1/3,z=\pm 1/\sqrt{2}\)

Solution:

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

\(A’={{A}^{-1}}\Leftrightarrow AA’=1\)

Now, \(AA’=\left[ \begin{matrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \\ \end{matrix} \right]\left[ \begin{matrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \\ \end{matrix} \right]=\left[ \begin{matrix} 4{{y}^{2}}+{{z}^{2}} & 2{{y}^{2}}-{{z}^{2}} & -2{{y}^{2}}+{{z}^{{}}} \\ 2{{y}^{2}}-{{z}^{2}} & {{x}^{2}}+{{y}^{2}}+{{z}^{2}} & {{x}^{2}}-{{y}^{2}}-{{z}^{2}} \\ -2{{y}^{2}}+{{z}^{2}} & {{x}^{2}}-{{y}^{2}}-{{z}^{2}} & {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \\ \end{matrix} \right]\)

Thus, \(AA’=I\;\;\;\;\;\;\; \Rightarrow 4{{y}^{2}}+{{z}^{2}}=1,2{{y}^{2}}-{{z}^{2}}=0, \;\;\;\;\;\;\; {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=1,{{x}^{2}}-{{y}^{2}}-{{z}^{2}}=0\)

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

Illustration 5: If A \(=\left[ \begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \\ \end{matrix} \right] \;\;and \;\;{{A}^{-1}}=\left[ \begin{matrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & y \\ 5/2 & -3/2 & 1/2 \\ \end{matrix} \right],\)

then

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

Solution:

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

We have

\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]=A{{A}^{-1}}=\left[ \begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & y \\ 5/2 & -3/2 & 1/2 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & y+1 \\ 0 & 1 & 2\left( y+1 \right) \\ 4\left( 1-x \right) & 3\left( x-1 \right) & 2+xy \\ \end{matrix} \right]\)

\(\Rightarrow \,\,\,1-x=0,x-1=0;y+1=0,y+1=0,2+xy=1\)

∴ x = 1, y = -1