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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.

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Adjoint of a Matrix

Let the determinant of a square matrix A be

$$\begin{array}{l}\left| A \right|\end{array}$$

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

The matrix formed by the cofactors of the elements is

$$\begin{array}{l}\left[ \begin{matrix} {{A}_{11}} & {{A}_{12}} & {{A}_{13}} \\ {{A}_{21}} & {{A}_{22}} & {{A}_{23}} \\ {{A}_{31}} & {{A}_{32}} & {{A}_{33}} \\ \end{matrix} \right]\end{array}$$

Where

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

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

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

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

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

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

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

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

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

$$\begin{array}{l}=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right]\times \left[ \begin{matrix} {{A}_{11}} & {{A}_{21}} & {{A}_{31}} \\ {{A}_{12}} & {{A}_{22}} & {{A}_{32}} \\ {{A}_{13}} & {{A}_{23}} & {{A}_{33}} \\ \end{matrix} \right]\end{array}$$

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

=

$$\begin{array}{l}\left[ \begin{matrix} \left| A \right| & 0 & 0 \\ 0 & \left| A \right| & 0 \\ 0 & 0 & \left| A \right| \\ \end{matrix} \right]=\left| A \right|\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]=\left| A \right|I.\end{array}$$

Example Problems on How to Find the Adjoint of a Matrix

Example 1: If AT = – A then the elements on the diagonal of the matrix are equal to

(a) 1 (b) -1 (c) 0 (d) none of these

Solution:

(c) AT = -A; A is skew-symmetric matrix; diagonal elements of A are zeros.

so option (c) is the answer.

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’ = 0. 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

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

then adj A is given by –

Solution:

Co-factors of the elements of any matrix are obtained by eliminating all the elements of the same row and column and calculating the determinant of the remaining elements.

$$\begin{array}{l}{{A}_{11}}=\left| \begin{matrix} 3 & 4 \\ 4 & 3 \\ \end{matrix} \right|=3\times 3-4\times 4=-7\end{array}$$

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

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

adj A = transpose of cofactor matrix.

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

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;

Solution:

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,

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

And

$$\begin{array}{l}A.{{A}^{-1}}=I;\end{array}$$
$$\begin{array}{l}{{A}^{-1}}=\frac{1}{\left| A \right|}\left( Adj.\,A \right)\end{array}$$

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

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

Solution:

By using the formula A-1

$$\begin{array}{l}=\frac{adj\,A}{\left| A \right|}\; we\; can\; obtain\; the\; value\; of \;{{A}^{-1}}\end{array}$$

We have

$$\begin{array}{l}{{A}_{11}}=\left[ \begin{matrix} 4 & 5 \\ -6 & -7 \\ \end{matrix} \right]=2\,\,\,{{A}_{12}}=-\left[ \begin{matrix} 3 & 5 \\ 0 & -7 \\ \end{matrix} \right]=21\end{array}$$

Similarly

$$\begin{array}{l}{{A}_{13}}=-18,{{A}_{31}}=4,{{A}_{32}}=-8,{{A}_{33}}=4,{{A}_{21}}=+6,{{A}_{22}}=-7,{{A}_{23}}=6\end{array}$$

cofactor matrix of A =

$$\begin{array}{l}\begin{bmatrix} 2 & 21 & -18\\ 6&-7 & 6\\ 4 & -8 & 4 \end{bmatrix}\end{array}$$

adj A = transpose of cofactor matrix

$$\begin{array}{l}=\left[ \begin{matrix} 2 & 6 & 4 \\ 21 & -7 & -8 \\ -18 & 6 & 4 \\ \end{matrix} \right]\end{array}$$

Also

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

=-28+30+18

=20

$$\begin{array}{l}{{A}^{-1}}=\frac{adj\,A}{\left| A \right|}=\frac{1}{20}\left[ \begin{matrix} 2 & 6 & 4 \\ 21 & -7 & -8 \\ -18 & 6 & 4 \\ \end{matrix} \right]\end{array}$$

Illustration 2: If the product of a matrix A and

$$\begin{array}{l}\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right] \;is\; the\; matrix \;\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \\ \end{matrix} \right],\end{array}$$

then A-1 is given by:

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

(d) None of these

Solution:

(a) We know if AB = C, then

$$\begin{array}{l}{{B}^{-1}}{{A}^{-1}}={{C}^{-1}}\Rightarrow {{A}^{-1}}=B{{C}^{-1}}\end{array}$$
by using this formula we will get value of A-1 in the above problem.

Here,

$$\begin{array}{l}A\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \\ \end{matrix} \right]\Rightarrow {{A}^{-1}}=\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]{{\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \\ \end{matrix} \right]}^{-1}}=\left[ \begin{matrix} 1 & 1 \\ 2 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -2 \\ -1 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 1 \\ 2 & -4 \\ \end{matrix} \right]\end{array}$$

Illustration 3:

Let

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

Solution:

By obtaining | AB | and adj AB we can obtain

$$\begin{array}{l}{{\left( AB \right)}^{-1}}\end{array}$$
by using the formula
$$\begin{array}{l}{{\left( AB \right)}^{-1}}=\frac{adj\,AB}{\left| AB \right|}.\end{array}$$
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,

$$\begin{array}{l}AB=\left[ \begin{matrix} 2 & 1 & -1 \\ 0 & 1 & 0 \\ 1 & 3 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 2+2+1 & 4+3-1 & 10+1-1 \\ 0+2+0 & 0+3+0 & 0+1+0 \\ 1+6+1 & 2+9-1 & 5+3-1 \\ \end{matrix} \right]=\left[ \begin{matrix} 5 & 6 & 10 \\ 2 & 3 & 1 \\ 8 & 10 & 7 \\ \end{matrix} \right]\end{array}$$

Now,

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

The matrix of cofactors of | AB | is equal to

$$\begin{array}{l}\left[ \begin{matrix} 3\left( 7 \right)-1\left( 10 \right) & -\left\{ 2\left( 7 \right)-8\left( 1 \right) \right\} & 2\left( 10 \right)-3\left( 8 \right) \\ -\left\{ 6\left( 7 \right)-10\left( 10 \right) \right\} & 5\left( 7 \right)-8\left( 10 \right) & -\left\{ 5\left( 10 \right)-6\left( 8 \right) \right\} \\ 6\left( 1 \right)-10\left( 3 \right) & -\left\{ 5\left( 1 \right)-2\left( 10 \right) \right\} & 5\left( 3 \right)-6\left( 2 \right) \\ \end{matrix} \right]=\left[ \begin{matrix} 11 & -6 & -4 \\ 58 & -45 & -2 \\ -24 & 15 & 3 \\ \end{matrix} \right]\end{array}$$

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

Next,

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

Cofactor matrix of B =

$$\begin{array}{l}\begin{bmatrix} 2 & -3 &5 \\ 3& 6 & -3\\ -13 & 9 & -1 \end{bmatrix}\end{array}$$

$$\begin{array}{l}\begin{bmatrix} 2 & 3 & -13\\ -3&6 &9 \\ 5 & -3 &-1 \end{bmatrix}\end{array}$$

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

Similarly cofactor matrix of A =

$$\begin{array}{l}\begin{bmatrix} -1 & 0 & -1\\ -2&-1 & -5\\ 1 & 0 & 2 \end{bmatrix}\end{array}$$

$$\begin{array}{l}\,\,{{A}^{-1}}=\frac{adj\,A}{\left| A \right|}=\frac{1}{-1}\left[ \begin{matrix} -1 & -2 & 1 \\ 0 & -1 & 0 \\ -1 & -5 & 2 \\ \end{matrix} \right]\end{array}$$

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

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

Illustration 4: If A

$$\begin{array}{l}=\left[ \begin{matrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \\ \end{matrix} \right] satisfies\; A’={{A}^{-1}},\end{array}$$

then

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

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

Solution:

(b) Given that

$$\begin{array}{l}A’={{A}^{-1}}\end{array}$$
and we know that
$$\begin{array}{l}A{{A}^{-1}}=I\end{array}$$
and therefore
$$\begin{array}{l}AA’=I.\end{array}$$
Using the multiplication method we can obtain values of x, y and z.

$$\begin{array}{l}A’={{A}^{-1}}\Leftrightarrow AA’=1\end{array}$$

Now,

$$\begin{array}{l}AA’=\left[ \begin{matrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \\ \end{matrix} \right]\left[ \begin{matrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \\ \end{matrix} \right]=\left[ \begin{matrix} 4{{y}^{2}}+{{z}^{2}} & 2{{y}^{2}}-{{z}^{2}} & -2{{y}^{2}}+{{z}^{{2}}} \\ 2{{y}^{2}}-{{z}^{2}} & {{x}^{2}}+{{y}^{2}}+{{z}^{2}} & {{x}^{2}}-{{y}^{2}}-{{z}^{2}} \\ -2{{y}^{2}}+{{z}^{2}} & {{x}^{2}}-{{y}^{2}}-{{z}^{2}} & {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \\ \end{matrix} \right]\end{array}$$

Thus,

$$\begin{array}{l}AA’=I\;\;\;\;\;\;\; \Rightarrow 4{{y}^{2}}+{{z}^{2}}=1,2{{y}^{2}}-{{z}^{2}}=0, \;\;\;\;\;\;\; {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=1,{{x}^{2}}-{{y}^{2}}-{{z}^{2}}=0\end{array}$$

$$\begin{array}{l}x=\pm 1/\sqrt{2},y=\pm 1/\sqrt{6},z=\pm 1/\sqrt{3}\end{array}$$

Illustration 5: If A

$$\begin{array}{l}=\left[ \begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \\ \end{matrix} \right] \;\;and \;\;{{A}^{-1}}=\left[ \begin{matrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & y \\ 5/2 & -3/2 & 1/2 \\ \end{matrix} \right],\end{array}$$

then

$$\begin{array}{l}(a) x=1,y=-1\;\;\;\; (b) x=-1,y=1\;\;\;\;\; (c)x=2,y=-1/2 \;\;\;\;(d) x=1/2,y=\frac{1}{2}\end{array}$$

Solution:

(a) We know

$$\begin{array}{l}A{{A}^{-1}}=I, \end{array}$$
hence by solving it we can obtain the values of x and y.

We have

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

$$\begin{array}{l}\Rightarrow \,\,\,1-x=0,x-1=0;y+1=0,y+1=0,2+xy=1\end{array}$$

∴ x = 1, y = -1

Give the formula to find the Inverse of a matrix A.

The Inverse of a matrix A is given by A-1=adj A/det A.

What do you mean by the Adjoint of a matrix?

The Adjoint of a matrix is the transpose of the cofactor matrix of that matrix.

Give the cofactor formula.

The cofactor formula is given by Aij = (-1)i+j det Mij. Here det Mij is the minor of aij.

What is A (adj A), if A is a square matrix of order n?

If A is a square matrix of order n, then A adj(A) = adj(A) A = |A|I, where I is the identity matrix of order n.

What do you mean by minor of a determinant?

Minor of an element aij of a determinant is calculated by deleting its ith row and jth column in which the element aij lies. We denote the minor of an element aij by Mij.

What do you mean by a non singular matrix?

A square matrix X is said to be non singular if |X| ≠ 0. i.e the determinant will be a non zero value.

What do you mean by a singular matrix?

A square matrix B is said to be singular if |B| = 0.

How to find the adjoint of a 2×2 matrix?

Interchange the elements on the main diagonal. (a11 and a22). Then give negative sign for the elements at a12 and a21 position. Resulting matrix is the adjoint of the given 2×2 matrix.

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