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 matrix A, gives an identity matrix. The inverse of a Matrix A is denoted by A-1.
Table of Contents
Adjoint of a Matrix
Let the determinant of a square matrix A be |A|.
\(\begin{array}{l}If A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right],\;\; then \;\;\left| A \right|=\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|\end{array} \)
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 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 the unit matrix multiplied by the determinant A.
Let A be a square matrix, then (Adjoint A). A = A. (Adjoint A) = | A |. I
Let
\(\begin{array}{l}A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right] \;\;and \;\;adj \;A\;=\;\left[ \begin{matrix} {{A}_{11}} & {{A}_{21}} & {{A}_{31}} \\ {{A}_{12}} & {{A}_{22}} & {{A}_{32}} \\ {{A}_{13}} & {{A}_{23}} & {{A}_{33}} \\ \end{matrix} \right]\end{array} \)
A. (adj. A)
\(\begin{array}{l}=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right]\times \left[ \begin{matrix} {{A}_{11}} & {{A}_{21}} & {{A}_{31}} \\ {{A}_{12}} & {{A}_{22}} & {{A}_{32}} \\ {{A}_{13}} & {{A}_{23}} & {{A}_{33}} \\ \end{matrix} \right]\end{array} \)
\(\begin{array}{l}=\left[ \begin{matrix} {{a}_{11}}{{A}_{11}}+{{a}_{12}}{{A}_{12}}+{{a}_{13}}{{A}_{13}} & {{a}_{11}}{{A}_{21}}+{{a}_{12}}{{A}_{22}}+{{a}_{13}}{{A}_{23}} & {{a}_{11}}{{A}_{31}}+{{a}_{12}}{{A}_{32}}+{{a}_{13}}{{A}_{33}} \\ {{a}_{21}}{{A}_{11}}+{{a}_{22}}{{A}_{12}}+{{a}_{23}}{{A}_{13}} & {{a}_{21}}{{A}_{21}}+{{a}_{22}}{{A}_{22}}+{{a}_{23}}{{A}_{23}} & {{a}_{21}}{{A}_{31}}+{{a}_{22}}{{A}_{32}}+{{a}_{23}}{{A}_{33}} \\ {{a}_{31}}{{A}_{11}}+{{a}_{32}}{{A}_{12}}+{{a}_{33}}{{A}_{13}} & {{a}_{31}}{{A}_{21}}+{{a}_{32}}{{A}_{22}}+{{a}_{33}}{{A}_{23}} & {{a}_{31}}{{A}_{31}}+{{a}_{32}}{{A}_{32}}+{{a}_{33}}{{A}_{33}} \\ \end{matrix} \right]\end{array} \)
\(\begin{array}{l}=\left[ \begin{matrix} \left| A \right| & 0 & 0 \\ 0 & \left| A \right| & 0 \\ 0 & 0 & \left| A \right| \\ \end{matrix} \right]=\left| A \right|\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]=\left| A \right|I.\end{array} \)
Solved 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:
AT = -A; A is a 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:
We are given A’ = -A and B’ = -B;
Now, (AB)’ = B’A’ = (-B) (-A) = BA = AB, if A and B commute.
Thus, the correct option is (c).
Example 3: Let A and B be two matrices such that AB’ + BA’ = 0. If A is skew-symmetric, then BA is
(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.
Thus, the correct option is (b).
Example 4: Let
\(\begin{array}{l}A=\left[ \begin{matrix} 1 & 2 & 3 \\ 1 & 3 & 4 \\ 1 & 4 & 3 \\ \end{matrix} \right],\end{array} \)
then find adj A.
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.
\(\begin{array}{l}\therefore Adj\,\,A=\left| \begin{matrix} -7 & 6 & -1 \\ 1 & 0 & -1 \\ 1 & -2 & 1 \\ \end{matrix} \right|\end{array} \)
Example 5: Which of the following statements is 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:
We have adj (AB) = adj (B) adj (A) and not adj (AB) = adj (A) adj (B).
Thus, the correct option is (d).
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 the 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 identity 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
\(\begin{array}{l}A=\left[ \begin{matrix} 1 & 0 & -1 \\ 3 & 4 & 5 \\ 0 & -6 & -7 \\ \end{matrix} \right].\end{array} \)
What is inverse of A ?
Solution:
From the formula,
\(\begin{array}{l}{A}^{-1}=\frac{adj\,A}{\left| A \right|}\end{array} \)
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 the cofactor matrix
\(\begin{array}{l}adj A =\left[ \begin{matrix} 2 & 6 & 4 \\ 21 & -7 & -8 \\ -18 & 6 & 4 \\ \end{matrix} \right]\end{array} \)
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 (AB)-1 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
\(\begin{array}{l}\left[ \begin{matrix} 3\left( 7 \right)-1\left( 10 \right) & -\left\{ 2\left( 7 \right)-8\left( 1 \right) \right\} & 2\left( 10 \right)-3\left( 8 \right) \\ -\left\{ 6\left( 7 \right)-10\left( 10 \right) \right\} & 5\left( 7 \right)-8\left( 10 \right) & -\left\{ 5\left( 10 \right)-6\left( 8 \right) \right\} \\ 6\left( 1 \right)-10\left( 3 \right) & -\left\{ 5\left( 1 \right)-2\left( 10 \right) \right\} & 5\left( 3 \right)-6\left( 2 \right) \\ \end{matrix} \right]=\left[ \begin{matrix} 11 & -6 & -4 \\ 58 & -45 & -2 \\ -24 & 15 & 3 \\ \end{matrix} \right]\end{array} \)
adj AB =
\(\begin{array}{l}\left[ \begin{matrix} 11 & 58 & -24 \\ -6 & -45 & 15 \\ -4 & -2 & 3 \\ \end{matrix} \right]\\ So, \,\,{{\left( AB \right)}^{-1}}=\frac{adj\,AB}{\left| AB \right|}=\frac{-1}{21}\left[ \begin{matrix} 11 & 58 & -24 \\ -6 & -45 & 15 \\ -4 & -2 & 3 \\ \end{matrix} \right]\end{array} \)
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
\(\begin{array}{l}=\begin{bmatrix} 2 & -3 &5 \\ 3& 6 & -3\\ -13 & 9 & -1 \end{bmatrix}\end{array} \)
\(\begin{array}{l}adj B =\begin{bmatrix} 2 & 3 & -13\\ -3&6 &9 \\ 5 & -3 &-1 \end{bmatrix}\end{array} \)
∴
\(\begin{array}{l}{{B}^{-1}}=\frac{adj\,B}{\left| B \right|}=\frac{1}{21}\left[ \begin{matrix} 2 & 3 & -13 \\ -3 & 6 & 9 \\ 5 & -3 & -1 \\ \end{matrix} \right]; \;\;\left| A \right|=\left[ \begin{matrix} 2 & 1 & -1 \\ 0 & 1 & 0 \\ 1 & 3 & -1 \\ \end{matrix} \right]=1\left( -2+1 \right)=-1\end{array} \)
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
\(\begin{array}{l}A=\left[ \begin{matrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \\ \end{matrix} \right] satisfies\; A’={{A}^{-1}},\end{array} \)
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
\(\begin{array}{l}A=\left[ \begin{matrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \\ \end{matrix} \right] \;\;and \;\;{{A}^{-1}}=\left[ \begin{matrix} 1/2 & -1/2 & 1/2 \\ -4 & 3 & y \\ 5/2 & -3/2 & 1/2 \\ \end{matrix} \right],\end{array} \)
then
(a) x = 1, y = -1
(b) x = -1, y = 1
(c) = x = 2, y = -1/2
(d) x = 1/2, y = 1/2
Solution:
(a) We know that
\(\begin{array}{l}A{{A}^{-1}}=I, \end{array} \)
; 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
Frequently Asked Questions
Q1
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.
Q2
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.
Q3
Give the cofactor formula.
The cofactor formula is given by Aij = (-1)i+j det Mij. Here, det Mij is the minor of aij.
Q4
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.
Q5
What do you mean by the minor of a determinant?
The 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 can denote the minor of an element aij by Mij.
Q6
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.
Q7
What do you mean by a singular matrix?
A square matrix B is said to be singular if |B| = 0.
Q8
How to find the adjoint of a 2×2 matrix?
Interchange the elements on the main diagonal (a11 and a22). Then, give the negative sign for the elements at a12 and a21 positions. The resulting matrix is the adjoint of the given 2×2 matrix.
