Adjoint and Inverse of Matrix: Overview, Questions, Preparation

Adjoint, also known as adjugate, is simply the transpose of its cofactor matrice. Let us understand this through a mathematical example.

Suppose there is a square matrix A = $\left[a_{ij}\right]$  of order n.

Adjoint of A can be represented as:

Adj. A = ${\left[A_{ij}\right]}^T$

Where $A_{ij}$ = cofactor of $a_{ij}$ in A.

If A = $\left[a_{ij}\right]$ is a square matrix and B = $\left[A_{ij}\right]$ (where $A_{ij}$  is the cofactor of the elements in matrix A), then the transpose B^T of matrix B is called adjoint of matrix A and it is denoted by “adj. A.”

For example, let A = $\left[\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right]$ , then

Adj. A = ${\left[\begin{array}{ccc}\hfill A_{11}\hfill & \hfill A_{12}\hfill & \hfill A_{13}\hfill \\ \hfill A_{21}\hfill & \hfill A_{22}\hfill & \hfill A_{23}\hfill \\ \hfill A_{31}\hfill & \hfill A_{32}\hfill & \hfill A_{33}\hfill \end{array}\right]}^T$ = $\left[\begin{array}{ccc}\hfill A_{11}\hfill & \hfill A_{21}\hfill & \hfill A_{31}\hfill \\ \hfill A_{12}\hfill & \hfill A_{22}\hfill & \hfill A_{32}\hfill \\ \hfill A_{13}\hfill & \hfill A_{23}\hfill & \hfill A_{33}\hfill \end{array}\right]$

Here are the steps that you need to follow in order to find the adjoint of a matrice:

• Initially, find the minor for each element in the matrice.
• Next, find the cofactor of each element using the formula Cij​=(−1)i+j⋅Mij​.
• Then, place all the cofactors in their respective positions in the matrice.
• Last step is to transpose the cofactor matrice.
• Final matrice obtained is the adjoint matrice.

Some crucial concepts of this chapter Determinants are given below for reference which the candidates need to be familiar with:

1. Minors

Minors of an element is referred to the determinant obtained by deleting its i^th row and j^th column in which lies.

$\text{Minor of}{\mathrm{ a}}_{ij}\text{ can be denoted by}{\mathrm{ M}}_{ij}$

Where,

I = row

J = column

2. Cofactors

A cofactor is a concept similar to that of minors, but with a simple difference of a sign. The cofactor of an element can be denoted by the equation:

$C_{ij}={(-1)}^{i+j}{M}_{ij}$

$\text{Where, }{M}_{ij}=\text{minor of}{\mathrm{ a}}_{ij}$

If you want to calculate the cofactor, simply calculate the minor and add the sign factor to compute the result, which will be the cofactor.

 

3. Area of a Triangle

Area of a triangle ABC whose vertices are $A (x_1,y_1),\mathrm{ B} (x_2,y_2)\text{, and }C (x_3,y_3)$can be calculated using the formula:

$\Delta =\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

• If the area of $△\text{ABC}=0$ , the points are known to be collinear.

4. Singular matrix

A singular matrix is referred to the square matrix whose determinant is zero. It can be denoted as $\left|\mathit{A}\right|=0$

 

5. Non-singular matrix

A non-singular matrix is referred to the square matrix whose determinant is not equal to zero i.e.    0. A square matrix A will be invertible if and only if A is a non-singular matrix.

Now, let us solve a question to clear our basic concepts.

Question: Find the adjoint of the given matrix A = $\left[\begin{array}{cc}\hfill 3\hfill & \hfill 4\hfill \\ \hfill 5\hfill & \hfill 2\hfill \end{array}\right]$ .

Answer: Given, A = $\left[\begin{array}{cc}\hfill 3\hfill & \hfill 4\hfill \\ \hfill 5\hfill & \hfill 2\hfill \end{array}\right]$

Let $A_{ij}$ denotes the cofactor of $a_{ij}$ in A = $\left[a_{ij}\right]$ , then

$A_{11}$ = (–1) ¹⁺¹(2) = 2                                  $A_{12}$ = (–1) ¹⁺²(5) = –5

$A_{21}$ = (–1) ²⁺¹(4) = – 4                                $A_{22}$ = (–1) ²⁺²(3) = 3      

Hence, adj. A = $\left[\begin{array}{cc}\hfill A_{11}\hfill & \hfill A_{21}\hfill \\ \hfill A_{12}\hfill & \hfill A_{22}\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill -4\hfill \\ \hfill -5\hfill & \hfill 3\hfill \end{array}\right]$

Let there be a square matrice A of order n.

Then, A is considered to be invertible only if there exists another inverse matrix B (having same order as that of A) such that AB = BA = $I$ .

Where,

B = A^-1

I = Identity Matrice

Formula for Inverse of a matrice can be given by:

A−1= adj(A)/ det(A)

Note: A matrice that has an inverse is called non-singular matrice and not every matrice will have an inverse.

The procedure mentioned below will help you calculate the inverse of a matrice using the determinant method:

STEP 1: First of all, find the determinant of A i.e. $\left|A\right|$ .

STEP 2: If $\left|A\right|$ = 0 then, stop the procedure and write “A is a singular matrix hence no inverse”. If $\left|A\right|$ =! 0, move further.

STEP 3: Calculate the cofactors of all the elements in matrix A. Formula: Cij​=(−1)i+j⋅Mij​

STEP 4: Arrange the cofactors in their respective positions in the matrice and then transpose the matrice to get adj. A (adjoint of A).

STEP 5: Find the inverse of A by using the equation: A^-1 = $\frac{1}{\left|A\right|}$ adj.A.

Now, Let us practice our base concepts through a solved example.

Example: Find the inverse of the matrix $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill 4\hfill \end{array}\right]$ .

Sol: Given, A = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill 4\hfill \end{array}\right]$ .

Then, $\left|A\right|$ = $\left|\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill 4\hfill \end{array}\right|$ = 8 – 15 = –7 $\ne$ 0

So, A is non - singular and hence invertible.

Let $A_{ij}$   denotes the cofactor of $a_{ij}$ in A = $\left[a_{ij}\right]$ , then

$A_{11}$ = (–1) ¹⁺¹(4) = 4                                    $A_{12}$ = (–1) ¹⁺²(5) = –5

$A_{21}$ = (–1) ²⁺¹(3) = – 3                               $A_{22}$ = (–1) ²⁺²(2) = 2      

$\therefore$ adj. A = $\left[\begin{array}{cc}\hfill A_{11}\hfill & \hfill A_{21}\hfill \\ \hfill A_{12}\hfill & \hfill A_{22}\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 4\hfill & \hfill -3\hfill \\ \hfill -5\hfill & \hfill 2\hfill \end{array}\right]$

Hence, A^-1 = $\frac{1}{\left|A\right|}$ (adj. A) = $-\frac{1}{7}$ $\left[\begin{array}{cc}\hfill 4\hfill & \hfill -3\hfill \\ \hfill -5\hfill & \hfill 2\hfill \end{array}\right]$