The adjoint of the matrix [(2,1,1) (1,1,1) (1,-1,2)] is

[(3, -3, 0) (-1, 3, -1) (-2, 3, 1)]

[(-3, 3, 1) (-1, 2, -1) (0, 3, 1)]

[(3, -3, -1) (-1, 3, -1) (-2, 0, 1)]

[(-3, -3, -1) (-1, 1, -1) (-2, 3, 1)]

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.

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Adjoint of a Matrix

Let the determinant of a square matrix A be |A|.

\begin{array}{l} I f A = [\begin{array}{c} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}], t h e n | A | = | \begin{array}{c} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} | \end{array}

The matrix formed by the cofactors of the elements is

\begin{array}{l} [\begin{array}{c} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array}] \end{array}

Where

\begin{array}{l} A_{11} = {(- 1)}^{1 + 1} | \begin{array}{c} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array} | = a_{22} a_{33} - a_{23} . a_{32} \end{array}

\begin{array}{l} A_{12} = {(- 1)}^{1 + 2} | \begin{array}{c} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array} | = - a_{21} . a_{33} + a_{23} . a_{31}; A_{13} = {(- 1)}^{1 + 3} | \begin{array}{c} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array} | = a_{21} a_{32} - a_{22} a_{31}; \end{array}

\begin{array}{l} A_{21} = {(- 1)}^{2 + 1} | \begin{array}{c} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array} | = - a_{12} a_{33} + a_{13} . a_{32}; A_{22} = {(- 1)}^{2 + 2} | \begin{array}{c} a_{11} & a_{13} \\ a_{31} & a_{33} \end{array} | = a_{11} a_{33} - a_{13} . a_{31}; \end{array}

\begin{array}{l} A_{23} = {(- 1)}^{2 + 3} | \begin{array}{c} a_{11} & a_{12} \\ a_{31} & a_{32} \end{array} | = - a_{11} a_{32} + a_{12} . a_{31}; A_{31} = {(- 1)}^{3 + 1} | \begin{array}{c} a_{12} & a_{13} \\ a_{22} & a_{23} \end{array} | = a_{12} a_{23} - a_{13} . a_{22}; \end{array}

\begin{array}{l} A_{32} = {(- 1)}^{3 + 2} | \begin{array}{c} a_{11} & a_{13} \\ a_{21} & a_{23} \end{array} | = - a_{11} a_{23} + a_{13} . a_{21}; A_{33} = {(- 1)}^{3 + 3} | \begin{array}{c} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} | = 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} a d j A = [\begin{array}{c} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}] \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 = [\begin{array}{c} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}] a n d a d j A = [\begin{array}{c} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}] \end{array}

A. (adj. A)

\begin{array}{l} = [\begin{array}{c} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}] \times [\begin{array}{c} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}] \end{array}

\begin{array}{l} = [\begin{array}{c} 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{array}] \end{array}

\begin{array}{l} = [\begin{array}{c} | A | & 0 & 0 \\ 0 & | A | & 0 \\ 0 & 0 & | A | \end{array}] = | A | [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = | A | I . \end{array}

How to Find the Adjoint of a Matrix and Its Inverse?

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Example Problems on How to Find the Adjoint of a Matrix

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

(a) 1

(b) -1

(d) None of these

Solution:

A^T = -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

(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

(d) None of these

Solution:

= -BA

BA is skew symmetric.

Thus, the correct option is (b).

Example 4: Let

\begin{array}{l} A = [\begin{array}{c} 1 & 2 & 3 \\ 1 & 3 & 4 \\ 1 & 4 & 3 \end{array}], \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} = | \begin{array}{c} 3 & 4 \\ 4 & 3 \end{array} | = 3 \times 3 - 4 \times 4 = - 7 \end{array}

\begin{array}{l} A_{12} = - | \begin{array}{c} 1 & 4 \\ 1 & 3 \end{array} | = 1, A_{13} = | \begin{array}{c} 1 & 3 \\ 1 & 4 \end{array} | = 1; A_{21} = - | \begin{array}{c} 2 & 3 \\ 4 & 3 \end{array} | = 6, A_{22} = | \begin{array}{c} 1 & 3 \\ 1 & 3 \end{array} | = 0 \end{array}

\begin{array}{l} A_{23} = - | \begin{array}{c} 1 & 2 \\ 1 & 4 \end{array} | = - 2, A_{31} = | \begin{array}{c} 2 & 3 \\ 3 & 4 \end{array} | = - 1; A_{32} = - | \begin{array}{c} 1 & 3 \\ 1 & 4 \end{array} | = - 1, A_{33} = | \begin{array}{c} 1 & 2 \\ 1 & 3 \end{array} | = 1 \end{array}

adj A = transpose of cofactor matrix.

\begin{array}{l} ∴ A d j A = | \begin{array}{c} - 7 & 6 & - 1 \\ 1 & 0 & - 1 \\ 1 & - 2 & 1 \end{array} | \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

(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 . (A d j A) = | A | I o r \frac{A . (A d j A)}{| A |} = I (P r o v i d e d | A | \neq 0) \end{array}

And

\begin{array}{l} A . A^{- 1} = I; \end{array}

\begin{array}{l} A^{- 1} = \frac{1}{| A |} (A d j . A) \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 = [\begin{array}{c} 1 & 0 & - 1 \\ 3 & 4 & 5 \\ 0 & - 6 & - 7 \end{array}] . \end{array}

What is inverse of A ?

Solution:

From the formula,

\begin{array}{l} A^{- 1} = \frac{a d j A}{| A |} \end{array}

We have

\begin{array}{l} A_{11} = [\begin{array}{c} 4 & 5 \\ - 6 & - 7 \end{array}] = 2 A_{12} = - [\begin{array}{c} 3 & 5 \\ 0 & - 7 \end{array}] = 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{array}{c} 2 & 21 & - 18 \\ 6 & - 7 & 6 \\ 4 & - 8 & 4 \end{array}] \end{array}

adj A = transpose of the cofactor matrix

\begin{array}{l} a d j A = [\begin{array}{c} 2 & 6 & 4 \\ 21 & - 7 & - 8 \\ - 18 & 6 & 4 \end{array}] \end{array}

Also

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

= -28 + 30 + 18

= 20

\begin{array}{l} A^{- 1} = \frac{a d j A}{| A |} = \frac{1}{20} [\begin{array}{c} 2 & 6 & 4 \\ 21 & - 7 & - 8 \\ - 18 & 6 & 4 \end{array}] \end{array}

Illustration 2: If the product of a matrix A and

\begin{array}{l} [\begin{array}{c} 1 & 1 \\ 2 & 0 \end{array}] i s t h e m a t r i x [\begin{array}{c} 3 & 2 \\ 1 & 1 \end{array}], \end{array}

then A^-1 is given by:

\begin{array}{l} (a) [\begin{array}{c} 0 & - 1 \\ 2 & - 4 \end{array}] \\ (b) [\begin{array}{c} 0 & - 1 \\ - 2 & - 4 \end{array}] \\ (c) [\begin{array}{c} 0 & 1 \\ 2 & - 4 \end{array}] \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 [\begin{array}{c} 1 & 1 \\ 2 & 0 \end{array}] = [\begin{array}{c} 3 & 2 \\ 1 & 1 \end{array}] \Rightarrow A^{- 1} = [\begin{array}{c} 1 & 1 \\ 2 & 0 \end{array}] {[\begin{array}{c} 3 & 2 \\ 1 & 1 \end{array}]}^{- 1} = [\begin{array}{c} 1 & 1 \\ 2 & 0 \end{array}] [\begin{array}{c} 1 & - 2 \\ - 1 & 3 \end{array}] = [\begin{array}{c} 0 & 1 \\ 2 & - 4 \end{array}] \end{array}

Illustration 3:

Let

\begin{array}{l} A = [\begin{array}{c} 2 & 1 & - 1 \\ 0 & 1 & 0 \\ 1 & 3 & - 1 \end{array}] a n d B = [\begin{array}{c} 1 & 2 & 5 \\ 2 & 3 & 1 \\ - 1 & 1 & 1 \end{array}] . P r o v e t h a t {(A B)}^{- 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} {(A B)}^{- 1} = \frac{a d j A B}{| A B |} . \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} A B = [\begin{array}{c} 2 & 1 & - 1 \\ 0 & 1 & 0 \\ 1 & 3 & - 1 \end{array}] [\begin{array}{c} 1 & 2 & 5 \\ 2 & 3 & 1 \\ - 1 & 1 & 1 \end{array}] = [\begin{array}{c} 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{array}] = [\begin{array}{c} 5 & 6 & 10 \\ 2 & 3 & 1 \\ 8 & 10 & 7 \end{array}] \end{array}

Now,

\begin{array}{l} | A B | = | \begin{array}{c} 5 & 6 & 10 \\ 2 & 3 & 1 \\ 8 & 10 & 7 \end{array} | = 5 (21 - 10) - 6 (14 - 8) + 10 (20 - 24) = 55 - 36 - 40 = - 21. \end{array}

The matrix of cofactors of | AB | is equal to

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

adj AB =

\begin{array}{l} [\begin{array}{c} 11 & 58 & - 24 \\ - 6 & - 45 & 15 \\ - 4 & - 2 & 3 \end{array}] \\ S o, {(A B)}^{- 1} = \frac{a d j A B}{| A B |} = \frac{- 1}{21} [\begin{array}{c} 11 & 58 & - 24 \\ - 6 & - 45 & 15 \\ - 4 & - 2 & 3 \end{array}] \end{array}

Next,

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

Cofactor matrix of B

\begin{array}{l} = [\begin{array}{c} 2 & - 3 & 5 \\ 3 & 6 & - 3 \\ - 13 & 9 & - 1 \end{array}] \end{array}

\begin{array}{l} a d j B = [\begin{array}{c} 2 & 3 & - 13 \\ - 3 & 6 & 9 \\ 5 & - 3 & - 1 \end{array}] \end{array}

∴

\begin{array}{l} B^{- 1} = \frac{a d j B}{| B |} = \frac{1}{21} [\begin{array}{c} 2 & 3 & - 13 \\ - 3 & 6 & 9 \\ 5 & - 3 & - 1 \end{array}]; | A | = [\begin{array}{c} 2 & 1 & - 1 \\ 0 & 1 & 0 \\ 1 & 3 & - 1 \end{array}] = 1 (- 2 + 1) = - 1 \end{array}

Similarly, cofactor matrix of A

\begin{array}{l} = [\begin{array}{c} - 1 & 0 & - 1 \\ - 2 & - 1 & - 5 \\ 1 & 0 & 2 \end{array}] \end{array}

\begin{array}{l} A^{- 1} = \frac{a d j A}{| A |} = \frac{1}{- 1} [\begin{array}{c} - 1 & - 2 & 1 \\ 0 & - 1 & 0 \\ - 1 & - 5 & 2 \end{array}] \end{array}

∴

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

$\begin{array}{l} = - \frac{1}{21} [\begin{array}{c} 11 & 58 & - 24 \\ - 6 & - 45 & 15 \\ - 4 & - 2 & 3 \end{array}] \\ T h u s, {(A B)}^{- 1} = B^{- 1} A^{- 1} \end{array}$

Illustration 4: If

\begin{array}{l} A = [\begin{array}{c} 0 & 2 y & z \\ x & y & - z \\ x & - y & z \end{array}] s a t i s f i e s 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} A A^{'} = I . \end{array}

. Using the multiplication method, we can obtain values of x, y and z.

\begin{array}{l} A^{'} = A^{- 1} \Leftrightarrow A A^{'} = 1 \end{array}

Now,

\begin{array}{l} A A^{'} = [\begin{array}{c} 0 & 2 y & z \\ x & y & - z \\ x & - y & z \end{array}] [\begin{array}{c} 0 & x & x \\ 2 y & y & - y \\ z & - z & z \end{array}] = [\begin{array}{c} 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{array}] \end{array}

Thus,

\begin{array}{l} A A^{'} = 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 = [\begin{array}{c} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \end{array}] a n d A^{- 1} = [\begin{array}{c} 1 / 2 & - 1 / 2 & 1 / 2 \\ - 4 & 3 & y \\ 5 / 2 & - 3 / 2 & 1 / 2 \end{array}], \end{array}

then

(a) x = 1, y = -1

(b) x = -1, y = 1

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

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

∴ x = 1, y = -1

Frequently Asked Questions

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 A_ij = (-1)^i+j det M_ij. Here, det M_ij is the minor of a_ij.

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 the minor of a determinant?

The minor of an element a_ij of a determinant is calculated by deleting its i^th row and j^th column in which the element a_ij lies. We can denote the minor of an element a_ij by M_ij.

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 the negative sign for the elements at a₁₂ and a₂₁ positions. The resulting matrix is the adjoint of the given 2×2 matrix.

Adjoint of a Matrix

How to Find the Adjoint of a Matrix and Its Inverse?

Example Problems on How to Find the Adjoint of a Matrix

Inverse of a Matrix

How to find the inverse of a matrix by using the adjoint matrix?

Properties of Inverse and Adjoint of a Matrix

Problems on Finding the Inverse of a Matrix

Frequently Asked Questions

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

What do you mean by the adjoint of a matrix?

Give the cofactor formula.

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

What do you mean by the minor of a determinant?

What do you mean by a non-singular matrix?

What do you mean by a singular matrix?

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

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