Mathematics Adjoint of Matrix , Inverse of a Matrix For CBSE-NCERT
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star Inverse of a Matrix

### Adjoint and Inverse of a Matrix

● To find inverse of a matrix A, i.e., A^(–1) we shall first define adjoint of a matrix.

\color{green} ✍️ The adjoint of a square matrix A = [a_(ij ) ]_(n × n) is defined as the transpose of the matrix [A_(ij) ]_(n × n),

 => color {red} {A_(ij) " is the cofactor of the element" a_(ij)} . Adjoint of the matrix A is denoted by adj A.

Let A= [ (a_11 , a_12, a_13), ( a_21 , a_22 , a_23 ), ( a_31,a_32, a_33) ]

Then color{orange}{"adj A =Transpose of" [ (A_11 , A_12, A_13), ( A_21, A_22, A_23), ( A_31, A_32, A_33) ]}

color{orange}{ = [ (A_11 , A_21, A_31), ( A_12, A_22, A_32), ( A_13, A_23, A_33) ] }

Q 3154691554

Find adj A for A = [ ( 2,3), (1,4 ) ]
Class 12 Chapter 4 Example 23
Solution:

We have A_11 = 4, A_12 = –1, A_21 = –3, A_22 = 2

Hence adj A= [ (A_11 , A_21), ( A_12, A_22) ] = [ (4,-3), ( -1 ,2) ]

### Remember :

For a square matrix of order 2, given by

A = [ (a_11, a_12 ), ( a_21, a_22) ]

Adj A = [(a_22,-a_12),(-a_21,a_11)]

color{orange}{"The adj A can also be obtained by interchanging" a_11 "and" a_22 "and by changing signs of" a_12 "and" a_21}

### Singular And Non-singular Matrix

"Singular Matrix :"

color{red}{"A square matrix A is said to be singular if" | A | = 0.}

=> For example, the determinant of matrix A = [ (1,2), (4,8) ] is zero

Hence A is a singular matrix.

"NonSingular Matrix : "

color{red}{"A square matrix A is said to be non-singular if "| A | ≠ 0}

Let A = [ (1,2), (3,4) ]. Then  |A | = | (1,2), (3,4) | = 4 -6 = -2 ≠ 0.

Hence A is a nonsingular matrix

### Theorem 1 : color{red}{A(adj A) = (adj A) A = | A | I}

If A be any given square matrix of order n, then

color{red}{A(adj A) = (adj A) A = | A | I}

where I is the identity matrix of order n

"Proof :"

Let A = [ (a_11 , a_12, a_13), ( a_21, a_22, a_23), ( a_31, a_32, a_33) ]  , then adj A = [ (A_11 , A_21, A_31), ( A_12, A_22, A_32), ( A_13, A_23, A_33) ]

=> "As we know Since sum of product of elements of a row (or a column)"
" with corresponding cofactors is equal to |A| and otherwise zero, we have"

A (adj A) = [ ( |A| , 0,0 ), ( 0, |A| , 0 ), ( 0,0, |A|) ] = | A | [ (1,0,0 ), ( 0,1,0 ), ( 0,0,1) ] = |A| I

Similarly, we can show color{orange}{(adj A) A = |A | I}

Hence color{orange}{A (adj A) = (adj A) A = | A | I

### Theorems

Therem 1 : color{red}{"If A and B are nonsingular matrices of the same order, then AB and BA are also nonsingular matrices of the same order. "}

Theorem 2 :color{red}{" The determinant of the product of matrices is equal to product of their respective determinants,"}
color{red}{" that is," | AB | = | A | | B | , "where A and B are square matrices of the same order"}

Theorem 3 : color{red}{"If A is a square matrix of order n, then" |adj(A)| = |A|^(n – 1)}.

"Proof :" We know that (adj A) A = |A| I = [ ( |A| , 0, 0 ), ( 0, |A| , 0 ), ( 0 ,0 , |A| ) ]

Writing determinants of matrices on both sides, we have

| (adj A) A | = | ( |A| , 0,0 ), (0 , |A| , 0 ), ( 0 ,0 , |A|) |

i.e.,  | (adj A) | |A | = | A^3 | | (1,0,0 ),(0,1,0 ),(0,0,1) |

i.e. |(adj A)| |A| = |A|^3 (1)

i.e. color{orange}{|(adj A)| = |A|^2}

In general, if A is a square matrix of order n, then |adj(A)| = |A|^(n – 1) .

### Theorem : Inverse Of Matrix : A^(–1) = 1/(|A|) adj A

color{red}{"A square matrix A is invertible if and only if A is nonsingular matrix. "}

"Proof :" Let A be invertible matrix of order n and I be the identity matrix of order n.
Then, there exists a square matrix B of order n such that AB = BA = I

=> Now AB = I. So | AB | = I or |A| | B | = 1 (since | I | =1,| AB| = |A| | B| )

=> This gives | A | ≠ 0. Hence A is nonsingular.

● Conversely, let A be nonsingular. Then | A | ≠ 0

=> Now A (adj A) = (adj A) A = | A | I (Theorem 1)

=> A (1/(|A| ) adj A) =(1/(|A|) adj A ) A= I

=> AB = BA =I, where B= 1/(|A|) adj A

Thus A is invertible and color{green}{A^(–1) = 1/(|A|) adj A}

So color{blue}{A^(-1) "exist iff" |A| ne 0}
Q 3174691556

If A = [ (1,3,3), (1,4,3), (1,3,4) ] , then verify that A adj A = |A| I. Also find A^(-1) .
Class 12 Chapter 4 Example 24
Solution:

We have A = 1 (16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1 ≠ 0

Now A_11 = 7, A_12 = –1, A_13 = –1, A_21 = –3, A_22 = 1,A_23 = 0, A_31 = –3, A_32 = 0,
A_33 = 1

Therefore  adj A = [ ( 7 , -3 , -3), ( -1,1 , 0 ), ( -1, 0 ,1 ) ]

Now  A (adj A) = [ (1,3,3), ( 1,4,3), ( 1,3,4) ] [ ( 7 , -3 , -3), ( -1,1, 0 ), (-1, 0 ,1) ]

= [ ( 7-3-3, -3 +3 + 0 , -3 +0 + 3 ), ( 7 -4-3 , -3 +4 +0 , -3 +0 +3), ( 7 -3-4, -3+3 +0 , -3+0 +4) ]

= [ (1,0,0 ), ( 0,1,0 ), ( 0 ,0, 1) ] = (1) [ (1, 0,0 ), ( 0,1,0 ), ( 0 ,0 , 1) ] = |A| *I

Also  |A|^(-1) =1/( |A|) adj A = 1/1 [ (7 ,-3,-3), ( -1, 1 , 0 ), ( -1, 0 ,1 ) ] = [ ( 7 , -3 , -3 ), ( -1,1, 0 ), ( -1 , 0, 1) ]
Q 3134191952

If A = [ ( 2,3 ), ( 1,-4) ] and B = [ (1,-2), ( -1,3) ], then verify that (AB)^(–1 ) = B^(–1) A^(–1).
Class 12 Chapter 4 Example 25
Solution:

We have AB = [ ( 2,3), (1,-4) ] [ (1,-2), ( -1,3) ] = [ (-1,5), (5,-14) ]

Since |AB | = -11 ≠ 0, (AB)^(–1) exists and is given by

(AB)^(-1) = 1/( |AB |) adj (AB) = -1/11 [ (-14, -5), ( -5 , -1) ] = 1/11 [ (14,5), (5,1) ]

Further ,  | A | = -11 ≠ 0 and B = 1 ≠ 0. Therefore, A^(–1) and B^(–1 ) both exist and are given by

A^(-1) = -1/11 [ (-4,-3), ( -1,2) ] , B^(-1) = [ ( 3,2), ( 1,1) ]

Therefore B^(-1) A^(-1)= -1/11 [ (3,2), ( 1,1) ] [ (-4,-3), ( -1,2) ] = -1/11 [ (-14, -5), ( -5,-1) ] = 1/11 [ (14,5), ( 5,1) ]

Hence  (AB)^(-1) = B^(-1) A^(-1)
Q 3104191958

Show that the matrix A = [ ( 2,3), (1,2) ] satisfies the equation A^2 – 4A + I = O,
where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix. Using this equation, find A^(–1) .
Class 12 Chapter 4 Example 26
Solution:

We have A^2 = A*A = [ ( 2,3), (1,2) ] [ (2,3), (1,2) ] = [ (7,12), (4,7) ]

Hence A^2 - 4A + I = [ (7 ,12), (4,7) ] - [ (8,12), (4,8) ] + [ (1,0 ), (0,1) ] = [ (0,0 ), (0,0) ] = O

Now A^2 – 4A + I = O

Therefore A A – 4A = – I

or A A (A^(–1) ) – 4 A A^(–1) = – I A^(–1) (Post multiplying by A^(–1 ) because |A| ≠ 0)
or A (A A^(–1) ) – 4I = – A^(–1)

or AI – 4I = – A^(–1)

or A^(–1) = 4I – A = [ (4,0), (0,4) ] - [ ( 2,3), ( 1,2) ] = [ (2,-3), (-1,2) ]

Hence A^(-1) = [ ( 2,-3), ( -1,2) ]`