`star` Adjoint of Matrix

`star` Inverse of a Matrix

`star` 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) ] }`

` => 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

Class 12 Chapter 4 Example 23

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) ]`

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}`

`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 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

`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

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`

`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`

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 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)` .

`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}`

`"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

Class 12 Chapter 4 Example 24

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

Class 12 Chapter 4 Example 25

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

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

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) ]`