Let’s learn how to find inverse of matrix using adjoint

But first, let us define adjoint.

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Where A ij cofactor

 

Now, we have a property for (adj 𝐴)

𝐴 (adj 𝐴) = (adj 𝐴) 𝐴 = |𝐴| I

 

If we divide the equation by |𝐴|

𝐴 (π‘Žπ‘‘π‘— 𝐴) / |𝐴| = (π‘Žπ‘‘π‘— 𝐴) 𝐴 / (|𝐴|) = |𝐴| I/ |𝐴|

𝐴 (1/|𝐴| × π‘Žπ‘‘π‘— 𝐴) = (1/|𝐴|×π‘Žπ‘‘π‘— 𝐴) 𝐴 = I

 

This is similar to
𝐴𝐴 −1 = 𝐴 −1 =I

Thus,

𝐴 −1 = 1/|𝐴| adj 𝐴

Note : A -1  is only possible if |A| ≠   0

Also,

       If |𝐴| = 0,

       the matrix 𝐴 is called singular matrix

       If |𝐴| ≠ 0,

       the matrix 𝐴 is called non-singular matrix

 

 

Thus,

inverse of A is possible only when matrix is non-singular

Let’s look at how to find inverse of a 2 × 2 & a 3 × 3 matrix using adjoint

 

Inverse of a 2 × 2 matrix

Let

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We need to find it’s inverse

First, we will check if |A| ≠ 0

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Since |A| ≠ 0,

Inverse of A is possible

 

Now, let’s find adj A

 

We have a shortcut method to find adjoint of a 2 × 2 matrix

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Now we know that

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Inverse of a 3 × 3 matrix

Let

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We need to find it’s inverse

First, we will check if |A| ≠ 0

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      = 9 ((−1) × (−2) − 0 × 6) − 2 (5 × (−2) − 4 × 6) + 1 (5 × 0 − 4 × (−1))
      = 9 (2 − 0) − 2 (−10 − 24) + 1 (0 + 4)
      = 9 × 2 − 2 × (−34) + 1 × 4

      = 18 + 68 + 4
      = 90

 

Since |A| ≠ 0,
Inverse of A is possible

 

Now, let’s find adj A

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Finding cofactors

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  1. Chapter 4 Class 12 Determinants
  2. Concept wise

Transcript

For matrix A, A = [β– 8(π‘Ž_11&π‘Ž_12&π‘Ž_13@π‘Ž_21&π‘Ž_22&π‘Ž_23@π‘Ž_31&π‘Ž_32&π‘Ž_33 )] Adjoint of A is, adj A = Transpose of [β– 8(𝐴_11&𝐴_12&𝐴_13@𝐴_21&𝐴_22&𝐴_23@𝐴_31&𝐴_32&𝐴_33 )] = [β– 8(𝐴_11&𝐴_21&𝐴_31@𝐴_12&𝐴_22&𝐴_32@𝐴_13&𝐴_23&𝐴_33 )] Where Aij cofactor Now, we have a property for (adj 𝐴) 𝐴 (adj 𝐴) = (adj 𝐴) 𝐴 = |𝐴| I If we divide the equation by |𝐴| (𝐴 (π‘Žπ‘‘π‘— 𝐴))/(|𝐴|) = ((π‘Žπ‘‘π‘— 𝐴) 𝐴)/(|𝐴|) = (|𝐴| I)/(|𝐴|) 𝐴 (1/(|𝐴|)Γ—π‘Žπ‘‘π‘— 𝐴) = (1/(|𝐴|)Γ—π‘Žπ‘‘π‘— 𝐴) 𝐴 = I A = [β– 8(3&2@1&4)] A = [β– 8(3&2@1&4)] |A| = |β– 8(3&2@1&4)| = 3 Γ— 4 βˆ’ 1 Γ— 2 = 12 βˆ’ 2 = 10 adj A = [β– 8(3&2@1&4)] Let’s check AA-1 = [β– 8(3&2@1&4)] [β– 8(2/5&(βˆ’1)/5@(βˆ’1)/10&3/10)] = [β– 8(3Γ—2/5+2Γ—((βˆ’1)/10)&3Γ—((βˆ’1)/5)+2Γ—(3/10)@1Γ—2/5+4Γ—((βˆ’1)/10)&1Γ—((βˆ’1)/5)+4Γ—(3/10) )] = [β– 8(6/5βˆ’2/10&(βˆ’3)/5+3/5@2/5βˆ’4/10&(βˆ’1)/5+12/10)] = [β– 8(6/5βˆ’1/5&0@2/5βˆ’2/5&(βˆ’1)/5+6/5)] = [β– 8(5/5&0@0&5/5)] = [β– 8(1&0@0&1)] = [β– 8(5/5&0@0&5/5)] = [β– 8(1&0@0&1)] Inverse of a 3 Γ— 3 matrix Let A = [β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)] We need to find it’s inverse First, we will check if |A| β‰  0 |A| = |β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)| = 9 ((βˆ’1) Γ— (βˆ’2) βˆ’ 0 Γ— 6) βˆ’ 2 (5 Γ— (βˆ’2) βˆ’ 4 Γ— 6) + 1 (5 Γ— 0 βˆ’ 4 Γ— (βˆ’1)) = 9 (2 βˆ’ 0) βˆ’ 2 (βˆ’10 βˆ’ 24) + 1 (0 + 4) = 9 Γ— 2 βˆ’ 2 Γ— (βˆ’34) + 1 Γ— 4 = 18 + 68 + 4 = 90 Since |A| β‰  0, Inverse of A is possible Now, let’s find adj A adj A = Transpose of [β– 8(𝐴_11&𝐴_12&𝐴_13@𝐴_21&𝐴_22&𝐴_23@𝐴_31&𝐴_32&𝐴_33 )] = [β– 8(𝐴_11&𝐴_21&𝐴_31@𝐴_12&𝐴_22&𝐴_32@𝐴_13&𝐴_23&𝐴_33 )] A11 = |β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)| = |β– 8(βˆ’1&6@0&βˆ’2)| = (βˆ’1) Γ— (βˆ’2) βˆ’ 0 Γ— 6 = 2 βˆ’ 0 = 2 A12 = βˆ’1Γ—|β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)| = βˆ’1Γ— |β– 8(5&6@4&βˆ’2)| = βˆ’(5 Γ— (βˆ’2) βˆ’ 4 Γ— 6) = 34 A13 = |β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)| = |β– 8(5&βˆ’1@4&0)| = 5 Γ— 0 βˆ’ 4 Γ— (βˆ’1) = 0 + 4 = 4 A21 = βˆ’1Γ— |β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)| = βˆ’1Γ— |β– 8(2&1@0&βˆ’2)| = βˆ’(2 Γ— (βˆ’2) βˆ’ 0 Γ— 1) = 4 A22 = |β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)| A22 = |β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)| = (9 Γ— (βˆ’2) βˆ’ 4 Γ— 1) = βˆ’22 A23 = βˆ’1Γ—|β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)| = βˆ’1Γ— |β– 8(9&2@4&0)| = βˆ’(9 Γ— 0 βˆ’ 4 Γ— 2) = 8 A31 = |β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)| = |β– 8(2&1@βˆ’1&6)| = 2 Γ— 6 βˆ’ (βˆ’1) Γ— 1 = 12 + 1 = 13 A32 = βˆ’1Γ—|β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)| = βˆ’1Γ— |β– 8(9&1@5&6)| = βˆ’(9 Γ— 6 βˆ’ 5 Γ— 1) = βˆ’49 A33 = |β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)| = |β– 8(9&2@5&βˆ’1)| = 9 Γ— (βˆ’1) βˆ’ 5 Γ— 2 = βˆ’9 βˆ’ 10 = βˆ’19 Thus, adj A = [β– 8(2&4&13@34&βˆ’22&βˆ’49@4&8&βˆ’19)] And, Aβˆ’1 = 1/(|𝐴|) adj A = 1/90 [β– 8(2&4&13@34&βˆ’22&βˆ’49@4&8&βˆ’19)] = [β– 8(2/90&4/90&13/90@34/90&(βˆ’22)/90&(βˆ’49)/90@4/90&8/90&(βˆ’19)/90)] = [β– 8(1/45&2/45&13/90@17/45&(βˆ’11)/45&(βˆ’49)/90@2/45&4/45&(βˆ’19)/90)] Let’s check AAβˆ’1 = [β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)] [β– 8(1/45&2/45&13/90@17/45&(βˆ’11)/45&(βˆ’49)/90@2/45&4/45&(βˆ’19)/90)] = [β– 8(9(1/45)+2(17/45)+1(2/45)&9(2/45)+2((βˆ’11)/45)+1(4/45)&9(13/90)+2((βˆ’49)/90)+1((βˆ’19)/90)@5(1/45)+(βˆ’1)(17/45)+6(2/45)&5(2/45)+(βˆ’1)((βˆ’11)/45)+6(4/45)&5(13/90)+(βˆ’1)((βˆ’49)/90)+6((βˆ’19)/90)@4(1/45)+0(17/45)+(βˆ’2)(2/45)&4(2/45)+0((βˆ’11)/45)+(βˆ’2)(4/45)&9(13/90)+0((βˆ’49)/90)+(βˆ’2)((βˆ’19)/90) )] = [β– 8(9/45+34/45+2/45&18/45βˆ’22/45+4/45&117/90βˆ’98/90βˆ’19/90@5/45βˆ’17/45+12/45&10/45+11/45+24/45&65/90+49/90βˆ’114/90@4/45βˆ’4/45&8/45βˆ’8/45&52/90+38/90)] = [β– 8(45/45&0&0@0&45/45&0@0&0&90/90)] = [β– 8(1&0&0@0&1&0@0&0&1)] = I Thus, AAβˆ’1 = I

About the Author

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.