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Inverse of a 2 × 2 matrix

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Inverse of a 3 × 3 matrix

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  1. Chapter 4 Class 12 Determinants
  2. Concept wise

Transcript

Finding inverse of matrix using adjoint Let’s learn how to find inverse of matrix using adjoint But first, let us define adjoint. 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 This is similar to 〖𝐴𝐴〗^(βˆ’1) = 𝐴^(βˆ’1)=I Thus, 𝐴^(βˆ’1) = 1/(|𝐴|) adj 𝐴 Also, If |𝐴| = 0, the matrix 𝐴 is called singular matrix If |𝐴| β‰  0, the matrix 𝐴 is called non-singular matrix Note : 𝐴^(βˆ’1) is only possible if |𝐴| β‰  0 Thus, inverse of A is possible only when matrix is non-singular Let’s look at how to find inverse of a 2 Γ— 2 & a 3 Γ— 3 matrix using adjoint Inverse of a 2 Γ— 2 matrix Let A = [β– 8(3&2@1&4)] We need to find it’s inverse First, we will check if |A| β‰  0 |A| = |β– 8(3&2@1&4)| = 3 Γ— 4 βˆ’ 1 Γ— 2 = 12 βˆ’ 2 = 10 Since |A| β‰  0, Inverse of A is possible Now, let’s find adj A adj A = [β– 8(3&2@1&4)] We have a shortcut method to find adjoint of a 2 Γ— 2 matrix adj A = [β– 8(3&2@1&4)] = [β– 8(4&βˆ’2@βˆ’1&3)] Now we know that 𝐴^(βˆ’1) = 1/(|𝐴|) adj A = 1/10 [β– 8(4&βˆ’2@βˆ’1&3)] = [β– 8(4/10&(βˆ’2)/10@(βˆ’1)/10&3/10)] = [β– 8(2/5&(βˆ’1)/5@(βˆ’1)/10&3/10)] 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)] Thus, AA-1 = I 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 )] Finding cofactors A11 = |β– 8(9&2&1@5&βˆ’1&6@4&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)| = 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’(2 Γ— (βˆ’2) βˆ’ 0 Γ— 1) = 4 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

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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.