


Inverse of a 2 × 2 matrix
Inverse of a 3 × 3 matrix
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Finding Inverse of a matrix
Last updated at May 29, 2023 by Teachoo
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Finding inverse of matrix using adjoint Letβs learn how to find inverse of matrix using adjoint But first, let us define adjoint. For matrix A, A = [β 8(π_11&π_12&π_13@π_21&π_22&π_23@π_31&π_32&π_33 )] Adjoint of A is, adj A = Transpose of [β 8(π΄_11&π΄_12&π΄_13@π΄_21&π΄_22&π΄_23@π΄_31&π΄_32&π΄_33 )] = [β 8(π΄_11&π΄_21&π΄_31@π΄_12&π΄_22&π΄_32@π΄_13&π΄_23&π΄_33 )] Where Aij cofactor Now, we have a property for (adj π΄) π΄ (adj π΄) = (adj π΄) π΄ = |π΄| I If we divide the equation by |π΄| (π΄ (πππ π΄))/(|π΄|) = ((πππ π΄) π΄)/(|π΄|) = (|π΄| I)/(|π΄|) π΄ (1/(|π΄|)Γπππ π΄) = (1/(|π΄|)Γπππ π΄) π΄ = I This is similar to γπ΄π΄γ^(β1) = π΄^(β1)=I Thus, π΄^(β1) = 1/(|π΄|) adj π΄ Also, If |π΄| = 0, the matrix π΄ is called singular matrix If |π΄| β 0, the matrix π΄ is called non-singular matrix Note : π΄^(β1) is only possible if |π΄| β 0 Thus, inverse of A is possible only when matrix is non-singular Letβs 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 itβs inverse First, we will check if |A| β 0 |A| = |β 8(3&2@1&4)| = 3 Γ 4 β 1 Γ 2 = 12 β 2 = 10 Since |A| β 0, Inverse of A is possible Now, letβs 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&β2@β1&3)] Now we know that π΄^(β1) = 1/(|π΄|) adj A = 1/10 [β 8(4&β2@β1&3)] = [β 8(4/10&(β2)/10@(β1)/10&3/10)] = [β 8(2/5&(β1)/5@(β1)/10&3/10)] Letβs check AA-1 = [β 8(3&2@1&4)] [β 8(2/5&(β1)/5@(β1)/10&3/10)] = [β 8(3Γ2/5+2Γ((β1)/10)&3Γ((β1)/5)+2Γ(3/10)@1Γ2/5+4Γ((β1)/10)&1Γ((β1)/5)+4Γ(3/10) )] = [β 8(6/5β2/10&(β3)/5+3/5@2/5β4/10&(β1)/5+12/10)] = [β 8(6/5β1/5&0@2/5β2/5&(β1)/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&β1&6@4&0&β2)] We need to find itβs inverse First, we will check if |A| β 0 |A| = |β 8(9&2&1@5&β1&6@4&0&β2)| = 9 ((β1) Γ (β2) β 0 Γ 6) β 2 (5 Γ (β2) β 4 Γ 6) + 1 (5 Γ 0 β 4 Γ (β1)) = 9 (2 β 0) β 2 (β10 β 24) + 1 (0 + 4) = 9 Γ 2 β 2 Γ (β34) + 1 Γ 4 = 18 + 68 + 4 = 90 Since |A| β 0, Inverse of A is possible Now, letβs find adj A adj A = Transpose of [β 8(π΄_11&π΄_12&π΄_13@π΄_21&π΄_22&π΄_23@π΄_31&π΄_32&π΄_33 )] = [β 8(π΄_11&π΄_21&π΄_31@π΄_12&π΄_22&π΄_32@π΄_13&π΄_23&π΄_33 )] Finding cofactors A11 = |β 8(9&2&1@5&β1&6@4&0&β2)| = (β1) Γ (β2) β 0 Γ 6 = 2 β 0 = 2 A12 = β1Γ|β 8(9&2&1@5&β1&6@4&0&β2)| = β1Γ |β 8(5&6@4&β2)| = β(5 Γ (β2) β 4 Γ 6) = 34 A13 = |β 8(9&2&1@5&β1&6@4&0&β2)| = 5 Γ 0 β 4 Γ (β1) = 0 + 4 = 4 A21 = β1Γ |β 8(9&2&1@5&β1&6@4&0&β2)| = β(2 Γ (β2) β 0 Γ 1) = 4 A22 = |β 8(9&2&1@5&β1&6@4&0&β2)| = (9 Γ (β2) β 4 Γ 1) = β22 A23 = β1Γ|β 8(9&2&1@5&β1&6@4&0&β2)| = β1Γ |β 8(9&2@4&0)| = β(9 Γ 0 β 4 Γ 2) = 8 A31 = |β 8(9&2&1@5&β1&6@4&0&β2)| = |β 8(2&1@β1&6)| = 2 Γ 6 β (β1) Γ 1 = 12 + 1 = 13 A32 = β1Γ|β 8(9&2&1@5&β1&6@4&0&β2)| = β1Γ |β 8(9&1@5&6)| = β(9 Γ 6 β 5 Γ 1) = β49 A33 = |β 8(9&2&1@5&β1&6@4&0&β2)| = |β 8(9&2@5&β1)| = 9 Γ (β1) β 5 Γ 2 = β9 β 10 = β19 Thus, adj A = [β 8(2&4&13@34&β22&β49@4&8&β19)] And, Aβ1 = 1/(|π΄|) adj A = 1/90 [β 8(2&4&13@34&β22&β49@4&8&β19)] = [β 8(2/90&4/90&13/90@34/90&(β22)/90&(β49)/90@4/90&8/90&(β19)/90)] = [β 8(1/45&2/45&13/90@17/45&(β11)/45&(β49)/90@2/45&4/45&(β19)/90)] Letβs check AAβ1 = [β 8(9&2&1@5&β1&6@4&0&β2)] [β 8(1/45&2/45&13/90@17/45&(β11)/45&(β49)/90@2/45&4/45&(β19)/90)] = [β 8(9(1/45)+2(17/45)+1(2/45)&9(2/45)+2((β11)/45)+1(4/45)&9(13/90)+2((β49)/90)+1((β19)/90)@5(1/45)+(β1)(17/45)+6(2/45)&5(2/45)+(β1)((β11)/45)+6(4/45)&5(13/90)+(β1)((β49)/90)+6((β19)/90)@4(1/45)+0(17/45)+(β2)(2/45)&4(2/45)+0((β11)/45)+(β2)(4/45)&9(13/90)+0((β49)/90)+(β2)((β19)/90) )] = [β 8(9/45+34/45+2/45&18/45β22/45+4/45&117/90β98/90β19/90@5/45β17/45+12/45&10/45+11/45+24/45&65/90+49/90β114/90@4/45β4/45&8/45β8/45&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, AAβ1 = I