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Finding Inverse of Matrix using adjoint - Both 2x2 and 3x3 - Teachoo

Finding inverse of matrix using adjoint - Part 2
Finding inverse of matrix using adjoint - Part 3 Finding inverse of matrix using adjoint - Part 4

Inverse of a 2 Γ— 2 matrix

Finding inverse of matrix using adjoint - Part 5 Finding inverse of matrix using adjoint - Part 6 Finding inverse of matrix using adjoint - Part 7 Finding inverse of matrix using adjoint - Part 8 Finding inverse of matrix using adjoint - Part 9

Inverse of a 3 Γ— 3 matrix

Finding inverse of matrix using adjoint - Part 10 Finding inverse of matrix using adjoint - Part 11 Finding inverse of matrix using adjoint - Part 12 Finding inverse of matrix using adjoint - Part 13 Finding inverse of matrix using adjoint - Part 14 Finding inverse of matrix using adjoint - Part 15 Finding inverse of matrix using adjoint - Part 16

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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[email protected]π‘Ž_21&π‘Ž_22&π‘Ž[email protected]π‘Ž_31&π‘Ž_32&π‘Ž_33 )] Adjoint of A is, adj A = Transpose of [β– 8(𝐴_11&𝐴_12&𝐴[email protected]𝐴_21&𝐴_22&𝐴[email protected]𝐴_31&𝐴_32&𝐴_33 )] = [β– 8(𝐴_11&𝐴_21&𝐴[email protected]𝐴_12&𝐴_22&𝐴[email protected]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&[email protected]&4)] We need to find it’s inverse First, we will check if |A| β‰  0 |A| = |β– 8(3&[email protected]&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&[email protected]&4)] We have a shortcut method to find adjoint of a 2 Γ— 2 matrix adj A = [β– 8(3&[email protected]&4)] = [β– 8(4&βˆ’[email protected]βˆ’1&3)] Now we know that 𝐴^(βˆ’1) = 1/(|𝐴|) adj A = 1/10 [β– 8(4&βˆ’[email protected]βˆ’1&3)] = [β– 8(4/10&(βˆ’2)/[email protected](βˆ’1)/10&3/10)] = [β– 8(2/5&(βˆ’1)/[email protected](βˆ’1)/10&3/10)] Let’s check AA-1 = [β– 8(3&[email protected]&4)] [β– 8(2/5&(βˆ’1)/[email protected](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/[email protected]/5βˆ’4/10&(βˆ’1)/5+12/10)] = [β– 8(6/5βˆ’1/5&[email protected]/5βˆ’2/5&(βˆ’1)/5+6/5)] = [β– 8(5/5&[email protected]&5/5)] = [β– 8(1&[email protected]&1)] Thus, AA-1 = I Inverse of a 3 Γ— 3 matrix Let A = [β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)] We need to find it’s inverse First, we will check if |A| β‰  0 |A| = |β– 8(9&2&[email protected]&βˆ’1&[email protected]&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[email protected]𝐴_21&𝐴_22&𝐴[email protected]𝐴_31&𝐴_32&𝐴_33 )] = [β– 8(𝐴_11&𝐴_21&𝐴[email protected]𝐴_12&𝐴_22&𝐴[email protected]𝐴_13&𝐴_23&𝐴_33 )] Finding cofactors A11 = |β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| = (βˆ’1) Γ— (βˆ’2) βˆ’ 0 Γ— 6 = 2 βˆ’ 0 = 2 A12 = βˆ’1Γ—|β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| = βˆ’1Γ— |β– 8(5&[email protected]&βˆ’2)| = βˆ’(5 Γ— (βˆ’2) βˆ’ 4 Γ— 6) = 34 A13 = |β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| = 5 Γ— 0 βˆ’ 4 Γ— (βˆ’1) = 0 + 4 = 4 A21 = βˆ’1Γ— |β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| = βˆ’(2 Γ— (βˆ’2) βˆ’ 0 Γ— 1) = 4 A22 = |β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| = (9 Γ— (βˆ’2) βˆ’ 4 Γ— 1) = βˆ’22 A23 = βˆ’1Γ—|β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| = βˆ’1Γ— |β– 8(9&[email protected]&0)| = βˆ’(9 Γ— 0 βˆ’ 4 Γ— 2) = 8 A31 = |β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| = |β– 8(2&[email protected]βˆ’1&6)| = 2 Γ— 6 βˆ’ (βˆ’1) Γ— 1 = 12 + 1 = 13 A32 = βˆ’1Γ—|β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| = βˆ’1Γ— |β– 8(9&[email protected]&6)| = βˆ’(9 Γ— 6 βˆ’ 5 Γ— 1) = βˆ’49 A33 = |β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| = |β– 8(9&[email protected]&βˆ’1)| = 9 Γ— (βˆ’1) βˆ’ 5 Γ— 2 = βˆ’9 βˆ’ 10 = βˆ’19 Thus, adj A = [β– 8(2&4&[email protected]&βˆ’22&βˆ’[email protected]&8&βˆ’19)] And, Aβˆ’1 = 1/(|𝐴|) adj A = 1/90 [β– 8(2&4&[email protected]&βˆ’22&βˆ’[email protected]&8&βˆ’19)] = [β– 8(2/90&4/90&13/[email protected]/90&(βˆ’22)/90&(βˆ’49)/[email protected]/90&8/90&(βˆ’19)/90)] = [β– 8(1/45&2/45&13/[email protected]/45&(βˆ’11)/45&(βˆ’49)/[email protected]/45&4/45&(βˆ’19)/90)] Let’s check AAβˆ’1 = [β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)] [β– 8(1/45&2/45&13/[email protected]/45&(βˆ’11)/45&(βˆ’49)/[email protected]/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/[email protected]/45βˆ’17/45+12/45&10/45+11/45+24/45&65/90+49/90βˆ’114/[email protected]/45βˆ’4/45&8/45βˆ’8/45&52/90+38/90)] = [β– 8(45/45&0&[email protected]&45/45&[email protected]&0&90/90)] = [β– 8(1&0&[email protected]&1&[email protected]&0&1)] = I Thus, AAβˆ’1 = I

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.