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Let’s look at what are minors & cofactor of a 2 Γ— 2 & a 3 Γ— 3 determinant

Β 

For a 2 Γ— 2 determinant

For

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We have elements,
Β  Β  π‘Ž 11 = 3
Β  Β  π‘Ž 12 = 2
Β  Β  π‘Ž 21 = 1
Β  Β  π‘Ž 22 = 4

Β 

Minor will be

𝑀 11 ,Β  𝑀 12 ,Β  𝑀 21 ,Β  𝑀 22

Minor and Cofactor of a determinant - Part 2

And cofactors will be

𝐴 11 ,  𝐴 12 ,  𝐴 21 ,  𝐴 22

Minor and Cofactor of a determinant - Part 3

Β 

For a 3 Γ— 3 matrix

Minor and Cofactor of a determinant - Part 4

Β 

Minor will be

M 11 , M 12 , M 13 , M 21 , M 22 , M 23 , M 31 , M 32 , M 33

Minor and Cofactor of a determinant - Part 5

Minor and Cofactor of a determinant - Part 6

Note : We can also calculate cofactors without calculating minors

If i + j is odd,

A ij = βˆ’1 Γ— M ij

If i + j is even,

A ij = M ij

But, why use cofactor?

Β 

Minor and Cofactor of a determinant - Part 7

Β 

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Transcript

For a 2 Γ— 2 determinant For βˆ† = |β– 8(3&[email protected]&4)| Minor will be 𝑀_11, 𝑀_12, 𝑀_21, 𝑀_22 M11 = |β– 8(3&[email protected]&4)| = 4 M12 = |β– 8(3&[email protected]&4)|= 1 M21 = |β– 8(3&[email protected]&4)|= 2 M22 = |β– 8(3&[email protected]&4)|= 3 And cofactors will be 𝐴_11, 𝐴_12, 𝐴_21, 𝐴_22 𝐴_11 = γ€–(βˆ’1)γ€—^(1 + 1) 𝑀_11 = γ€–(βˆ’1)γ€—^2 𝑀_11 = 𝑀_11 = 4 𝐴_12 = γ€–(βˆ’1)γ€—^(1 +2) 𝑀_12 = γ€–(βˆ’1)γ€—^3 𝑀_12 = βˆ’1 Γ— 𝑀_12 = βˆ’1 Γ— 1 = βˆ’1 𝐴_21 = γ€–(βˆ’1)γ€—^(2 + 1) 𝑀_21 = γ€–(βˆ’1)γ€—^3 𝑀_21 = βˆ’1 Γ— 𝑀_11 = βˆ’1 Γ— 2 = βˆ’2 𝐴_22 = γ€–(βˆ’1)γ€—^(2 +2) 𝑀_22 = γ€–(βˆ’1)γ€—^4 𝑀_22 = 1 Γ— 𝑀_22 = 3 For a 3 Γ— 3 matrix For βˆ† = |β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| We have elements, π‘Ž_11 = 9 π‘Ž_12 = 2 π‘Ž_13 = 1 π‘Ž_21 = 5 π‘Ž_22 = βˆ’1 π‘Ž_23 = 6 π‘Ž_31 = 1 π‘Ž_32 = 6 π‘Ž_33 = βˆ’2 . Minor will be 𝑀_11, 𝑀_12, 𝑀_13, 𝑀_21, 𝑀_22, 𝑀_23, 𝑀_31, 𝑀_32, 𝑀_33 M11 = |β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| = |β– 8(βˆ’1&[email protected]&βˆ’2)| = (βˆ’1) Γ— (βˆ’2) βˆ’ 0 Γ— 6 = 2 βˆ’ 0 = 2 M12 = |β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| = |β– 8(5&[email protected]&βˆ’2)| = 5 Γ— (βˆ’2) βˆ’ 4 Γ— 6 = βˆ’10 βˆ’ 24 = βˆ’34 M13 = |β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| = |β– 8(5&βˆ’[email protected]&0)| = 5 Γ— 0 βˆ’ 4 Γ— (βˆ’1) = 0 + 4 = 4 M21 = |β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| = |β– 8(2&[email protected]&βˆ’2)| = 2 Γ— (βˆ’2) βˆ’ 0 Γ— 1 = βˆ’4 βˆ’ 0 = βˆ’4 M22 = |β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| = |β– 8(9&[email protected]&βˆ’2)| = 9 Γ— (βˆ’2) βˆ’ 4 Γ— 1 = βˆ’18 βˆ’ 4 = βˆ’ 22 M23 = |β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| = |β– 8(9&[email protected]&0)| = 9 Γ— 0 βˆ’ 4 Γ— 2 = 0 βˆ’ 8 = βˆ’ 8 M31 = |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 M32 = |β– 8(9&2&[email protected]&βˆ’1&[email protected]&0&βˆ’2)| = |β– 8(9&[email protected]&6)| = 9 Γ— 6 βˆ’ 5 Γ— 1 = 54 βˆ’ 5 = 49 M33 = |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 And, cofactors will be 𝐴_11 = γ€–(βˆ’1)γ€—^(1 + 1) 𝑀_11 = 𝑀_11 = 2 𝐴_12 = γ€–(βˆ’1)γ€—^(1 + 2) 𝑀_12 = βˆ’1 Γ— 𝑀_12 = βˆ’1 Γ— βˆ’34 = 34 𝐴_13 = γ€–(βˆ’1)γ€—^(1 + 3) 𝑀_13 = 𝑀_13 = 4 𝐴_21 = γ€–(βˆ’1)γ€—^(2 + 1) 𝑀_21 = βˆ’1 Γ— 𝑀_21 = βˆ’1 Γ— βˆ’4 = 4 𝐴_22 = γ€–(βˆ’1)γ€—^(2 + 2) 𝑀_22 = 𝑀_22 = βˆ’22 𝐴_23 = γ€–(βˆ’1)γ€—^(2 +3) 𝑀_23 = βˆ’1 Γ— 𝑀_23 = βˆ’1 Γ— βˆ’8 = 8 𝐴_31 = γ€–(βˆ’1)γ€—^(3 + 1) 𝑀_31 = 𝑀_31 = 13 𝐴_32 = γ€–(βˆ’1)γ€—^(3 + 2) 𝑀_32 = βˆ’1 Γ— 𝑀_32 = βˆ’1 Γ— 49 = βˆ’49 𝐴_33 = γ€–(βˆ’1)γ€—^(3 +3) 𝑀_33 = 𝑀_33 = βˆ’19 Note : We can also calculate cofactors without calculating minors If i + j is odd, Aij = βˆ’1 Γ— Mij If i + j is even, Aij = Mij But, why use cofactor? Let’s take a general determinant, βˆ† = |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 )| βˆ† = π‘Ž_11 |β– 8(π‘Ž_22&π‘Ž[email protected]π‘Ž_32&π‘Ž_33 )| βˆ’ π‘Ž_12 |β– 8(π‘Ž_21&π‘Ž[email protected]π‘Ž_31&π‘Ž_33 )| βˆ’ π‘Ž_13 |β– 8(π‘Ž_21&π‘Ž[email protected]π‘Ž_31&π‘Ž_32 )| We can also write it as βˆ† = π‘Ž_11 Γ— γ€–(βˆ’1)γ€—^(1+1) |β– 8(π‘Ž_22&π‘Ž[email protected]π‘Ž_32&π‘Ž_33 )| + π‘Ž_12 Γ— γ€–(βˆ’1)γ€—^(1+2) |β– 8(π‘Ž_22&π‘Ž[email protected]π‘Ž_32&π‘Ž_33 )| + π‘Ž_13 Γ— γ€–(βˆ’1)γ€—^(1+3) |β– 8(π‘Ž_22&π‘Ž[email protected]π‘Ž_32&π‘Ž_33 )| βˆ† = π‘Ž_11 𝐴_11 + π‘Ž_12 𝐴_12 + π‘Ž_13 𝐴_13

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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.