Finding Minors and cofactors

Chapter 4 Class 12 Determinants
Concept wise

Letβ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

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

Β

Minor will be

π 11 ,Β  π 12 ,Β  π 21 ,Β  π 22

And cofactors will be

π΄ 11 ,Β  π΄ 12 ,Β  π΄ 21 ,Β  π΄ 22

For a 3 Γ 3 matrix

Β

Minor will be

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

Note : We can also calculate cofactors without calculating minors

If i + j is odd,

A ij = β1 Γ M ij

If i + j is even,

A ij = M ij

But, why use cofactor?

Β

Β

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Transcript

For a 2 Γ 2 determinant For β = |β 8(3&[email protected]&4)| Minor will be π_11, π_12, π_21, π_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 π΄_11, π΄_12, π΄_21, π΄_22 π΄_11 = γ(β1)γ^(1 + 1) π_11 = γ(β1)γ^2 π_11 = π_11 = 4 π΄_12 = γ(β1)γ^(1 +2) π_12 = γ(β1)γ^3 π_12 = β1 Γ π_12 = β1 Γ 1 = β1 π΄_21 = γ(β1)γ^(2 + 1) π_21 = γ(β1)γ^3 π_21 = β1 Γ π_11 = β1 Γ 2 = β2 π΄_22 = γ(β1)γ^(2 +2) π_22 = γ(β1)γ^4 π_22 = 1 Γ π_22 = 3 For a 3 Γ 3 matrix For β = |β 8(9&2&[email protected]&β1&[email protected]&0&β2)| We have elements, π_11 = 9 π_12 = 2 π_13 = 1 π_21 = 5 π_22 = β1 π_23 = 6 π_31 = 1 π_32 = 6 π_33 = β2 . Minor will be π_11, π_12, π_13, π_21, π_22, π_23, π_31, π_32, π_33 M11 = |β 8(9&2&[email protected]&β1&[email protected]&0&β2)| = |β 8(β1&[email protected]&β2)| = (β1) Γ (β2) β 0 Γ 6 = 2 β 0 = 2 M12 = |β 8(9&2&[email protected]&β1&[email protected]&0&β2)| = |β 8(5&[email protected]&β2)| = 5 Γ (β2) β 4 Γ 6 = β10 β 24 = β34 M13 = |β 8(9&2&[email protected]&β1&[email protected]&0&β2)| = |β 8(5&β[email protected]&0)| = 5 Γ 0 β 4 Γ (β1) = 0 + 4 = 4 M21 = |β 8(9&2&[email protected]&β1&[email protected]&0&β2)| = |β 8(2&[email protected]&β2)| = 2 Γ (β2) β 0 Γ 1 = β4 β 0 = β4 M22 = |β 8(9&2&[email protected]&β1&[email protected]&0&β2)| = |β 8(9&[email protected]&β2)| = 9 Γ (β2) β 4 Γ 1 = β18 β 4 = β 22 M23 = |β 8(9&2&[email protected]&β1&[email protected]&0&β2)| = |β 8(9&[email protected]&0)| = 9 Γ 0 β 4 Γ 2 = 0 β 8 = β 8 M31 = |β 8(9&2&[email protected]&β1&[email protected]&0&β2)| = |β 8(2&[email protected]β1&6)| = 2 Γ 6 β (β1) Γ 1 = 12 + 1 = 13 M32 = |β 8(9&2&[email protected]&β1&[email protected]&0&β2)| = |β 8(9&[email protected]&6)| = 9 Γ 6 β 5 Γ 1 = 54 β 5 = 49 M33 = |β 8(9&2&[email protected]&β1&[email protected]&0&β2)| = |β 8(9&[email protected]&β1)| = 9 Γ (β1) β 5 Γ 2 = β9 β 10 = β 19 And, cofactors will be π΄_11 = γ(β1)γ^(1 + 1) π_11 = π_11 = 2 π΄_12 = γ(β1)γ^(1 + 2) π_12 = β1 Γ π_12 = β1 Γ β34 = 34 π΄_13 = γ(β1)γ^(1 + 3) π_13 = π_13 = 4 π΄_21 = γ(β1)γ^(2 + 1) π_21 = β1 Γ π_21 = β1 Γ β4 = 4 π΄_22 = γ(β1)γ^(2 + 2) π_22 = π_22 = β22 π΄_23 = γ(β1)γ^(2 +3) π_23 = β1 Γ π_23 = β1 Γ β8 = 8 π΄_31 = γ(β1)γ^(3 + 1) π_31 = π_31 = 13 π΄_32 = γ(β1)γ^(3 + 2) π_32 = β1 Γ π_32 = β1 Γ 49 = β49 π΄_33 = γ(β1)γ^(3 +3) π_33 = π_33 = β19 Note : We can also calculate cofactors without calculating minors If i + j is odd, Aij = β1 Γ Mij If i + j is even, Aij = Mij But, why use cofactor? Letβs take a general determinant, β = |β 8(π_11&π_12&π[email protected]π_21&π_22&π[email protected]π_31&π_32&π_33 )| β = π_11 |β 8(π_22&π[email protected]π_32&π_33 )| β π_12 |β 8(π_21&π[email protected]π_31&π_33 )| β π_13 |β 8(π_21&π[email protected]π_31&π_32 )| We can also write it as β = π_11 Γ γ(β1)γ^(1+1) |β 8(π_22&π[email protected]π_32&π_33 )| + π_12 Γ γ(β1)γ^(1+2) |β 8(π_22&π[email protected]π_32&π_33 )| + π_13 Γ γ(β1)γ^(1+3) |β 8(π_22&π[email protected]π_32&π_33 )| β = π_11 π΄_11 + π_12 π΄_12 + π_13 π΄_13