ਦਿੱਤੇ ਗਏ ਡਿਟਰਮੀਨੈਂਟ ਦੇ ਹਰੇਕ ਐਲੀਮੈਂਟ ਦੇ ਮਾਈਨਰ (Minor) ਅਤੇ ਕੋਫੈਕਟਰ (Cofactor) ਲੱਭਣ ਲਈ, ਅਸੀਂ ਹੇਠਾਂ ਦਿੱਤੇ ਕਦਮਾਂ ਦੀ ਪਾਲਣਾ ਕਰਾਂਗੇ।
ਦਿੱਤਾ ਗਿਆ ਡਿਟਰਮੀਨੈਂਟ ਹੈ:
∣ 1 0 4 3 5 − 1 0 1 2 ∣ \begin{vmatrix}
1 & 0 & 4 \\
3 & 5 & -1 \\
0 & 1 & 2
\end{vmatrix} 1 3 0 0 5 1 4 − 1 2
ਇੱਕ ਐਲੀਮੈਂਟ a i j a_{ij} a ij ਦਾ ਮਾਈਨਰ M i j M_{ij} M ij ਉਸ ਸਬ-ਡਿਟਰਮੀਨੈਂਟ ਦਾ ਮੁੱਲ ਹੁੰਦਾ ਹੈ ਜੋ i i i -ਵੀਂ ਕਤਾਰ ਅਤੇ j j j -ਵੇਂ ਕਾਲਮ ਨੂੰ ਹਟਾਉਣ ਨਾਲ ਪ੍ਰਾਪਤ ਹੁੰਦਾ ਹੈ।
ਇੱਕ ਐਲੀਮੈਂਟ a i j a_{ij} a ij ਦਾ ਕੋਫੈਕਟਰ C i j C_{ij} C ij ਫਾਰਮੂਲੇ C i j = ( − 1 ) i + j M i j C_{ij} = (-1)^{i+j} M_{ij} C ij = ( − 1 ) i + j M ij ਦੁਆਰਾ ਦਿੱਤਾ ਜਾਂਦਾ ਹੈ।
Step 1: ਹਰੇਕ ਐਲੀਮੈਂਟ ਦੇ ਮਾਈਨਰ (Minors) ਲੱਭੋ।
ਐਲੀਮੈਂਟ a 11 = 1 a_{11} = 1 a 11 = 1 ਲਈ:
M 11 = ∣ 5 − 1 1 2 ∣ = ( 5 ) ( 2 ) − ( − 1 ) ( 1 ) = 10 − ( − 1 ) = 11 M_{11} = \begin{vmatrix} 5 & -1 \\ 1 & 2 \end{vmatrix} = (5)(2) - (-1)(1) = 10 - (-1) = 11 M 11 = 5 1 − 1 2 = ( 5 ) ( 2 ) − ( − 1 ) ( 1 ) = 10 − ( − 1 ) = 11
ਐਲੀਮੈਂਟ a 12 = 0 a_{12} = 0 a 12 = 0 ਲਈ:
M 12 = ∣ 3 − 1 0 2 ∣ = ( 3 ) ( 2 ) − ( − 1 ) ( 0 ) = 6 − 0 = 6 M_{12} = \begin{vmatrix} 3 & -1 \\ 0 & 2 \end{vmatrix} = (3)(2) - (-1)(0) = 6 - 0 = 6 M 12 = 3 0 − 1 2 = ( 3 ) ( 2 ) − ( − 1 ) ( 0 ) = 6 − 0 = 6
ਐਲੀਮੈਂਟ a 13 = 4 a_{13} = 4 a 13 = 4 ਲਈ:
M 13 = ∣ 3 5 0 1 ∣ = ( 3 ) ( 1 ) − ( 5 ) ( 0 ) = 3 − 0 = 3 M_{13} = \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = (3)(1) - (5)(0) = 3 - 0 = 3 M 13 = 3 0 5 1 = ( 3 ) ( 1 ) − ( 5 ) ( 0 ) = 3 − 0 = 3
ਐਲੀਮੈਂਟ a 21 = 3 a_{21} = 3 a 21 = 3 ਲਈ:
M 21 = ∣ 0 4 1 2 ∣ = ( 0 ) ( 2 ) − ( 4 ) ( 1 ) = 0 − 4 = − 4 M_{21} = \begin{vmatrix} 0 & 4 \\ 1 & 2 \end{vmatrix} = (0)(2) - (4)(1) = 0 - 4 = -4 M 21 = 0 1 4 2 = ( 0 ) ( 2 ) − ( 4 ) ( 1 ) = 0 − 4 = − 4
ਐਲੀਮੈਂਟ a 22 = 5 a_{22} = 5 a 22 = 5 ਲਈ:
M 22 = ∣ 1 4 0 2 ∣ = ( 1 ) ( 2 ) − ( 4 ) ( 0 ) = 2 − 0 = 2 M_{22} = \begin{vmatrix} 1 & 4 \\ 0 & 2 \end{vmatrix} = (1)(2) - (4)(0) = 2 - 0 = 2 M 22 = 1 0 4 2 = ( 1 ) ( 2 ) − ( 4 ) ( 0 ) = 2 − 0 = 2
ਐਲੀਮੈਂਟ a 23 = − 1 a_{23} = -1 a 23 = − 1 ਲਈ:
M 23 = ∣ 1 0 0 1 ∣ = ( 1 ) ( 1 ) − ( 0 ) ( 0 ) = 1 − 0 = 1 M_{23} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0) = 1 - 0 = 1 M 23 = 1 0 0 1 = ( 1 ) ( 1 ) − ( 0 ) ( 0 ) = 1 − 0 = 1
ਐਲੀਮੈਂਟ a 31 = 0 a_{31} = 0 a 31 = 0 ਲਈ:
M 31 = ∣ 0 4 5 − 1 ∣ = ( 0 ) ( − 1 ) − ( 4 ) ( 5 ) = 0 − 20 = − 20 M_{31} = \begin{vmatrix} 0 & 4 \\ 5 & -1 \end{vmatrix} = (0)(-1) - (4)(5) = 0 - 20 = -20 M 31 = 0 5 4 − 1 = ( 0 ) ( − 1 ) − ( 4 ) ( 5 ) = 0 − 20 = − 20
ਐਲੀਮੈਂਟ a 32 = 1 a_{32} = 1 a 32 = 1 ਲਈ:
M 32 = ∣ 1 4 3 − 1 ∣ = ( 1 ) ( − 1 ) − ( 4 ) ( 3 ) = − 1 − 12 = − 13 M_{32} = \begin{vmatrix} 1 & 4 \\ 3 & -1 \end{vmatrix} = (1)(-1) - (4)(3) = -1 - 12 = -13 M 32 = 1 3 4 − 1 = ( 1 ) ( − 1 ) − ( 4 ) ( 3 ) = − 1 − 12 = − 13
ਐਲੀਮੈਂਟ a 33 = 2 a_{33} = 2 a 33 = 2 ਲਈ:
M 33 = ∣ 1 0 3 5 ∣ = ( 1 ) ( 5 ) − ( 0 ) ( 3 ) = 5 − 0 = 5 M_{33} = \begin{vmatrix} 1 & 0 \\ 3 & 5 \end{vmatrix} = (1)(5) - (0)(3) = 5 - 0 = 5 M 33 = 1 3 0 5 = ( 1 ) ( 5 ) − ( 0 ) ( 3 ) = 5 − 0 = 5
Step 2: ਹਰੇਕ ਐਲੀਮੈਂਟ ਦੇ ਕੋਫੈਕਟਰ (Cofactors) ਲੱਭੋ।
ਐਲੀਮੈਂਟ a 11 = 1 a_{11} = 1 a 11 = 1 ਲਈ:
C 11 = ( − 1 ) 1 + 1 M 11 = ( − 1 ) 2 ( 11 ) = 1 × 11 = 11 C_{11} = (-1)^{1+1} M_{11} = (-1)^2 (11) = 1 \times 11 = 11 C 11 = ( − 1 ) 1 + 1 M 11 = ( − 1 ) 2 ( 11 ) = 1 × 11 = 11
ਐਲੀਮੈਂਟ a 12 = 0 a_{12} = 0 a 12 = 0 ਲਈ:
C 12 = ( − 1 ) 1 + 2 M 12 = ( − 1 ) 3 ( 6 ) = − 1 × 6 = − 6 C_{12} = (-1)^{1+2} M_{12} = (-1)^3 (6) = -1 \times 6 = -6 C 12 = ( − 1 ) 1 + 2 M 12 = ( − 1 ) 3 ( 6 ) = − 1 × 6 = − 6
ਐਲੀਮੈਂਟ a 13 = 4 a_{13} = 4 a 13 = 4 ਲਈ:
C 13 = ( − 1 ) 1 + 3 M 13 = ( − 1 ) 4 ( 3 ) = 1 × 3 = 3 C_{13} = (-1)^{1+3} M_{13} = (-1)^4 (3) = 1 \times 3 = 3 C 13 = ( − 1 ) 1 + 3 M 13 = ( − 1 ) 4 ( 3 ) = 1 × 3 = 3
ਐਲੀਮੈਂਟ a 21 = 3 a_{21} = 3 a 21 = 3 ਲਈ:
C 21 = ( − 1 ) 2 + 1 M 21 = ( − 1 ) 3 ( − 4 ) = − 1 × ( − 4 ) = 4 C_{21} = (-1)^{2+1} M_{21} = (-1)^3 (-4) = -1 \times (-4) = 4 C 21 = ( − 1 ) 2 + 1 M 21 = ( − 1 ) 3 ( − 4 ) = − 1 × ( − 4 ) = 4
ਐਲੀਮੈਂਟ a 22 = 5 a_{22} = 5 a 22 = 5 ਲਈ:
C 22 = ( − 1 ) 2 + 2 M 22 = ( − 1 ) 4 ( 2 ) = 1 × 2 = 2 C_{22} = (-1)^{2+2} M_{22} = (-1)^4 (2) = 1 \times 2 = 2 C 22 = ( − 1 ) 2 + 2 M 22 = ( − 1 ) 4 ( 2 ) = 1 × 2 = 2
ਐਲੀਮੈਂਟ a 23 = − 1 a_{23} = -1 a 23 = − 1 ਲਈ:
C 23 = ( − 1 ) 2 + 3 M 23 = ( − 1 ) 5 ( 1 ) = − 1 × 1 = − 1 C_{23} = (-1)^{2+3} M_{23} = (-1)^5 (1) = -1 \times 1 = -1 C 23 = ( − 1 ) 2 + 3 M 23 = ( − 1 ) 5 ( 1 ) = − 1 × 1 = − 1
ਐਲੀਮੈਂਟ a 31 = 0 a_{31} = 0 a 31 = 0 ਲਈ:
C 31 = ( − 1 ) 3 + 1 M 31 = ( − 1 ) 4 ( − 20 ) = 1 × ( − 20 ) = − 20 C_{31} = (-1)^{3+1} M_{31} = (-1)^4 (-20) = 1 \times (-20) = -20 C 31 = ( − 1 ) 3 + 1 M 31 = ( − 1 ) 4 ( − 20 ) = 1 × ( − 20 ) = − 20
ਐਲੀਮੈਂਟ a 32 = 1 a_{32} = 1 a 32 = 1 ਲਈ:
C 32 = ( − 1 ) 3 + 2 M 32 = ( − 1 ) 5 ( − 13 ) = − 1 × ( − 13 ) = 13 C_{32} = (-1)^{3+2} M_{32} = (-1)^5 (-13) = -1 \times (-13) = 13 C 32 = ( − 1 ) 3 + 2 M 32 = ( − 1 ) 5 ( − 13 ) = − 1 × ( − 13 ) = 13
ਐਲੀਮੈਂਟ a 33 = 2 a_{33} = 2 a 33 = 2 ਲਈ:
C 33 = ( − 1 ) 3 + 3 M 33 = ( − 1 ) 6 ( 5 ) = 1 × 5 = 5 C_{33} = (-1)^{3+3} M_{33} = (-1)^6 (5) = 1 \times 5 = 5 C 33 = ( − 1 ) 3 + 3 M 33 = ( − 1 ) 6 ( 5 ) = 1 × 5 = 5
ਅੰਤਿਮ ਜਵਾਬ:
ਹਰੇਕ ਐਲੀਮੈਂਟ ਦੇ ਮਾਈਨਰ ਅਤੇ ਕੋਫੈਕਟਰ ਇਸ ਪ੍ਰਕਾਰ ਹਨ:
a 11 = 1 a_{11} = 1 a 11 = 1 : M 11 = 11 M_{11} = \text{11} M 11 = 11 , C 11 = 11 C_{11} = \text{11} C 11 = 11
a 12 = 0 a_{12} = 0 a 12 = 0 : M 12 = 6 M_{12} = \text{6} M 12 = 6 , C 12 = -6 C_{12} = \text{-6} C 12 = -6
a 13 = 4 a_{13} = 4 a 13 = 4 : M 13 = 3 M_{13} = \text{3} M 13 = 3 , C 13 = 3 C_{13} = \text{3} C 13 = 3
a 21 = 3 a_{21} = 3 a 21 = 3 : M 21 = -4 M_{21} = \text{-4} M 21 = -4 , C 21 = 4 C_{21} = \text{4} C 21 = 4
a 22 = 5 a_{22} = 5 a 22 = 5 : M 22 = 2 M_{22} = \text{2} M 22 = 2 , C 22 = 2 C_{22} = \text{2} C 22 = 2
a 23 = − 1 a_{23} = -1 a 23 = − 1 : M 23 = 1 M_{23} = \text{1} M 23 = 1 , C 23 = -1 C_{23} = \text{-1} C 23 = -1
a 31 = 0 a_{31} = 0 a 31 = 0 : M 31 = -20 M_{31} = \text{-20} M 31 = -20 , C 31 = -20 C_{31} = \text{-20} C 31 = -20
a 32 = 1 a_{32} = 1 a 32 = 1 : M 32 = -13 M_{32} = \text{-13} M 32 = -13 , C 32 = 13 C_{32} = \text{13} C 32 = 13
a 33 = 2 a_{33} = 2 a 33 = 2 : M 33 = 5 M_{33} = \text{5} M 33 = 5 , C 33 = 5 C_{33} = \text{5} C 33 = 5
That's 2 down. 3 left today — send the next one.