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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:

∣10435−1012∣\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}

ਇੱਕ ਐਲੀਮੈਂਟ aija_{ij} ਦਾ ਮਾਈਨਰ MijM_{ij} ਉਸ ਸਬ-ਡਿਟਰਮੀਨੈਂਟ ਦਾ ਮੁੱਲ ਹੁੰਦਾ ਹੈ ਜੋ ii-ਵੀਂ ਕਤਾਰ ਅਤੇ jj-ਵੇਂ ਕਾਲਮ ਨੂੰ ਹਟਾਉਣ ਨਾਲ ਪ੍ਰਾਪਤ ਹੁੰਦਾ ਹੈ।
ਇੱਕ ਐਲੀਮੈਂਟ aija_{ij} ਦਾ ਕੋਫੈਕਟਰ CijC_{ij} ਫਾਰਮੂਲੇ Cij=(−1)i+jMijC_{ij} = (-1)^{i+j} M_{ij} ਦੁਆਰਾ ਦਿੱਤਾ ਜਾਂਦਾ ਹੈ।

  1. ਹਰੇਕ ਐਲੀਮੈਂਟ ਦੇ ਮਾਈਨਰ (Minors) ਲੱਭੋ।

    • ਐਲੀਮੈਂਟ a11=1a_{11} = 1 ਲਈ: M11=∣5−112∣=(5)(2)−(−1)(1)=10−(−1)=11M_{11} = \begin{vmatrix} 5 & -1 \\ 1 & 2 \end{vmatrix} = (5)(2) - (-1)(1) = 10 - (-1) = 11

    • ਐਲੀਮੈਂਟ a12=0a_{12} = 0 ਲਈ: M12=∣3−102∣=(3)(2)−(−1)(0)=6−0=6M_{12} = \begin{vmatrix} 3 & -1 \\ 0 & 2 \end{vmatrix} = (3)(2) - (-1)(0) = 6 - 0 = 6

    • ਐਲੀਮੈਂਟ a13=4a_{13} = 4 ਲਈ: M13=∣3501∣=(3)(1)−(5)(0)=3−0=3M_{13} = \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = (3)(1) - (5)(0) = 3 - 0 = 3

    • ਐਲੀਮੈਂਟ a21=3a_{21} = 3 ਲਈ: M21=∣0412∣=(0)(2)−(4)(1)=0−4=−4M_{21} = \begin{vmatrix} 0 & 4 \\ 1 & 2 \end{vmatrix} = (0)(2) - (4)(1) = 0 - 4 = -4

    • ਐਲੀਮੈਂਟ a22=5a_{22} = 5 ਲਈ: M22=∣1402∣=(1)(2)−(4)(0)=2−0=2M_{22} = \begin{vmatrix} 1 & 4 \\ 0 & 2 \end{vmatrix} = (1)(2) - (4)(0) = 2 - 0 = 2

    • ਐਲੀਮੈਂਟ a23=−1a_{23} = -1 ਲਈ: M23=∣1001∣=(1)(1)−(0)(0)=1−0=1M_{23} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0) = 1 - 0 = 1

    • ਐਲੀਮੈਂਟ a31=0a_{31} = 0 ਲਈ: M31=∣045−1∣=(0)(−1)−(4)(5)=0−20=−20M_{31} = \begin{vmatrix} 0 & 4 \\ 5 & -1 \end{vmatrix} = (0)(-1) - (4)(5) = 0 - 20 = -20

    • ਐਲੀਮੈਂਟ a32=1a_{32} = 1 ਲਈ: M32=∣143−1∣=(1)(−1)−(4)(3)=−1−12=−13M_{32} = \begin{vmatrix} 1 & 4 \\ 3 & -1 \end{vmatrix} = (1)(-1) - (4)(3) = -1 - 12 = -13

    • ਐਲੀਮੈਂਟ a33=2a_{33} = 2 ਲਈ: M33=∣1035∣=(1)(5)−(0)(3)=5−0=5M_{33} = \begin{vmatrix} 1 & 0 \\ 3 & 5 \end{vmatrix} = (1)(5) - (0)(3) = 5 - 0 = 5

  2. ਹਰੇਕ ਐਲੀਮੈਂਟ ਦੇ ਕੋਫੈਕਟਰ (Cofactors) ਲੱਭੋ।

    • ਐਲੀਮੈਂਟ a11=1a_{11} = 1 ਲਈ: C11=(−1)1+1M11=(−1)2(11)=1×11=11C_{11} = (-1)^{1+1} M_{11} = (-1)^2 (11) = 1 \times 11 = 11

    • ਐਲੀਮੈਂਟ a12=0a_{12} = 0 ਲਈ: C12=(−1)1+2M12=(−1)3(6)=−1×6=−6C_{12} = (-1)^{1+2} M_{12} = (-1)^3 (6) = -1 \times 6 = -6

    • ਐਲੀਮੈਂਟ a13=4a_{13} = 4 ਲਈ: C13=(−1)1+3M13=(−1)4(3)=1×3=3C_{13} = (-1)^{1+3} M_{13} = (-1)^4 (3) = 1 \times 3 = 3

    • ਐਲੀਮੈਂਟ a21=3a_{21} = 3 ਲਈ: C21=(−1)2+1M21=(−1)3(−4)=−1×(−4)=4C_{21} = (-1)^{2+1} M_{21} = (-1)^3 (-4) = -1 \times (-4) = 4

    • ਐਲੀਮੈਂਟ a22=5a_{22} = 5 ਲਈ: C22=(−1)2+2M22=(−1)4(2)=1×2=2C_{22} = (-1)^{2+2} M_{22} = (-1)^4 (2) = 1 \times 2 = 2

    • ਐਲੀਮੈਂਟ a23=−1a_{23} = -1 ਲਈ: C23=(−1)2+3M23=(−1)5(1)=−1×1=−1C_{23} = (-1)^{2+3} M_{23} = (-1)^5 (1) = -1 \times 1 = -1

    • ਐਲੀਮੈਂਟ a31=0a_{31} = 0 ਲਈ: C31=(−1)3+1M31=(−1)4(−20)=1×(−20)=−20C_{31} = (-1)^{3+1} M_{31} = (-1)^4 (-20) = 1 \times (-20) = -20

    • ਐਲੀਮੈਂਟ a32=1a_{32} = 1 ਲਈ: C32=(−1)3+2M32=(−1)5(−13)=−1×(−13)=13C_{32} = (-1)^{3+2} M_{32} = (-1)^5 (-13) = -1 \times (-13) = 13

    • ਐਲੀਮੈਂਟ a33=2a_{33} = 2 ਲਈ: C33=(−1)3+3M33=(−1)6(5)=1×5=5C_{33} = (-1)^{3+3} M_{33} = (-1)^6 (5) = 1 \times 5 = 5

    ਅੰਤਿਮ ਜਵਾਬ:
    ਹਰੇਕ ਐਲੀਮੈਂਟ ਦੇ ਮਾਈਨਰ ਅਤੇ ਕੋਫੈਕਟਰ ਇਸ ਪ੍ਰਕਾਰ ਹਨ:

    • a11=1a_{11} = 1: M11=11M_{11} = \boxed{\text{11}}, C11=11C_{11} = \boxed{\text{11}}
    • a12=0a_{12} = 0: M12=6M_{12} = \boxed{\text{6}}, C12=-6C_{12} = \boxed{\text{-6}}
    • a13=4a_{13} = 4: M13=3M_{13} = \boxed{\text{3}}, C13=3C_{13} = \boxed{\text{3}}
    • a21=3a_{21} = 3: M21=-4M_{21} = \boxed{\text{-4}}, C21=4C_{21} = \boxed{\text{4}}
    • a22=5a_{22} = 5: M22=2M_{22} = \boxed{\text{2}}, C22=2C_{22} = \boxed{\text{2}}
    • a23=−1a_{23} = -1: M23=1M_{23} = \boxed{\text{1}}, C23=-1C_{23} = \boxed{\text{-1}}
    • a31=0a_{31} = 0: M31=-20M_{31} = \boxed{\text{-20}}, C31=-20C_{31} = \boxed{\text{-20}}
    • a32=1a_{32} = 1: M32=-13M_{32} = \boxed{\text{-13}}, C32=13C_{32} = \boxed{\text{13}}
    • a33=2a_{33} = 2: M33=5M_{33} = \boxed{\text{5}}, C33=5C_{33} = \boxed{\text{5}}
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