Question

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plz solve it​

Answer 1

Question :

Find the minors and cofactors of elements of the determinant

$\tt \left | \begin{array}{ccc} -1 & 0 & 4 \\ -2 & 1 & 3 \\ 0 & -4 & 2 \end{array} \right |$

Solution:

Minors

$\tt {M}_{11} = \left | \begin{array}{cc} 1 & 3 \\ -4 & 2 \end{array} \right |$

$\tt \implies {M}_{11}$ = (1 ×2) - ( -4 ×3)

$\tt \implies {M}_{11}$ = 2 - (-12)

$\tt \implies {M}_{11}$ = 2 + 12

$\tt \therefore {M}_{11}$ = 14

$\tt {M}_{12} = \left | \begin{array}{cc} -2 & 3 \\ 0 & 2 \end{array} \right |$

$\tt \implies {M}_{12}$ = ( 2 × -2) - (0 ×3)

$\tt \implies {M}_{12}$ = -4 - 0

$\tt \implies {M}_{12}$ = -4

$\tt \therefore {M}_{12}$ = -4

$\tt {M}_{13} = \left | \begin{array}{cc} -2& 1 \\ 0 & -4 \end{array} \right |$

$\tt \implies {M}_{13}$ = (-2×-4)-(0×1)

$\tt \implies {M}_{13}$ = 8 - 0

$\tt \implies {M}_{13}$ = 8

$\tt \therefore {M}_{13}$ = 8

$\tt {M}_{21} = \left | \begin{array}{cc} 0& 4 \\ -4 & 2 \end{array} \right |$

$\tt \implies {M}_{21}$ = (0×2)-(4×-4)

$\tt \implies {M}_{21}$ = 0 - (-16)

$\tt \implies {M}_{21}$ = 16

$\tt \therefore {M}_{21}$ = 16

$\tt {M}_{22} = \left | \begin{array}{cc} -1& 4 \\ 0 & 2 \end{array} \right |$

$\tt \implies {M}_{22}$ = (-1×2)-(0×4)

$\tt \implies {M}_{22}$ = -2 - 0

$\tt \implies {M}_{22}$ = -2

$\tt \therefore {M}_{22}$ = -2

$\tt {M}_{23} = \left | \begin{array}{cc} -1& 0 \\ 0 & -4 \end{array} \right |$

$\tt \implies {M}_{23}$= (0×-1) -(0×-4)

$\tt \implies {M}_{23}$=0

$\tt \therefore {M}_{23}$=0

$\tt {M}_{31} = \left | \begin{array}{cc} 0& 1 \\ 4 & 3 \end{array} \right |$

$\tt \implies {M}_{31}$ = (0×3) - (1×4)

$\tt \implies {M}_{31}$ = 0 - 4

$\tt \implies {M}_{31}$ =.-4

$\tt \therefore {M}_{31}$ = -4

$\tt {M}_{32} = \left | \begin{array}{cc} -1& 4\\ -2& 3 \end{array} \right |$

$\tt \implies {M}_{32}$ = (-1×3) - (4 ×-2)

$\tt \implies {M}_{32}$ = -3 +8

$\tt \implies {M}_{31}$ = 5

$\tt \therefore {M}_{32}$ = 5

$\tt {M}_{32} = \left | \begin{array}{cc} -1& 0\\ -2& 1 \end{array} \right |$

$\tt \implies {M}_{33}$ = (-1×-2)-(1×0)

$\tt \implies {M}_{33}$ = 2 - 0

$\tt \implies {M}_{32}$ = 2

$\tt \therefore {M}_{33}$ = 2

Cofactors :

$\tt {C}_{ij} = {-1}^{i+j}{M}_{ij}$

$\tt {C}_{11} ={-1}^{1+1}{M}_{11}$

$\tt \implies {C}_{11} ={-1}^{2}(14)$

$\tt \implies {C}_{11} = 14$

$\tt \therefore {C}_{11} = 14$

$\tt {C}_{12} ={-1}^{1+2}{M}_{12}$

$\tt \implies {C}_{12} = {-1}^{3}(-4)$

$\tt \implies {C}_{12} = (-1)(-4)$

$\tt \implies {C}_{12} = 4$

$\tt \therefore {C}_{12} = 4$

$\tt {C}_{13} ={-1}^{1+3}{M}_{13}$

$\tt \implies {C}_{13} = {-1}^{4}(8)$

$\tt \implies {C}_{13} = 8$

$\tt \therefore {C}_{13} = 8$

$\tt {C}_{21} ={-1}^{2+1}{M}_{21}$

$\tt \implies {C}_{21} = {-1}^{3}(16)$

$\tt \implies {C}_{21} = (-1)(16)$

$\tt \implies {C}_{21} = -16$

$\tt \therefore {C}_{21} = -16$

$\tt {C}_{22} ={-1}^{2+2}{M}_{22}$

$\tt \implies {C}_{22} = {-1}^{4}(-2)$

$\tt \implies {C}_{21} = 2$

$\tt \therefore {C}_{21} = 2$

$\tt {C}_{23} ={-1}^{2+3}{M}_{23}$

$\tt \implies {C}_{22} = {-1}^{5}(0)$

$\tt \implies {C}_{21} = 0$

$\tt \therefore {C}_{21} = 0$

$\tt {C}_{31} ={-1}^{3+1}{M}_{31}$

$\tt \implies {C}_{22} = {-1}^{4}(-4)$

$\tt \implies {C}_{21} = -4$

$\tt \therefore {C}_{21} = -4$

$\tt {C}_{32} ={-1}^{3+2}{M}_{32}$

$\tt \implies {C}_{22} = {-1}^{5}(2)$

$\tt \implies {C}_{21} =- 2$

$\tt \therefore {C}_{21} =- 2$

$\tt {C}_{33} ={-1}^{3+3}{M}_{33}$

$\tt \implies {C}_{22} = {-1}^{6}(2)$

$\tt \implies {C}_{21} = 2$

$\tt \therefore {C}_{21} = 2$

Answer 2

Answer

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