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Exercise:
forall Ain M_ntimes nK we have the following n different formulas for textdetA: textdetA_i^n-^i+jA_ijtextdetAi|jquad j...n. The scalars C_ij:-^i+jtextdetAi|j are called the ij cofactor of A. We have textdetA_i^n A_ij C_ij expansion of textdetA by the cofactors of the j'th col. Show that if we replace A_ij in the formula above by A_ik k is fixed always get . In other words forall Ain M_ntimes nK forall jneq k we have _i^n A_ik C_ij.

Solution:
Proof. Let Bin M_ntimes nK be the matrix obtained from A by replacing col# j of A by col# k of A. So in B col# j col# k. Clearly textdetB since B has two equal columns iff B^T has two equal rows textdetB_i^n -^i+jB_ijtextdetBi|j _i^n -^i+jA_ik textdetAi|j _i^n A_ik C_ij
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Exercise:
forall Ain M_ntimes nK we have the following n different formulas for textdetA: textdetA_i^n-^i+jA_ijtextdetAi|jquad j...n. The scalars C_ij:-^i+jtextdetAi|j are called the ij cofactor of A. We have textdetA_i^n A_ij C_ij expansion of textdetA by the cofactors of the j'th col. Show that if we replace A_ij in the formula above by A_ik k is fixed always get . In other words forall Ain M_ntimes nK forall jneq k we have _i^n A_ik C_ij.

Solution:
Proof. Let Bin M_ntimes nK be the matrix obtained from A by replacing col# j of A by col# k of A. So in B col# j col# k. Clearly textdetB since B has two equal columns iff B^T has two equal rows textdetB_i^n -^i+jB_ijtextdetBi|j _i^n -^i+jA_ik textdetAi|j _i^n A_ik C_ij
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determinant, eth, fs23, lineare algebra, proof
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(3, default)
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Language
ENG (English)
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Proof
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