- #1
Winzer
- 597
- 0
So I am looking at the proof for this in a linear algebra book and I half way get it:
Theorem:
If the all elements of the jth column of a determinate D are linear combinations of two columns of numbers, i.e., if
D=\lambda b_{i}+uc{i} where lambda and mu are fixed numbers, then D is equalto a linear comination of the two determinates:
D=D_{1}\lambda+D{2}u
Here both determinates D1 and D2 have the same columns as the determinate D except for the jth column; the jth colum of D1 consists of the numbers b_{i} wile the jth column of D2 consists of the numbers c_{i}
Theorem:
If the all elements of the jth column of a determinate D are linear combinations of two columns of numbers, i.e., if
D=\lambda b_{i}+uc{i} where lambda and mu are fixed numbers, then D is equalto a linear comination of the two determinates:
D=D_{1}\lambda+D{2}u
Here both determinates D1 and D2 have the same columns as the determinate D except for the jth column; the jth colum of D1 consists of the numbers b_{i} wile the jth column of D2 consists of the numbers c_{i}
