Question #236809

You are given the following matrix

 A=(abcdefghi).A=\left(\begin{array}{cccc}a & b & c\\ d & e & f \\g & h & i\end{array} \right).

Which of the following is the determinant of A?


  1. aei+bfgcdhcegafhbdiaei+bfg−cdh−ceg−afh−bdi
  2. aei+bfg+cdhcegafh+bdiaei+bfg+cdh−ceg−afh+bdi
  3. aeibfg+cdhceg+afhbdiaei−bfg+cdh−ceg+afh−bdi
  4. aei+bfg+cdhcegafhbdiaei+bfg+cdh−ceg−afh−bdi
  5. aeibfgcdh+ceg+afh+bdi−aei−bfg−cdh+ceg+afh+bdi
1
Expert's answer
2021-09-20T04:51:19-0400

There is a square matrix A=A=(a11a12a13a21a22a23a31a32a33)\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}. By defenition det(A)det(A) is a sum all of products of 3 elements. The elements should be taken one at a time from each row and column. A sign of product depending on the permutation.

Take a products, in each of them indexes by rows have no inversion:

1) a11a22a33a_{11}a_{22}a_{33}

Indexes by row: 1,2,3: no inversion -- 0

Indexes by column: 1,2,3: no inversions -- 0.

0 + 0 = 0, it's even number, sign is plus.

2) a12a23a31a_{12}a_{23}a_{31}

Indexes by row: 1,2,3: no inversions -- 0

Indexes by column: 2,3,1

2 and 1 are inversed, 3 and 1 are inversed -- 2 inversions.

0 + 2 = 2, it's even number, sign is plus.

3) a13a21a32a_{13}a_{21}a_{32}

Indexes by row: 1,2,3: no inversions -- 0

Indexes by column: 3,1,2

3 and 1 are inversed, 3 and 2 are inversed -- 2 inversions.

0 + 2 = 2, it's even number, sign is plus.

4) a13a22a31a_{13}a_{22}a_{31}

Indexes by row: 1,2,3: no inversions -- 0

Indexes by column: 3,2,1

3 and 2 are inversed, 3 and 1 are inversed, 2 and 1 are inversed -- 3 inversions.

0 + 3 = 3, it's odd number, sign is minus.

5) a11a23a32a_{11}a_{23}a_{32}

Indexes by row: 1,2,3: no inversions -- 0

Indexes by column: 1,3,2

3 and 2 are inversed -- 1 inversion.

0 + 1 = 1, it's odd number, sign is minus.

6) a12a21a33a_{12}a_{21}a_{33}

Indexes by row: 1,2,3: no inversions -- 0

Indexes by column: 2,1,3

2 and 1 are inversed -- 1 inversion.

0 + 1 = 1, it's odd number, sign is minus.


So, the det(A)=a11a22a33+a12a23a31+a13a21a32a13a22a31a11a23a32a12a21a33det(A)=a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32} -a_{12}a_{21}a_{33}


If there is A=(abcdefghi)A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} , using the defenition of determinant for matrix 3x3 and calculete it:

det(A)=aei+bfg+cdhcegafhbdidet(A)=aei+bfg+cdh-ceg-afh-bdi

It is answer number 4.


Answer: 4. aei+bfg+cdhcegafhbdiaei+bfg+cdh-ceg-afh-bdi



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