submitdep14902920 For each given matrix A first compute detA

submit?dep-14902920 For each given matrix A, first compute det(A). Then interchange two rows of your choosing and compute the determinant of the resulting matrix A 0-1 1 (a) A-1 3 1 2 0 5 det(A) det(A) = 4 2 11 (b) A=101 o 0-2 det(A) det(A\') = Form a conjecture about the effect of row interchanges on determinants Interchanging rows doubles the determinant. Interchanging rows halves the determinant. erchanging rows changes the sign of the determinant. Interchanging rows does not change the determinant. Interchanging rows has no consistent effect on the determinant.

Solution

Sol:

let A=

0 -1 1

-1 3 1

2 0 5

detA=0 det[3 1 +1 det[ -1 1 +1 det [-1 3

0 5] 2 5] 2 0]

=0+1[-1*5-1*2]+1[-1*0-3*2]

=0+1[-5-2]+1[0-6]

=-7-6

detA=-13

interchange two rows: R2 AND R3

A^\'

= 0 -1 1

2 0 5

-1 3 1

=0 det[0 5 +1 det[2 5 +1 det[ 2 0

3 1] -1 1] -1 3]

=0+1(2+5)+1(6-0]

=0+7+6

detA^\'=13

interchanging rows changes sign of the determinant from negative 13 to positive 13.value remains constant.

previously it was positive value now negative after interchanging two rows.

Solutionb:

let A= 4 2 1

0 1 1

0 0 -2

detA=4det[1 1 -2det[ 0 1 +1det[0 1

0 -2] 0 -2] 0 0]

=4[1*-2-1*0]-2[0*-2-1*0]+1[0*0-1*0]

=4[-2-0]-2[0-0]+1[0-0]

detA=-8

interchange first and second rows

A^\'= 0 1 1

4 2 1

0 0 -2

det A\'

=0 det[2 1 -1det[ 4 1 +1 det[ 4 2

0 -2] 0 -2] 0 0]

=0[2*-2-1*0]-1(4*-2-1*0)+1(4*0-2*0]

=0-1(-8-0)+1(0)

=-1(-8)

=8

interchanging rows changes sign

here negative 8 to positive 8

marked option is right