Question

Assume that A is a 3 by 3 matrix such that det(A) = −10. Let B be a matrix obtained from A using the following elementary row operations:

R3 + 2R1 → R3,

5R1 → R1,

−2R2 → R2

R2 ↔ R3.

Find the determinant of B obtained from the resulting operations, i.e., det(B)

Answer 1

Answer:-

$$R_3+2R_1\to R_3$$

The value of a determinant is not changed by adding the elements of one column multiplied by an arbitrary number to the corresponding elements of another column.

If we multiply one row of a matrix by $t,$ the determinant is multiplied by $t$

$$5R_1\to R_1=>t_1=\dfrac{1}{5}$$ $$-2R_2\to R_2=>t_2=-\dfrac{1}{2}$$

If you exchange two rows of a matrix, you reverse the sign of its determinant from positive to negative or from negative to positive. 

$$R_2\leftrightarrow R_3$$

Then

$$\det(B)=\dfrac{1}{5}(-\dfrac{1}{2})(-1)\det (A)=\dfrac{1}{10}(\det A)$$ $$=\dfrac{1}{10}(-10)=-1$$