Question #198533

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).


1
Expert's answer
2021-05-28T10:23:19-0400


R3+2R1R3R_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,t, the determinant is multiplied by tt


5R1R1=>t1=155R_1\to R_1=>t_1=\dfrac{1}{5}

2R2R2=>t2=12-2R_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. 


R2R3R_2\leftrightarrow R_3

Then

det(B)=15(12)(1)det(A)=110(detA)\det(B)=\dfrac{1}{5}(-\dfrac{1}{2})(-1)\det (A)=\dfrac{1}{10}(\det A)

=110(10)=1=\dfrac{1}{10}(-10)=-1



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