If a 4x4 matrix A with rows v1 , v2 , v3 , and v4 has determinant detA=−4, then what is the det of:
4v1+6v4
v2
v3
6v1+6v4
can someone help me with this?
I thought it was 4, but I have tried to put this into the computer many times and it is wrong
4v1+6v4
v2
v3
6v1+6v4
can someone help me with this?
I thought it was 4, but I have tried to put this into the computer many times and it is wrong

well if u applied the column transformationa and ended up wit this then the answer will be 4
but if u multiplied the fist column with 4 and the fouth with 6 and then in the next step added them , you should have divided the determinant with 4*6
similarly for the fourth column!
but if u ended up using ONLY column transformations and no scalar multiplication or ay of the rows or colums then yes the answer is 4!
take care!
but if u multiplied the fist column with 4 and the fouth with 6 and then in the next step added them , you should have divided the determinant with 4*6
similarly for the fourth column!
but if u ended up using ONLY column transformations and no scalar multiplication or ay of the rows or colums then yes the answer is 4!
take care!

If you multiply a row by a constant and add it to another row, the determinant will not change.
But if a row is multiplied by a constant k, then the determinant also becomes k times.
so if the matrix had been
v1+6v4
v2
v3
6v1+v4
The det would have been the same.
But in
4v1+6v4
v2
v3
6v1+6v4
the first row is multiplied by 4 and the fourth row by 6. Thus the det will be 4 X 4 X 6 = 96
But if a row is multiplied by a constant k, then the determinant also becomes k times.
so if the matrix had been
v1+6v4
v2
v3
6v1+v4
The det would have been the same.
But in
4v1+6v4
v2
v3
6v1+6v4
the first row is multiplied by 4 and the fourth row by 6. Thus the det will be 4 X 4 X 6 = 96