I keep getting a multiple of the actual value of the determinant of matrices after performing row operations to reduce it to an upper triangular matrix. Any ideas why this might happen?For an easy 3x3 for example

  • $\begingroup$ Welcome to stackexchange. Please edit your question to show us just what you did. Then we might be able to help. Use mathjax: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Ethan Bolker Mar 6 '18 at 11:53
  • $\begingroup$ It would be good if you include an example for a $2\times 2$-matrix. $\endgroup$ – Dietrich Burde Mar 6 '18 at 12:22

Each and every Row/Column operations will affect the value of the determinant depends on the constant that we are multiplying for the row.

For example,

$R_1 -> R_1 - R_2$ will not affect the value of the determinant.

But, $R_1 -> 2R_1 - R_2$ will double the value of the determinant as we are multiplying the $R_1$ by 2.

Even the operation $R_1 -> R_2 - R_1$, will actually affect the determinant value as $-1 * Determinant$

So, At each steps whenever we are doing such operations, we should ensure that we have the Same Co-efficient with sign on both the side. If you want to do the operations by multiplying with some constants, then we should take the constant outside as a divisor.

For example,

$R_1 -> R_1 - R_2$ --> $Actual\,Determinant = Determinant\,\,after\,Row/Column\,operation$

$R_1 -> R_2 - R_1$ --> $Actual\,Determinant = \frac{Determinant\,\,after\,Row/Column\,operation}{-1}$

$R_2 -> -3R_2 - 5R_1$ --> $Actual\,Determinant = \frac{Determinant\,\,after\,Row/Column\,operation}{-3}$

In Your attached example,

At first step you have done $C_1 -> 5C_2 - 6C_1$ --> $Actual\,Determinant = (1/-6)\,(Det)$

At Second step you have done $R_3 -> R_1 + R_3$ --> $Actual\,Determinant = (1/1)\,(Det)$

At Third step you have done $R_2 -> 2R_2 - R_3$ --> $Actual\,Determinant = (1/-1)\,(Det)$

You got your determinant value as 162 after these operations.

So, $Actual\,Determinant = \frac{162}{(-6)(1)(-1)}\,=27$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.