If the matrix $$ A = \pmatrix{row1 \\ row2\\row3}\ and \left|\begin{array}[ccc]\\ A \end{array}\right| =10$$

And matrix

$$ B = \pmatrix{2row1+row2-row3 \\ 2row3\\5row2}\ \ $$ then find $$ \\\left|\begin{array}[ccc]\\ B \end{array}\right| = ?$$

iam stuck at this, i know that subtracting a multiple of one row from another row does not change determinate (A), and also if we do permutation of rows 1 time the sign will be negative, but i do not know if this information is useful here or not, i need help with this one and

after reading StackTD hints B, will be: $$ B = \pmatrix{2row1\\2row3\\5row2}\ \ $$ and i know the determinant of B will in negative sign because of rows swap but i couldn't obtain B determinant


Use properties of determinants:

  • the determinant is linear in each row/column;
  • a determinant with two identical rows is $0$;
  • swapping two rows changes the sign of the determinant.

Now start with linearity and follow up (I write $A_i$ for the $i$th row of the original matrix $A$): $$\begin{vmatrix} 2A_1+A_2-A_3 \\ 2A_3 \\ 5A_2 \end{vmatrix} = \begin{vmatrix} 2A_1 \\ 2A_3 \\ 5A_2 \end{vmatrix}+\begin{vmatrix} A_2 \\ 2A_3 \\ 5A_2 \end{vmatrix}+\begin{vmatrix} -A_3 \\ 2A_3 \\ 5A_2 \end{vmatrix} = \ldots$$

Addition after comment: $$\begin{vmatrix} \color{blue}{2}A_1 \\ \color{green}{2}A_3 \\ \color{red}{5}A_2 \end{vmatrix}=\color{blue}{2}\cdot\color{green}{2}\cdot\color{red}{5}\cdot\begin{vmatrix} A_1 \\ \color{purple}{A_3} \\ \color{purple}{A_2} \end{vmatrix}=\ldots$$

  • $\begingroup$ thanks for explaining that to me, now i updated the subject with matrix B but i couldn't obtain B determinant is that right ?, and how we obtain b determinant ? $\endgroup$ – The Beard Dec 12 '18 at 12:50
  • $\begingroup$ Linearity also allows you to 'extract' the multiples (in each row!); then notice the swap of two rows. See updated answer. $\endgroup$ – StackTD Dec 12 '18 at 12:53
  • $\begingroup$ thank you again for explaining to me now i have more understading of determinant properties because of your help, and from what you have told me the determinant of B will be -20 right ? $\endgroup$ – The Beard Dec 12 '18 at 13:00
  • $\begingroup$ Almost, $-20$ times the determinant $A$, so...? $\endgroup$ – StackTD Dec 12 '18 at 13:27
  • $\begingroup$ so then the determinant of b will be -10 ? $\endgroup$ – The Beard Dec 12 '18 at 13:56

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.