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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

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

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  • $\begingroup$ 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

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