Question

Suppose A and B are 3x3 matrices. Show that If B is obtained from A by adding 2 times the first row of A to the last row of A, then det(A) =det(B).

Answer 1

$\textsf{Let}\hspace{0.1cm}A = \begin{pmatrix} a_1 & a_2 & a_3\\ a_4 & a_5 & a_6\\ a_7 & a_8 & a_9 \end{pmatrix}\\ \mathrm{det} A = a_1(a_5(a_9 + 2a_3) - a_6(a_8 + 2a_2)) - a_2(a_4(a_9 + 2a_3) - a_6(a_7 + 2a_1)) + a_3(a_4(a_8 + 2a_2) - a_5(a_7 + 2a_1)) = a_1(a_5(a_9) - a_6(a_8)) - a_2(a_4(a_9) - a_6(a_7)) + a_3(a_4(a_8) - a_5(a_7))\\ B = \begin{pmatrix} a_1 & a_2 & a_3\\ a_4 & a_5 & a_6\\ a_7 + 2a_1 & a_8 + 2a_2 & a_9 + 2a_3 \end{pmatrix} = \\a_1(a_5(a_9 + 2a_3) - a_6(a_8 + 2a_2)) - a_2(a_4(a_9 + 2a_3) - a_6(a_7 + 2a_1)) + a_3(a_4(a_8 + 2a_2) - a_5(a_7 + 2a_1)) \\= a_1(a_5(a_9) - a_6(a_8)) - a_2(a_4(a_9) - a_6(a_7)) + a_3(a_4(a_8) - a_5(a_7)) + 2(a_1a_3a_5 - a_1a_2a_6 - a_2a_3a_4 + a_1a_2a_6 + a_2a_3a_4 - a_1a_3a_5) \\= a_1(a_5(a_9) - a_6(a_8)) - a_2(a_4(a_9) - a_6(a_7)) + a_3(a_4(a_8) - a_5(a_7)) + 0 = a_1(a_5(a_9) - a_6(a_8)) - a_2(a_4(a_9) - a_6(a_7)) + a_3(a_4(a_8) - a_5(a_7)) = \mathrm{det} A$