Question #134728
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)
1
Expert's answer
2020-09-23T17:35:51-0400

LetA=(a1a2a3a4a5a6a7a8a9)detA=a1(a5(a9)−a6(a8))−a2(a4(a9)−a6(a7))+a3(a4(a8)−a5(a7))B=(a1a2a3a4a5a6a7+2a1a8+2a2a9+2a3)detB=a1(a5(a9+2a3)−a6(a8+2a2))−a2(a4(a9+2a3)−a6(a7+2a1))+a3(a4(a8+2a2)−a5(a7+2a1))=a1(a5(a9)−a6(a8))−a2(a4(a9)−a6(a7))+a3(a4(a8)−a5(a7))+2(a1a3a5−a1a2a6−a2a3a4+a1a2a6+a2a3a4−a1a3a5)=a1(a5(a9)−a6(a8))−a2(a4(a9)−a6(a7))+a3(a4(a8)−a5(a7))+2(0)=a1(a5(a9)−a6(a8))−a2(a4(a9)−a6(a7))+a3(a4(a8)−a5(a7))=detA∴detB=detA\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) - 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}\\ \mathrm{det}B = 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)) + 2(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\\ \therefore \mathrm{det}B= \mathrm{det}A


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