Answer on Question#39416, Math, Matrix
Let us multiply ABABAB and BABABA. AB=(3a+63b+104a+34b+5)AB = \begin{pmatrix} 3a + 6 & 3b + 10 \\ 4a + 3 & 4b + 5 \end{pmatrix}AB=(3a+64a+33b+104b+5), BA=(3a+4b2a+b2911)BA = \begin{pmatrix} 3a + 4b & 2a + b \\ 29 & 11 \end{pmatrix}BA=(3a+4b292a+b11). Using the second row of identity AB=BAAB = BAAB=BA, obtain 4a+3=29;4b+5=114a + 3 = 29; 4b + 5 = 114a+3=29;4b+5=11, thus a=6;b=32a = 6; b = \frac{3}{2}a=6;b=23.
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