Verified answer

Se calculează:

[tex]B(a) \cdot A = \begin{pmatrix} 0 & a - 2\\ 1 & 3a \end{pmatrix} \cdot \begin{pmatrix} 3 & -6\\ 2 & -3 \end{pmatrix} = \begin{pmatrix} 0 + 2(a - 2) & 0 -3(a - 2)\\ 1 \cdot 3 + 2 \cdot 3a & 1 \cdot (-6) - 3 \cdot 3a \end{pmatrix}\\[/tex]

[tex]= \begin{pmatrix} 2a - 4 & 6 - 3a)\\ 6a + 3 & - 9a - 6 \end{pmatrix}\\[/tex]

[tex]B(3a) = \begin{pmatrix} 0 & 3a - 2\\ 1 & 9a \end{pmatrix}[/tex]

[tex]B(a) \cdot A + B(3a) = \begin{pmatrix} 2a - 4 & 6 - 3a)\\ 6a + 3 & - 9a - 6 \end{pmatrix} + \begin{pmatrix} 0 & 3a - 2\\ 1 & 9a \end{pmatrix} = \begin{pmatrix} 2(a - 2) & 4 \\ 2(3x + 2) & -6 \end{pmatrix}\\[/tex]

[tex]det \Big(B(a) \cdot A + B(3a)\Big) = 4 \iff \begin{vmatrix} 2(a - 2) & 4 \\ 2(3x + 2) & -6 \end{vmatrix} = 4[/tex]

[tex]-6 \cdot 2(a - 2) - 4 \cdot 2(3x + 2) = 4 \iff -12a + 24 -24a -16 = 4\\[/tex]

[tex]-36a + 8 = 4 \iff 36a = 4 \implies \boldsymbol{a = \dfrac{1}{9}}\\[/tex]