1. **State the problem:** We need to evaluate the matrix product and simplify the expression: $$\left(\begin{array}{cc} b^{\alpha} & a_{b_{\alpha}}^{-1} \end{array}\right) \left(\begin{array}{c} 4 \ln c - 3 \end{array}\right) = -3 \ln e \cdot -1 \ln \left(\frac{q}{b_s}\right)$$ 2. **Understand the matrix multiplication:** The first matrix is a 1x2 row vector and the second is a 1x1 column vector, which is not conformable for multiplication. Likely, the second matrix should be a 2x1 column vector for multiplication to be valid. Assuming the second matrix is a 2x1 vector with entries $4 \ln c$ and $-3$: $$\left(\begin{array}{cc} b^{\alpha} & a_{b_{\alpha}}^{-1} \end{array}\right) \left(\begin{array}{c} 4 \ln c \\ -3 \end{array}\right)$$ 3. **Matrix multiplication formula:** $$\text{Result} = b^{\alpha} \cdot (4 \ln c) + a_{b_{\alpha}}^{-1} \cdot (-3)$$ 4. **Simplify the expression:** $$= 4 b^{\alpha} \ln c - 3 a_{b_{\alpha}}^{-1}$$ 5. **Simplify the right side:** Recall $\ln e = 1$, so: $$-3 \ln e \cdot -1 \ln \left(\frac{q}{b_s}\right) = -3 \cdot 1 \cdot -1 \ln \left(\frac{q}{b_s}\right) = 3 \ln \left(\frac{q}{b_s}\right)$$ 6. **Equate both sides:** $$4 b^{\alpha} \ln c - 3 a_{b_{\alpha}}^{-1} = 3 \ln \left(\frac{q}{b_s}\right)$$ This is the simplified form of the given expression. **Final answer:** $$4 b^{\alpha} \ln c - 3 a_{b_{\alpha}}^{-1} = 3 \ln \left(\frac{q}{b_s}\right)$$