Let alpha, beta, gamma be real numbers. If be | vedclass

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(A) Performing the matrix multiplication on the left side: $\begin{bmatrix} 7(1) + 5(3) + \alpha(2) \ \beta(1) + 2(3) + 11(2) \ 3(1) + \gamma(3) + 1

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$27$

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(A) Performing the matrix multiplication on the left side: $\begin{bmatrix} 7(1) + 5(3) + \alpha(2) \ \beta(1) + 2(3) + 11(2) \ 3(1) + \gamma(3) + 1(2) \end{bmatrix} = \begin{bmatrix} 22 + 2\alpha \ \beta + 28 \ 5 + 3\gamma \end{bmatrix}$. Equating this to the right side matrix: $1) \ 22 + 2\alpha = \alpha + \beta \implies \alpha - \beta = -22$ $2) \ \beta + 28 = -2\alpha + \beta - 2\gamma \implies 2\alpha + 2\gamma = -28 \implies \alpha + \gamma = -14$ $3) \ 5 + 3\gamma = \alpha + 2\beta + 3\gamma \implies \alpha + 2\beta = 5$ From $(1)$, $\beta = \alpha + 22$. Substituting into $(3)$: $\alpha + 2(\alpha + 22) = 5 \implies 3\alpha + 44 = 5 \implies 3\alpha = -39 \implies \alpha = -13$. Then $\beta = -13 + 22 = 9$. From $(2)$, $\gamma = -14 - \alpha = -14 - (-13) = -1$. Now, calculate $100 + \frac{2\alpha + 11\beta}{\gamma} = 100 + \frac{2(-13) + 11(9)}{-1} = 100 + \frac{-26 + 99}{-1} = 100 + \frac{73}{-1} = 100 - 73 = 27$.

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