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(A) Given the matrix equation: $\begin{bmatrix} \alpha & \beta & \gamma \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \ 2 & 3 & -5 \ 1 & 2 & 5 \end{bmatri

Vedclass · Accepted Answer

$8$

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(A) Given the matrix equation: $\begin{bmatrix} \alpha & \beta & \gamma \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \ 2 & 3 & -5 \ 1 & 2 & 5 \end{bmatrix} = \begin{bmatrix} 3 & 5 & 2 \end{bmatrix}$ Performing matrix multiplication, we get: $\begin{bmatrix} \alpha + 2\beta + \gamma & 2\alpha + 3\beta + 2\gamma & 3\alpha - 5\beta + 5\gamma \end{bmatrix} = \begin{bmatrix} 3 & 5 & 2 \end{bmatrix}$ Equating the corresponding elements, we obtain the system of linear equations: $1) \alpha + 2\beta + \gamma = 3$ $2) 2\alpha + 3\beta + 2\gamma = 5$ $3) 3\alpha - 5\beta + 5\gamma = 2$ Subtracting twice the first equation from the second: $(2\alpha + 3\beta + 2\gamma) - 2(\alpha + 2\beta + \gamma) = 5 - 2(3) \Rightarrow -\beta = -1 \Rightarrow \beta = 1$. Substituting $\beta = 1$ into equations $(1)$ and $(3)$: $\alpha + \gamma = 3 - 2(1) = 1 \Rightarrow \alpha + \gamma = 1$ $3\alpha + 5\gamma = 2 + 5(1) = 7 \Rightarrow 3\alpha + 5\gamma = 7$ Solving these two equations: $3(1 - \gamma) + 5\gamma = 7 \Rightarrow 3 - 3\gamma + 5\gamma = 7 \Rightarrow 2\gamma = 4 \Rightarrow \gamma = 2$. Then $\alpha = 1 - 2 = -1$. Thus, $\alpha = -1, \beta = 1, \gamma = 2$. Finally, $\alpha^3 + \beta^3 + \gamma^3 = (-1)^3 + (1)^3 + (2)^3 = -1 + 1 + 8 = 8$.

If $\begin{bmatrix} \alpha & \beta & \gamma \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & -5 \\ 1 & 2 & 5 \end{bmatrix} = \begin{bmatrix} 3 & 5 & 2 \end{bmatrix}$, then $\alpha^3 + \beta^3 + \gamma^3 = $

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