If (x, y, z)=(alpha, beta, gamma) is the uniq | vedclass

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(D) The given system of equations is: $3x - 4y + z = -7$ $2x + 3y - z = 10$ $x - 2y - 3z = 3$ Representing in matrix form $AX = B$: $A = \begin{bmatri

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(D) The given system of equations is: $3x - 4y + z = -7$ $2x + 3y - z = 10$ $x - 2y - 3z = 3$ Representing in matrix form $AX = B$: $A = \begin{bmatrix} 3 & -4 & 1 \ 2 & 3 & -1 \ 1 & -2 & -3 \end{bmatrix}, X = \begin{bmatrix} x \ y \ z \end{bmatrix}, B = \begin{bmatrix} -7 \ 10 \ 3 \end{bmatrix}$ Calculating the determinant $|A|$: $|A| = 3(-9 - 2) + 4(-6 + 1) + 1(-4 - 3) = 3(-11) + 4(-5) + 1(-7) = -33 - 20 - 7 = -60$ Since $|A| 
eq 0$, the system has a unique solution. Using Cramer's Rule for $x = \alpha$: $D_x = \begin{vmatrix} -7 & -4 & 1 \ 10 & 3 & -1 \ 3 & -2 & -3 \end{vmatrix} = -7(-9 - 2) + 4(-30 + 3) + 1(-20 - 9) = -7(-11) + 4(-27) + 1(-29) = 77 - 108 - 29 = -60$ Thus, $\alpha = \frac{D_x}{|A|} = \frac{-60}{-60} = 1$.

If $(x, y, z)=(\alpha, \beta, \gamma)$ is the unique solution of the system of simultaneous linear equations $3x - 4y + z + 7 = 0$, $2x + 3y - z = 10$, and $x - 2y - 3z = 3$, then $\alpha = $

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