The system of equations a + b - 2c = 0,2a - 3 | vedclass

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(B) The given system of equations is:$a + b - 2c = 0$$2a - 3b + c = 0$$a - 5b + 4c = \alpha$For a system of linear equations to be consistent,the dete

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(B) The given system of equations is:$a + b - 2c = 0$$2a - 3b + c = 0$$a - 5b + 4c = \alpha$For a system of linear equations to be consistent,the determinant of the coefficient matrix $D$ must be zero,and the augmented determinants $D_1, D_2, D_3$ must also be zero.First,calculate the determinant of the coefficient matrix $D$:$D = \begin{vmatrix} 1 & 1 & -2 \ 2 & -3 & 1 \ 1 & -5 & 4 \end{vmatrix}$$D = 1((-3)(4) - (1)(-5)) - 1((2)(4) - (1)(1)) - 2((2)(-5) - (-3)(1))$$D = 1(-12 + 5) - 1(8 - 1) - 2(-10 + 3)$$D = 1(-7) - 1(7) - 2(-7) = -7 - 7 + 14 = 0$Since $D = 0$,the system is consistent if the determinant $D_1$ (replacing the first column with the constants) is zero:$D_1 = \begin{vmatrix} 0 & 1 & -2 \ 0 & -3 & 1 \ \alpha & -5 & 4 \end{vmatrix} = 0$Expanding along the first column:$D_1 = \alpha \begin{vmatrix} 1 & -2 \ -3 & 1 \end{vmatrix} = \alpha(1 - 6) = -5\alpha$Setting $D_1 = 0$,we get $-5\alpha = 0$,which implies $\alpha = 0$.

The system of equations $a + b - 2c = 0$,$2a - 3b + c = 0$,and $a - 5b + 4c = \alpha$ is consistent for $\alpha$ equal to

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