The equations x-y+2z=4,3x+y+4z=6,and x+y+z=1 | vedclass

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(B) Given system of equations: $x-y+2z=4$ $3x+y+4z=6$ $x+y+z=1$ Let $\Delta = \begin{vmatrix} 1 & -1 & 2 \ 3 & 1 & 4 \ 1 & 1 & 1 \end{vmatrix}$ $= 1

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infinitely many solutions

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(B) Given system of equations: $x-y+2z=4$ $3x+y+4z=6$ $x+y+z=1$ Let $\Delta = \begin{vmatrix} 1 & -1 & 2 \ 3 & 1 & 4 \ 1 & 1 & 1 \end{vmatrix}$ $= 1(1-4) - (-1)(3-4) + 2(3-1)$ $= 1(-3) + 1(-1) + 2(2) = -3 - 1 + 4 = 0$ Since $\Delta = 0$,we check $\Delta_1, \Delta_2, \Delta_3$: $\Delta_1 = \begin{vmatrix} 4 & -1 & 2 \ 6 & 1 & 4 \ 1 & 1 & 1 \end{vmatrix} = 4(1-4) + 1(6-4) + 2(6-1) = 4(-3) + 1(2) + 2(5) = -12 + 2 + 10 = 0$ $\Delta_2 = \begin{vmatrix} 1 & 4 & 2 \ 3 & 6 & 4 \ 1 & 1 & 1 \end{vmatrix} = 1(6-4) - 4(3-4) + 2(3-6) = 1(2) - 4(-1) + 2(-3) = 2 + 4 - 6 = 0$ $\Delta_3 = \begin{vmatrix} 1 & -1 & 4 \ 3 & 1 & 6 \ 1 & 1 & 1 \end{vmatrix} = 1(1-6) + 1(3-6) + 4(3-1) = 1(-5) + 1(-3) + 4(2) = -5 - 3 + 8 = 0$ Since $\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$,the system has infinitely many solutions.

The equations $x-y+2z=4$,$3x+y+4z=6$,and $x+y+z=1$ have

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