The system of equations x_1 - x_2 + x_3 = 2,3 | vedclass

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(C) The system of equations is given by:$x_1 - x_2 + x_3 = 2$$3x_1 - x_2 + 2x_3 = -6$$3x_1 + x_2 + x_3 = -18$First,we calculate the determinant $D$ of

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(C) The system of equations is given by:$x_1 - x_2 + x_3 = 2$$3x_1 - x_2 + 2x_3 = -6$$3x_1 + x_2 + x_3 = -18$First,we calculate the determinant $D$ of the coefficient matrix:$D = \begin{vmatrix} 1 & -1 & 1 \ 3 & -1 & 2 \ 3 & 1 & 1 \end{vmatrix} = 1(-1 - 2) - (-1)(3 - 6) + 1(3 + 3) = 1(-3) + 1(-3) + 1(6) = -3 - 3 + 6 = 0$Next,we calculate $D_1$ by replacing the first column with the constants:$D_1 = \begin{vmatrix} 2 & -1 & 1 \ -6 & -1 & 2 \ -18 & 1 & 1 \end{vmatrix} = 2(-1 - 2) - (-1)(-6 + 36) + 1(-6 - 18) = 2(-3) + 1(30) - 24 = -6 + 30 - 24 = 0$Similarly,calculating $D_2$ and $D_3$:$D_2 = \begin{vmatrix} 1 & 2 & 1 \ 3 & -6 & 2 \ 3 & -18 & 1 \end{vmatrix} = 1(-6 + 36) - 2(3 - 6) + 1(-54 + 18) = 30 - 2(-3) - 36 = 30 + 6 - 36 = 0$$D_3 = \begin{vmatrix} 1 & -1 & 2 \ 3 & -1 & -6 \ 3 & 1 & -18 \end{vmatrix} = 1(18 + 6) - (-1)(-54 + 18) + 2(3 + 3) = 24 - 36 + 12 = 0$Since $D = D_1 = D_2 = D_3 = 0$,the system is consistent and has infinitely many solutions.

The system of equations $x_1 - x_2 + x_3 = 2$,$3x_1 - x_2 + 2x_3 = -6$ and $3x_1 + x_2 + x_3 = -18$ has

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