If the system of linear equations 2x + y - z | vedclass

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(B) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix must be zero,and the augmented matrix

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$3$

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(B) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix must be zero,and the augmented matrix must satisfy the consistency condition.First,we set the determinant of the coefficient matrix to zero:$\Delta = \begin{vmatrix} 2 & 1 & -1 \ 1 & -3 & 2 \ 1 & 4 & \delta \end{vmatrix} = 0$$2(-3\delta - 8) - 1(\delta - 2) - 1(4 + 3) = 0$$-6\delta - 16 - \delta + 2 - 7 = 0$$-7\delta - 21 = 0 \Rightarrow \delta = -3$Now,for the system to have infinitely many solutions,the augmented matrix must also have a determinant of zero when replacing a column with the constants:$\Delta_x = \begin{vmatrix} 7 & 1 & -1 \ 1 & -3 & 2 \ k & 4 & -3 \end{vmatrix} = 0$$7(9 - 8) - 1(-3 - 2k) - 1(4 + 3k) = 0$$7(1) + 3 + 2k - 4 - 3k = 0$$6 - k = 0 \Rightarrow k = 6$Thus,$\delta + k = -3 + 6 = 3$.

If the system of linear equations $2x + y - z = 7$,$x - 3y + 2z = 1$,and $x + 4y + \delta z = k$,where $\delta, k \in R$,has infinitely many solutions,then $\delta + k$ is equal to

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