If the system of equations kx + y + 2z = 1,3x | vedclass

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(A) The given system of equations is:$1) kx + y + 2z = 1$$2) 3x - y - 2z = 2$$3) -2x - 2y - 4z = 3$For a system of linear equations to have infinitely

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$21$

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(A) The given system of equations is:$1) kx + y + 2z = 1$$2) 3x - y - 2z = 2$$3) -2x - 2y - 4z = 3$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 specific consistency conditions.Let the coefficient matrix be $A = \begin{bmatrix} k & 1 & 2 \ 3 & -1 & -2 \ -2 & -2 & -4 \end{bmatrix}$.Setting $|A| = 0$:$|A| = k((-1)(-4) - (-2)(-2)) - 1((3)(-4) - (-2)(-2)) + 2((3)(-2) - (-1)(-2))$$|A| = k(4 - 4) - 1(-12 - 4) + 2(-6 - 2)$$|A| = k(0) - 1(-16) + 2(-8) = 16 - 16 = 0$.Since the determinant is always $0$ regardless of $k$,we check for consistency using the augmented matrix $[A|B] = \begin{bmatrix} k & 1 & 2 & | & 1 \ 3 & -1 & -2 & | & 2 \ -2 & -2 & -4 & | & 3 \end{bmatrix}$.Adding equations $(2)$ and $(3)$:$(3x - y - 2z) + (-2x - 2y - 4z) = 2 + 3$$x - 3y - 6z = 5 \Rightarrow x = 3y + 6z + 5$.For infinitely many solutions,the system must be consistent. By observing the rows,we find that for the system to be consistent,the value of $k$ must satisfy the dependency of the planes. Specifically,adding the first two equations gives $(k+3)x = 3$. Solving for $k$ in the context of the augmented matrix consistency leads to $k = 21$.

If the system of equations $kx + y + 2z = 1$,$3x - y - 2z = 2$,and $-2x - 2y - 4z = 3$ has infinitely many solutions,then $k$ is equal to ..........

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