Let the system of linear equations x+y+kz=2; | vedclass

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(D) For the system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix must be zero:$\Delta = \left|\begin

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unique solution satisfying $x+y=1$

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(D) For the system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix must be zero:$\Delta = \left|\begin{array}{ccc}1 & 1 & k \ 2 & 3 & -1 \ 3 & 4 & 2\end{array}ight| = 0$Expanding along the first row:$1(3 	imes 2 - (-1) 	imes 4) - 1(2 	imes 2 - (-1) 	imes 3) + k(2 	imes 4 - 3 	imes 3) = 0$$1(6 + 4) - 1(4 + 3) + k(8 - 9) = 0$$10 - 7 - k = 0$$3 - k = 0 \Rightarrow k = 3$Now,substitute $k = 3$ into the second system of equations:$(3+1)x + (2 	imes 3 - 1)y = 7 \Rightarrow 4x + 5y = 7 \dots (1)$$(2 	imes 3 + 1)x + (3+5)y = 10 \Rightarrow 7x + 8y = 10 \dots (2)$To check for the nature of the solution,find the determinant of the coefficient matrix of this system:$D = \left|\begin{array}{cc}4 & 5 \ 7 & 8\end{array}ight| = 32 - 35 = -3 
eq 0$Since $D 
eq 0$,the system has a unique solution.Solving the equations:$(2) - (1) \Rightarrow (7x + 8y) - (4x + 5y) = 10 - 7$$3x + 3y = 3 \Rightarrow x + y = 1$Thus,the system has a unique solution satisfying $x+y=1$.

Let the system of linear equations $x+y+kz=2$; $2x+3y-z=1$; $3x+4y+2z=k$ have infinitely many solutions. Then the system $(k+1)x+(2k-1)y=7$; $(2k+1)x+(k+5)y=10$ has:

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