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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

unique solution satisfying $x+y=1$

Vedclass · Answer

(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:

Similar Questions

If $AX=D$ represents the system of linear equations $3x-4y+7z+6=0$, $5x+2y-4z+9=0$ and $8x-6y-z+5=0$, then

For real numbers $\alpha$ and $\beta$,consider the following system of linear equations:
$x+y-z=2, x+2y+\alpha z=1, 2x-y+z=\beta$. If the system has infinite solutions,then $\alpha+\beta$ is equal to $.....$

The number of $3 \times 3$ matrices $A$ whose entries are either $0$ or $1$ and for which the system $A\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ has exactly two distinct solutions,is

For $\alpha, \beta \in R$,suppose the system of linear equations $x-y+z=5$,$2x+2y+\alpha z=8$,and $3x-y+4z=\beta$ has infinitely many solutions. Then $\alpha$ and $\beta$ are the roots of:

If $x=\alpha, y=\beta, z=\gamma$ is the unique solution of the system of linear equations $2x-3y+5z=12$, $5x+2y+3z=11$, and $x+2y-3z=-3$, then $2\alpha+5\beta+3\gamma=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module