If the unique solution of the simultaneous li | vedclass

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

Question

(B) Given the system of equations: $3x - 2y + z = 5k$ $2x + 3y - 2z = -5k$ $x + 4y + 3z = k$ First, calculate the determinant $D$ of the coefficient m

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given the system of equations: $3x - 2y + z = 5k$ $2x + 3y - 2z = -5k$ $x + 4y + 3z = k$ First, calculate the determinant $D$ of the coefficient matrix: $D = \begin{vmatrix} 3 & -2 & 1 \ 2 & 3 & -2 \ 1 & 4 & 3 \end{vmatrix} = 3(9 + 8) + 2(6 + 2) + 1(8 - 3) = 3(17) + 2(8) + 1(5) = 51 + 16 + 5 = 72$. Next, calculate $D_3$ using Cramer's rule: $D_3 = \begin{vmatrix} 3 & -2 & 5k \ 2 & 3 & -5k \ 1 & 4 & k \end{vmatrix} = k \begin{vmatrix} 3 & -2 & 5 \ 2 & 3 & -5 \ 1 & 4 & 1 \end{vmatrix} = k [3(3 + 20) + 2(2 + 5) + 5(8 - 3)] = k [3(23) + 2(7) + 5(5)] = k [69 + 14 + 25] = 108k$. Since $z = \frac{D_3}{D} = 3$, we have: $\frac{108k}{72} = 3$ $\frac{3k}{2} = 3$ $3k = 6$ $k = 2$.

If the unique solution of the simultaneous linear equations $3x - 2y + z = 5k$, $2x + 3y - 2z = -5k$, and $x + 4y + 3z = k$ is $x = \alpha, y = \beta, z = 3$, then $k =$

Similar Questions

The number of non-trivial solutions of the system $x-y+z=0, x+2y-z=0, 2x+y+3z=0$ is

The system of equations $kx + 2y - z = 1$,$(k - 1)y - 2z = 2$,and $(k + 2)z = 3$ has a unique solution if $k$ is equal to:

If the system of equations $x + y - z = 0, 3x - \alpha y - 3z = 0, x - 3y + z = 0$ has a non-zero solution,then $\alpha = $

If the system of equations $3x + y + 4z = 3$, $2x + ay - z = -3$, $x + 2y + z = 4$ has no solution, then the value of $a$ is equal to:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module