If the system of equations x - ky - z = 0,kx | vedclass

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

Question

(D) For a system of linear equations to have a non-zero (non-trivial) solution,the determinant of the coefficient matrix must be equal to zero,i.e.,$\

Vedclass · Accepted Answer

$-1, 1$

Vedclass · Answer

(D) For a system of linear equations to have a non-zero (non-trivial) solution,the determinant of the coefficient matrix must be equal to zero,i.e.,$\Delta = 0$.The system is given by:$x - ky - z = 0$$kx - y - z = 0$$x + y - z = 0$The determinant $\Delta$ is:$\Delta = \begin{vmatrix} 1 & -k & -1 \ k & -1 & -1 \ 1 & 1 & -1 \end{vmatrix} = 0$Expanding along the first row:$1((-1)(-1) - (-1)(1)) - (-k)((k)(-1) - (-1)(1)) + (-1)((k)(1) - (-1)(1)) = 0$$1(1 + 1) + k(-k + 1) - 1(k + 1) = 0$$2 - k^2 + k - k - 1 = 0$$1 - k^2 = 0$$k^2 = 1$$k = \pm 1$Thus,the possible values of $k$ are $1$ and $-1$.

If the system of equations $x - ky - z = 0$,$kx - y - z = 0$ and $x + y - z = 0$ has a non-zero solution,then the possible values of $k$ are:

Similar Questions

If $\left|\begin{array}{ccc}0 & ab^2 & ac^2 \\ a^2b & 0 & bc^2 \\ a^2c & b^2c & 0\end{array}\right|=m(abc)^k$,then $m+k=$ . . . . . . .

Find the values of $k$ if the area of the triangle is $4$ square units and the vertices are $(-2, 0), (0, 4), (0, k)$.

The solution$(s)$ of the equation $\left| \begin{matrix} x & a & b \\ a & x & a \\ b & b & x \end{matrix} \right| = 0$ is/are:

If $\left|\begin{array}{ccc}\cos (A+B) & -\sin (A+B) & \cos 2 B \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B\end{array}\right|=0$, then $B$ is equal to

If $\left| \begin{array}{ccc} 3x - 8 & 3 & 3 \\ 3 & 3x - 8 & 3 \\ 3 & 3 & 3x - 8 \end{array} \right| = 0$,then the values of $x$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module