If the system of equations kx + (k+1)y + (k-1 | vedclass

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

Question

(B) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero. Let $D$ be the

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(B) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero. Let $D$ be the determinant of the coefficient matrix: $D = \begin{vmatrix} k & k+1 & k-1 \ k-1 & k+2 & k \ k+1 & k & k+2 \end{vmatrix} = 0$. Applying column operations $C_1 	o C_1 + C_2 + C_3$: $D = \begin{vmatrix} 3k & k+1 & k-1 \ 3k+1 & k+2 & k \ 3k+3 & k & k+2 \end{vmatrix} = 0$. This simplifies to $D = (3k+1) \begin{vmatrix} 1 & k+1 & k-1 \ 1 & k+2 & k \ 1 & k & k+2 \end{vmatrix} = 0$. Subtracting $R_1$ from $R_2$ and $R_3$: $D = (3k+1) \begin{vmatrix} 1 & k+1 & k-1 \ 0 & 1 & 1 \ 0 & -1 & 3 \end{vmatrix} = 0$. Expanding the determinant: $(3k+1) [1(3 - (-1))] = 0$. $(3k+1)(4) = 0$. $3k+1 = 0 \implies k = -\frac{1}{3}$. Wait,re-evaluating the determinant: $D = k((k+2)(k+2) - k^2) - (k+1)((k-1)(k+2) - k(k+1)) + (k-1)(k(k-1) - (k+2)(k+1)) = 0$. $D = k(k^2+4k+4-k^2) - (k+1)(k^2+k-2-k^2-k) + (k-1)(k^2-k-k^2-3k-2) = 0$. $D = k(4k+4) - (k+1)(-2) + (k-1)(-4k-2) = 0$. $4k^2+4k + 2k+2 - 4k^2-2k+4k+2 = 0$. $8k + 4 = 0 \implies 8k = -4 \implies k = -\frac{1}{2}$. Thus,the only possible value for $k$ is $-\frac{1}{2}$,and the sum of all possible values is $-\frac{1}{2}$.

If the system of equations $kx + (k+1)y + (k-1)z = 0$,$(k-1)x + (k+2)y + kz = 0$,and $(k+1)x + ky + (k+2)z = 0$ has a non-trivial solution,then the sum of all possible values of $k$ is:

Similar Questions

If $A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$,then $A^{2} + xA + yI = 0$ for $(x, y)$ is

The system of equations $3x + 2y + z = 6$, $3x + 4y + 3z = 14$ and $6x + 10y + 8z = a$ has an infinite number of solutions if $a$ is equal to

Let $A = \begin{bmatrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{bmatrix}$. If $A^3 = 4A^2 - A - 21I$,where $I$ is the identity matrix of order $3 \times 3$,then $2a + 3b$ is equal to:

Consider the system of equations:
$x - 2y + 3z = -1$; $-x + y - 2z = k$; $x - 3y + 4z = 1$
$STATEMENT-1$: The system of equations has no solution for $k \neq 3$.
$STATEMENT-2$: The determinant $\left|\begin{array}{ccc}1 & -2 & 3 \\ -1 & 1 & -2 \\ 1 & -3 & 4\end{array}\right| = 0$.

The number of solutions of the equations $x + y - z = 0$,$3x - y - z = 0$,and $x - 3y + z = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module