Let $S$ be the set of all $\lambda \in R$ for which the system of linear equations $$2x - y + 2z = 2$$ $$x - 2y + \lambda z = -4$$ $$x + \lambda y + z = 4$$ has no solution. Then the set $S$ (1) Contains more than two elements (2) Is an empty set (3) Is a singleton (4) Contains exactly two elements
Let $A = \left\{ X = ( x , y , z ) ^ { T } : P X = 0 \text{ and } x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 1 \right\}$ where $P = \left[ \begin{array} { c c c } 1 & 2 & 1 \\ - 2 & 3 & - 4 \\ 1 & 9 & - 1 \end{array} \right]$ then the set $A$ (1) Is a singleton. (2) Is an empty set. (3) Contains more than two elements (4) Contains exactly two elements
If the system of equations $x + y + z = 2$ $2 x + 4 y - z = 6$ $3 x + 2 y + \lambda z = \mu$ has infinitely many solutions, then: (1) $\lambda + 2 \mu = 14$ (2) $2 \lambda - \mu = 5$ (3) $\lambda - 2 \mu = - 5$ (4) $2 \lambda + \mu = 14$
If the system of linear equations $$x + y + 3z = 0$$ $$x + 3y + k^2z = 0$$ $$3x + y + 3z = 0$$ has a non-zero solution $(x, y, z)$ for some $k \in \mathrm{R}$, then $x + \left(\frac{y}{z}\right)$ is equal to: (1) $-3$ (2) $9$ (3) $3$ (4) $-9$
The values of $\lambda$ and $\mu$ for which the system of linear equations $x + y + z = 2 , x + 2 y + 3 z = 5$, $x + 3 y + \lambda z = \mu$ has infinitely many solutions, are respectively (1) 6 and 8 (2) 5 and 7 (3) 5 and 8 (4) 4 and 9
For the system of linear equations: $$x - 2 y = 1 , x - y + k z = - 2 , k y + 4 z = 6 , k \in R$$ Consider the following statements: (A) The system has unique solution if $k \neq 2 , k \neq - 2$. (B) The system has unique solution if $k = - 2$. (C) The system has unique solution if $k = 2$. (D) The system has no-solution if $k = 2$. (E) The system has infinitely many solutions if $k = - 2$.
Consider the following system of equations: $$\begin{aligned}
& x + 2 y - 3 z = a \\
& 2 x + 6 y - 11 z = b \\
& x - 2 y + 7 z = c
\end{aligned}$$ where $a , b$ and $c$ are real constants. Then the system of equations : (1) has a unique solution when $5 a = 2 b + c$ (2) has no solution for all $a , b$ and $c$ (3) has infinite number of solutions when $5 a = 2 b + c$ (4) has a unique solution for all $a , b$ and $c$
The value of $k \in R$, for which the following system of linear equations $3 x - y + 4 z = 3$ $x + 2 y - 3 z = - 2$ $6 x + 5 y + k z = - 3$ has infinitely many solutions, is: (1) 3 (2) - 5 (3) 5 (4) - 3
Let the system of linear equations $x + y + a z = 2$ $3 x + y + z = 4$ $x + 2 z = 1$ have a unique solution $\left( x ^ { * } , y ^ { * } , z ^ { * } \right)$. If $\left( \left( a , x ^ { * } \right) , \left( y ^ { * } , \alpha \right) \right.$ and $\left( x ^ { * } , - y ^ { * } \right)$ are collinear points, then the sum of absolute values of all possible values of $\alpha$ is: (1) 4 (2) 3 (3) 2 (4) 1
The number of values of $\alpha$ for which the system of equations $x + y + z = \alpha$ $\alpha x + 2 \alpha y + 3 z = - 1$ $x + 3 \alpha y + 5 z = 4$ is inconsistent, is (1) 0 (2) 1 (3) 2 (4) 3
Let $A$ be a $3 \times 3$ real matrix such that $A \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$; $A \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ and $A \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$. If $X = \begin{pmatrix} x _ { 1 } \\ x _ { 2 } \\ x _ { 3 } \end{pmatrix}$ and $I$ is an identity matrix of order 3, then the system $( A - 2I ) X = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ has (1) no solution (2) infinitely many solutions (3) unique solution (4) exactly two solutions
If the system of equations $\alpha x + y + z = 5 , x + 2 y + 3 z = 4 , x + 3 y + 5 z = \beta$ has infinitely many solutions, then the ordered pair $( \alpha , \beta )$ is equal to (1) $( 1 , - 3 )$ (2) $( - 1,3 )$ (3) $( 1,3 )$ (4) $( - 1 , - 3 )$
The system of equations $-kx + 3y - 14z = 25$ $-15x + 4y - kz = 3$ $-4x + y + 3z = 4$ is consistent for all $k$ in the set (1) $R$ (2) $R - \{-11, 13\}$ (3) $R - \{-13\}$ (4) $R - \{-11, 11\}$
If the system of linear equations $2x + 3y - z = -2$ $x + y + z = 4$ $x - y + |\lambda|z = 4\lambda - 4$ where $\lambda \in \mathbb{R}$, has no solution, then (1) $\lambda = 7$ (2) $\lambda = -7$ (3) $\lambda = 8$ (4) $\lambda^2 = 1$
If the system of linear equations $2 x + y - z = 7$ $x - 3 y + 2 z = 1$ $x + 4 y + \delta z = k$, where $\delta , k \in R$ has infinitely many solutions, then $\delta + k$ is equal to (1) $- 3$ (2) 3 (3) 6 (4) 9
The number of real values of $\lambda$, such that the system of linear equations $2x - 3y + 5z = 9$ $x + 3y - z = -18$ $3x - y + (\lambda^2 - |\lambda|)z = 16$ has no solutions, is (1) 0 (2) 1 (3) 2 (4) 4
If the system of equations $x + y + z = 6$ $2x + 5y + \alpha z = \beta$ $x + 2y + 3z = 14$ has infinitely many solutions, then $\alpha + \beta$ is equal to (1) 8 (2) 36 (3) 44 (4) 48
Let $N$ denote the number that turns up when a fair die is rolled. If the probability that the system of equations $x + y + z = 1$, $2 x + N y + 2 z = 2$, $3 x + 3 y + N z = 3$ has unique solution is $\frac { k } { 6 }$, then the sum of value of $k$ and all possible values of $N$ is (1) 18 (2) 19 (3) 20 (4) 21
For the system of equations $x + y + z = 6$ $x + 2y + \alpha z = 10$ $x + 3y + 5z = \beta$, which one of the following is NOT true? (1) System has no solution for $\alpha = 3, \beta = 24$ (2) System has a unique solution for $\alpha = -3, \beta = 14$ (3) System has infinitely many solutions for $\alpha = 3, \beta = 14$ (4) System has a unique solution for $\alpha = 3, \beta = 14$
Let $S$ denote the set of all real values of $\lambda$ such that the system of equations $$\lambda x + y + z = 1$$ $$x + \lambda y + z = 1$$ $$x + y + \lambda z = 1$$ is inconsistent, then $\sum_{\lambda \in S} (\lambda^2 + \lambda)$ is equal to (1) 2 (2) 12 (3) 4 (4) 6
Let A be a symmetric matrix such that $| A | = 2$ and $\left[ \begin{array} { l l } 2 & 1 \\ 3 & \frac { 3 } { 2 } \end{array} \right] A = \left[ \begin{array} { l l } 1 & 2 \\ \alpha & \beta \end{array} \right]$. If the sum of the diagonal elements of A is $s$, then $\frac { \beta s } { \alpha ^ { 2 } }$ is equal to $\_\_\_\_$.