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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 infinite number of solutions if $k \neq-2$
Which of the following statements are correct?
$(C)$ and $(D)$ only
$(B)$ and $(E)$ only
$(A)$ and $(E)$ only
$(A)$ and $(D)$ only
Solution
$D =\left|\begin{array}{ccc}1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4\end{array}\right|=4- k ^{2}$
so, $A$ is correct and $B, C, E$ are incorrect. If $k =2$
$D_{1}=\left|\begin{array}{ccc}1 & -2 & 0 \\ -2 & -1 & 2 \\ 6 & 2 & 4\end{array}\right|=-48 \neq 0$
So no solution
$D$ is correct.
Similar Questions
Let $\alpha$ and $\beta$ be the distinct roots of the equation $x^2+x-1=0$. Consider the set $T=\{1, \alpha, \beta\}$. For a $3 \times 3$ matrix $M=\left(a_{\ell}\right) 3 \times 3_3$, define $R_l=a_{l 1}+a_{l 2}+a_\beta$ and $C_j=a_{1 j}+a_{2 l}+a_{3 j}$ for $i=1,2,3$ and $j=1,2,3$
Match each entry in $List-I$ to the correct entry in $List-II$.
$List-I$ | $List-II$ |
($P$) The number of matrices $M=\left(a_{i j}\right)_3 \times 3$ with all entries in $T$ such that $R_i=C_j=0$ for all $i, j$ is | ($1$) ($1$) |
($Q$) The number of symmetric matrices $M=\left(a_{i j}\right) 3 \times 3$ with all entries in $T$ such that $C_j=0$ for all $j$ is | ($2$) ($2$) |
($R$) Let $M=\left(a_{i j}\right) 3 \times 3$ be a skew symmetric matrix such that $a_{i j} \in T$ for $i>j$. Then the number of elements in the set $\left\{\left(\begin{array}{l}x \\ y \\ z\end{array}\right): x, y \cdot z \in R, M\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{c}a_{12} \\ 0 \\ -a_{23}\end{array}\right)\right\}$ is is | ($3$) Infinite |
($S$) Let $M=\left(a_{i j}\right)_3 \times 3$ be a matrix with all entries in $T$ such that $R_i=0$ for all $i$. Then the absolute value of the determinant of $M$ is | ($4$) ($6$) |
($5$) ($0$) |
The correct option is