Consider the system of equations
$ x-2 y+3 z=-1 $ ; $ -x+y-2 z=k $ ; $ x-3 y+4 z=1$
$STATEMENT -1$ : The system of equations has no solution for $\mathrm{k} \neq 3$. and
$STATEMENT - 2$ : The determinant $\left|\begin{array}{ccc}1 & 3 & -1 \\ -1 & -2 & \mathrm{k} \\ 1 & 4 & 1\end{array}\right| \neq 0$, for $\mathrm{k} \neq 3$.
Consider the system of equations
$ x-2 y+3 z=-1 $ ; $ -x+y-2 z=k $ ; $ x-3 y+4 z=1$
$STATEMENT -1$ : The system of equations has no solution for $\mathrm{k} \neq 3$. and
$STATEMENT - 2$ : The determinant $\left|\begin{array}{ccc}1 & 3 & -1 \\ -1 & -2 & \mathrm{k} \\ 1 & 4 & 1\end{array}\right| \neq 0$, for $\mathrm{k} \neq 3$.
- [IIT 2008]
- A
Statement-$1$ is True, Statement -$2$ is True; Statement-$2$ is a correct explanation for Statement-$1$
- B
Statement -$1$ is True, Statement -$2$ is True; Statement-$2$ is $NOT$ a correct explanation for Statement-$1$
- C
Statement -$1$ is True, Statement -$2$ is False
- D
Statement -$1$ is False, Statement - $2$ is True
Similar Questions
The ordered pair $(a, b)$, for which the system of linear equations $3 x-2 y+z=b$ ; $5 x-8 y+9 z=3$ ; $2 x+y+a z=-1$ has no solution, is
The ordered pair $(a, b)$, for which the system of linear equations $3 x-2 y+z=b$ ; $5 x-8 y+9 z=3$ ; $2 x+y+a z=-1$ has no solution, is
- [JEE MAIN 2022]
If $A = \left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|,B = \left| {\,\begin{array}{*{20}{c}}1&1&1\\{{a^2}}&{{b^2}}&{{c^2}}\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|,C = \left| {\,\begin{array}{*{20}{c}}a&b&c\\{{a^2}}&{{b^2}}&{{c^2}}\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|,$ then which relation is correct
If $A = \left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|,B = \left| {\,\begin{array}{*{20}{c}}1&1&1\\{{a^2}}&{{b^2}}&{{c^2}}\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|,C = \left| {\,\begin{array}{*{20}{c}}a&b&c\\{{a^2}}&{{b^2}}&{{c^2}}\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|,$ then which relation is correct
The values of $a$ and $b$, for which the system of equations $2 x+3 y+6 z=8$ ; $x+2 y+a z=5$ ; $3 x+5 y+9 z=b$ has no solution, are:
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- [JEE MAIN 2021]
If $A = \left| {\,\begin{array}{*{20}{c}}{\sin (\theta + \alpha )}&{\cos (\theta + \alpha )}&1\\{\sin (\theta + \beta )}&{\cos (\theta + \beta )}&1\\{\sin (\theta + \gamma )}&{\cos (\theta + \gamma )}&1\end{array}\,} \right|$ ,then
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If $f\left( x \right) = \left| {\begin{array}{*{20}{c}}
{\sin \left( {x + \alpha } \right)}&{\sin \left( {x + \beta } \right)}&{\sin \left( {x + \gamma } \right)} \\
{\cos \left( {x + \alpha } \right)}&{\cos \left( {x + \beta } \right)}&{\cos \left( {x + \gamma } \right)} \\
{\sin \left( {\alpha + \beta } \right)}&{\sin \left( {\beta + \gamma } \right)}&{\sin \left( {\gamma + \alpha } \right)}
\end{array}} \right|$ and $f(10) = 10$ then $f(\pi)$ is equal to
If $f\left( x \right) = \left| {\begin{array}{*{20}{c}}
{\sin \left( {x + \alpha } \right)}&{\sin \left( {x + \beta } \right)}&{\sin \left( {x + \gamma } \right)} \\
{\cos \left( {x + \alpha } \right)}&{\cos \left( {x + \beta } \right)}&{\cos \left( {x + \gamma } \right)} \\
{\sin \left( {\alpha + \beta } \right)}&{\sin \left( {\beta + \gamma } \right)}&{\sin \left( {\gamma + \alpha } \right)}
\end{array}} \right|$ and $f(10) = 10$ then $f(\pi)$ is equal to