Gujarati
Hindi
3 and 4 .Determinants and Matrices
normal

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 $\mathrm{k} \neq 3$.

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

(IIT-2008)

Solution

$D=\left|\begin{array}{ccc}1 & -2 & 3 \\ -1 & 1 & -2 \\ 1 & -3 & 4\end{array}\right|=0$

and $D_1=\left|\begin{array}{ccc}-1 & -2 & 3 \\ k & 1 & -2 \\ 1 & -3 & 4\end{array}\right|=(3-k)=0$ if $k=3$

$D_2=\left|\begin{array}{ccc}1 & -1 & 3 \\ -1 & k & -2 \\ 1 & 1 & 4\end{array}\right|=(k-3)=0$, if $k=3$

$D_3=\left|\begin{array}{ccc}1 & -2 & -1 \\ -1 & 1 & k \\ 1 & -3 & 1\end{array}\right|=(k-3)=0$, if $k=3$

$\Rightarrow$ system of equations has no solution for $\mathrm{k} \neq 3$.

Standard 12
Mathematics

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