Consider the system of equations:x - 2y + 3z | vedclass

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(A) The system of equations is given by $AX = B$,where $A = \begin{bmatrix} 1 & -2 & 3 \ -1 & 1 & -2 \ 1 & -3 & 4 \end{bmatrix}$.First,calculate the

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Statement-$1$ is True,Statement-$2$ is True; Statement-$2$ is a correct explanation for Statement-$1$.

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(A) The system of equations is given by $AX = B$,where $A = \begin{bmatrix} 1 & -2 & 3 \ -1 & 1 & -2 \ 1 & -3 & 4 \end{bmatrix}$.First,calculate the determinant of the coefficient matrix $D = |A| = 1(4-6) + 2(-4+2) + 3(3-1) = 1(-2) + 2(-2) + 3(2) = -2 - 4 + 6 = 0$.Since $D = 0$,the system either has no solution or infinitely many solutions.Now,calculate $D_1 = \left|\begin{array}{ccc}-1 & -2 & 3 \ k & 1 & -2 \ 1 & -3 & 4\end{array}ight| = -1(4-6) + 2(4k+2) + 3(3k-1) = 2 + 8k + 4 + 9k - 3 = 17k + 3$.Wait,let's re-evaluate $D_1$: $D_1 = -1(4-6) + 2(4k+2) + 3(3k-1) = 2 + 8k + 4 + 9k - 3 = 17k + 3$. Actually,let's use row operations: $R_3 	o R_3 - R_1$ and $R_2 	o R_2 + R_1$: $\left|\begin{array}{ccc}1 & -2 & 3 \ 0 & -1 & 1 \ 0 & -1 & 1\end{array}ight| = 0$.For the system to have no solution,at least one of $D_1, D_2, D_3$ must be non-zero. $D_1 = \left|\begin{array}{ccc}-1 & -2 & 3 \ k & 1 & -2 \ 1 & -3 & 4\end{array}ight| = -1(4-6) + 2(4k+2) + 3(-3k-1) = 2 + 8k + 4 - 9k - 3 = 3 - k$.$D_1 = 0$ only if $k = 3$. If $k 
eq 3$,$D_1 
eq 0$,so the system has no solution.Thus,Statement-$1$ is True. Statement-$2$ is True as it correctly identifies $D=0$ which is the condition for the system to have no solution or infinite solutions.

List-$I$	List-$II$
$(P)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma=28$,then the system has	$(1)$ a unique solution
$(Q)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma \neq 28$,then the system has	$(2)$ no solution
$(R)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$,then the system has	$(3)$ infinitely many solutions
$(S)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma=28$,then the system has	$(4)$ $x=11, y=-2$ and $z=0$ as a solution
	$(5)$ $x=-15, y=4$ and $z=0$ as a solution

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

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