If A is a matrix such that left[begin{array}{ | vedclass

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(D) Let $A = \left[\begin{array}{l} x_1 \ x_2 \end{array}ight]$.Given the equation $\left[\begin{array}{ll} 2 & 1 \ 3 & 2 \end{array}ight] \left

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$\left[\begin{array}{r} 2 \ -3 \end{array}ight]$

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(D) Let $A = \left[\begin{array}{l} x_1 \ x_2 \end{array}ight]$.Given the equation $\left[\begin{array}{ll} 2 & 1 \ 3 & 2 \end{array}ight] \left[\begin{array}{l} x_1 \ x_2 \end{array}ight] = \left[\begin{array}{l} 1 \ 0 \end{array}ight]$.Multiplying the matrices on the left side:$\left[\begin{array}{c} 2x_1 + x_2 \ 3x_1 + 2x_2 \end{array}ight] = \left[\begin{array}{l} 1 \ 0 \end{array}ight]$.Equating the corresponding elements,we get the system of linear equations:$2x_1 + x_2 = 1$ (Equation $1$)$3x_1 + 2x_2 = 0$ (Equation $2$)From Equation $1$,$x_2 = 1 - 2x_1$.Substituting this into Equation $2$:$3x_1 + 2(1 - 2x_1) = 0$$3x_1 + 2 - 4x_1 = 0$$-x_1 + 2 = 0 \Rightarrow x_1 = 2$.Now,finding $x_2$:$x_2 = 1 - 2(2) = 1 - 4 = -3$.Therefore,$A = \left[\begin{array}{r} 2 \ -3 \end{array}ight]$.

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

If $A$ is a matrix such that $\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] A \left[\begin{array}{l} 1 \\ 1 \end{array}\right] = \left[\begin{array}{l} 1 \\ 0 \end{array}\right]$,then $A$ is equal to

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