Given question is incorrect and all the other answers are also incorrect because question says one option is correct but both (C) and (D) are correct here.
You can decompose the coefficient matrix in two ways as following:
$(1)$
$\begin{pmatrix}
1 & 1 & -2 \\
1 & 3 & -1 \\
2 & 1 & -5
\end{pmatrix}=\begin{pmatrix}
1 & 0 & 0 \\
1 & 1 & 0 \\
2 & \frac{-1}{2} & 1
\end{pmatrix}\begin{pmatrix}
1 & 1 & -2 \\
0 & 2 & 1 \\
0 & 0 & \frac{-1}{2}
\end{pmatrix}$
So, $L_{32}=\frac{-1}{2}$ and $U_{33}=\frac{-1}{2}$
$(2)$
$\begin{pmatrix}
1 & 1 & -2 \\
1 & 3 & -1 \\
2 & 1 & -5
\end{pmatrix}=\begin{pmatrix}
1 & 0 & 0 \\
1 & 1 & 0 \\
2 & \frac{-1}{2} & \frac{-1}{4}
\end{pmatrix}\begin{pmatrix}
1 & 1 & -2 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{pmatrix}$
So, $L_{32}=\frac{-1}{2}$ and $U_{33}=2$
From the edit part at the end of this answer:
When $x=y=z=1,$ you get the first $LU$ decomposition and when $x=y=1$ and $z=\frac{-1}{4}$, you get the second $LU$ decomposition.
Now you can apply the method $(1)$ as given below and get the first $LU$ decomposition and to obtain the second $LU$ decomposition, you have to multiply third column of matrix $L$ in first $LU$ decomposition by $\frac{-1}{4}$ and divide the third row of matrix $U$ in first $LU$ decomposition by $\frac{-1}{4}$ as it is given in the below comment.
All the remaining things you can get from the rest of this answer.
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$\textbf{Method 1 (Shortcut)}:$
Here, we do $3$ elementary row operations on the following coefficient matrix $A$ to get $U$ as:
$ A = \begin{pmatrix}
1 & 1 & -2 \\
1 & 3 & -1 \\
2 & 1 & -5
\end{pmatrix} \xrightarrow[]{\text{$R_2\leftarrow\boldsymbol{-1}R_1 + R_2 $}} \begin{pmatrix}
1 & 1 & -2 \\
0 & 2 & 1 \\
2 & 1 & -5
\end{pmatrix} \xrightarrow[]{\text{$R_3\leftarrow\boldsymbol{-2}R_1 + R_3 $}} \begin{pmatrix}
1 & 1 & -2 \\
0 & 2 & 1 \\
0 & -1 & -1
\end{pmatrix}$
$ \xrightarrow[]{\text{$R_3\leftarrow\boldsymbol{\frac{1}{2}}R_2 + R_3 $}} \begin{pmatrix}
1 & 1 & -2 \\
0 & 2 & 1 \\
0 & 0 & \frac{-1}{2}
\end{pmatrix}= \textbf{
U}
$
Now, to get $L,$ we just have to reverse the sign of the bold text which we have used in above row operations and put in that position of $L$ for which we have used the above row operations.
For Example, we have used $\boldsymbol{-1}R_1+R_2$ to make the position of second row and first column as zero. So, $L_{21} = 1.$ Similarly, $\boldsymbol{-2}R_1+R_3$ is used to make third row and first column as zero and so, $L_{31} = 2$ and $\boldsymbol{\frac{1}{2}}R_2+R_3$ is used to make third row and second column as zero and so, $L_{32} = \frac{-1}{2}.$ Since, diagonal elements of $U$ are not all $1s$ and hence, all diagonal elements of $L$ will be $1.$
Hence, $L = \begin{pmatrix}
1 & 0 & 0 \\
1 & 1 & 0 \\
2 & \frac{-1}{2} & 1
\end{pmatrix}$ and $ U = \begin{pmatrix}
1 & 1 & -2 \\
0 & 2 & 1 \\
0 & 0 & \frac{-1}{2}
\end{pmatrix}$
Therefore, $\textbf{(D)}$
If anyone wants to know why this works then reason lies in the third method below and you can see the results of $E_{21}^{-1},E_{31}^{-1},E_{32}^{-1}$ to know why it works.
$\textbf{Method 2:}$
$x_1 + x_2 -2x_3 = 4 \qquad (1)$
$x_1 + 3x_2 -x_3 = 7 \qquad (2)$
$2x_1 + x_2 -5x_3 = 7\qquad(3)$
$2 eq(2) – eq(3)$ gives $5x_2 + 3x_3 = 7 $
$2 eq(1) – eq(3)$ gives $x_2 + x_3 = 1 \implies 5x_2 + 5x_3 = 5$
Solving $5x_2 + 3x_3 = 7 $ and $5x_2 + 5x_3 = 5$ gives $x_3=-1, x_2=2$ and by putting it in $eq(1),$ we get, $x_1=0$
So, $x_1=0,$ it means either option $(C)$ is correct or $(D)$
Now, we only need to find $U_{33}$ and get the answer for the given question.
To find $LU$ decomposition (if it is possible), we only need to convert the given matrix into $U$ by elementary row operations in Gaussian Elimination method (described at the end of the answer to find the complete solution for the given questions).
So, here, $A= \begin{bmatrix} 1 &1 &-2 \\ 1 &3 &-1 \\ 2 &1 &-5 \end{bmatrix}$
After $R_2 \leftarrow R_2 – R_1,$ $A$ becomes $\begin{bmatrix} 1 &1 &-2 \\ 0 &2 &1 \\ 2 &1 &-5 \end{bmatrix}$
After $R_3 \leftarrow R_3 – 2R_1,$ $A$ becomes $\begin{bmatrix} 1 &1 &-2 \\ 0 &2 &1 \\ 0 &-1 &-1 \end{bmatrix}$
After $R_3 \leftarrow R_3 +\frac{1}{2}R_2,$ $A$ becomes $\begin{bmatrix} 1 &1 &-2 \\ 0 &2 &1 \\ 0 &0 &\frac{-1}{2} \end{bmatrix}$
So, $U= \begin{bmatrix} 1 &1 &-2 \\ 0 &2 &1 \\ 0 &0 &\frac{-1}{2} \end{bmatrix}$
It means $U_{33} = \frac{-1}{2}$
Hence, $\textbf{Answer: D}$
$\textbf{Method 3:}$
Now, to find the complete solution for the given question, we can use this previous year question to find the $LU$ decomposition and then find $x_1$
After Applying $R_2 \leftarrow R_2 – R_1,$ $A$ becomes $\begin{bmatrix} 1 &1 &-2 \\ 0 &2 &1 \\ 2 &1 &-5 \end{bmatrix}$ and $E_{21}$ becomes $\begin{bmatrix} 1 &0 &0 \\ -1 &1 &0 \\ 0 &0 &1 \end{bmatrix}$
After Applying $R_3 \leftarrow R_3 – 2R_1,$ $A$ becomes $\begin{bmatrix} 1 &1 &-2 \\ 0 &2 &1 \\ 0 &-1 &-1 \end{bmatrix}$ and $E_{31}$ becomes $\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ -2 &0 &1 \end{bmatrix}$
After Applying $R_3 \leftarrow R_3 + \frac{1}{2}R_2,$ $A$ becomes $\begin{bmatrix} 1 &1 &-2 \\ 0 &2 &1 \\ 0 &0 &\frac{-1}{2} \end{bmatrix}$ and $E_{32}$ becomes $\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 &\frac{1}{2} &1 \end{bmatrix}$
So, $U= \begin{bmatrix} 1 &1 &-2 \\ 0 &2 &1 \\ 0 &0 &\frac{-1}{2} \end{bmatrix}$
and $E_{32}E_{31}E_{21}(A) = U$
$A= E_{21}^{-1} E_{31}^{-1}E_{32}^{-1}U $
$A= \begin{bmatrix} 1 &0 &0 \\ 1 &1 &0 \\ 0 &0 &1 \end{bmatrix} \begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 2 &0 &1 \end{bmatrix} \begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 &\frac{-1}{2} &1 \end{bmatrix} U$
$A= \begin{bmatrix} 1 &0 &0 \\ 1 &1 &0 \\ 2 &\frac{-1}{2} &1 \end{bmatrix} U$
Hence, $L=\begin{bmatrix} 1 &0 &0 \\ 1 &1 &0 \\ 2 &\frac{-1}{2} &1 \end{bmatrix}$ and $U= \begin{bmatrix} 1 &1 &-2 \\ 0 &2 &1 \\ 0 &0 &\frac{-1}{2} \end{bmatrix}$
Now, to find $x_1,x_2,x_3,$ we can write:
$AX=B \implies LUX = B$
let $UX=Y, $ So, $LY=B$
we can write $LY=B$ in matrix notation as: $\begin{bmatrix} 1 &0 &0 \\ 1 &1 &0 \\ 2 &\frac{-1}{2} &1 \end{bmatrix}$$\begin{bmatrix} y_1\\y_2 \\ y_3 \end{bmatrix} = \begin{bmatrix} 4\\7 \\ 7 \end{bmatrix}$
So, $y_1=4,y_1+y_2=7 \implies y_2=3, 8 – \frac{3}{2} + y_3 = 7 \implies y_3=\frac{1}{2}$
So, $y_1=4, y_2=3,y_3=\frac{1}{2}$
Now, on solving $UX=Y$ i.e. $\begin{bmatrix} 1 &1 &-2 \\ 0 &2 &1 \\ 0 &0 &\frac{-1}{2} \end{bmatrix} \begin{bmatrix} x_1\\x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 4\\3 \\ \frac{1}{2} \end{bmatrix} $
It means $\frac{-x_3}{2}=\frac{1}{2} \implies x_3=-1, 2x_2-1=3 \implies x_2 =2, x_1+x_2-2x_3 = 4 \implies x_1+2+2=4\implies x_1=0$
Hence, $x_1=0,x_2=2,x_3=-1$
Edit:
$LU$ decomposition is not necessarily unique if it exists.
Here, you can make infinitely many $A=LU$ decomposition as:
$A = \begin{pmatrix}
x & 0 & 0 \\
x & y & 0 \\
2x & \frac{-y}{2} & z
\end{pmatrix} \begin{pmatrix}
\frac{1}{x} & \frac{1}{x} & \frac{-2}{x} \\
0 & \frac{2}{y} & \frac{1}{y} \\
0 & 0 & \frac{-1}{2z}
\end{pmatrix}$
where $x,y,z \in \mathbb{R}$ for which $\det(L)$ and $\det(U)$ are non-zero because $\det(A)\neq 0.$
Now you can take $x=1,y=2,z=3$ or $x=3,y=-1,z=5$ etc, you get the same coefficient matrix.
The reason is that you can make LU decomposition unique by imposing the constraints on either the diagonal entries of $L$ or $U$.
You can make the above $LU$ decomposition by assuming $L_{11}=x,L_{22}=y,L_{33}=z$ and rest of the entries of $L$ and $U$ will automatically be obtained when you solve by solving equations after writing $A=LU.$
Another way is to assume $U_{11}=x,U_{22}=y,U_{33}=z$ and you get a different $LU$ decomposition.