Match List-I with List-II

List-I	List-II
Matrix/equations	Values
(A) $\begin{bmatrix} 2x+1 & 3y \\ 0 & y^2-5y \end{bmatrix} = \begin{bmatrix} x+3 & y^2+2 \\ 0 & -6 \end{bmatrix}$	(I) $x = 2, y = -1$
(B) $\begin{bmatrix} 1 & 2 & -1 \\ x & 0 & 3 \\ y & 3 & 4 \end{bmatrix}$ is symmetric	(II) $x = 2, y = 2$
(C) $[x \ \ 1]\begin{bmatrix} 1 & 0 \\ -2 & -3 \end{bmatrix}\begin{bmatrix} 5 & 2 \\ 0 & y \end{bmatrix} = O$	(III) $x = -2, y = 2$
(D) $\begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ -1 & y/2 \end{bmatrix}$	(IV) $x = 2, y = 0$

Choose the correct answer from the options given below:

$\begin{bmatrix} 2x+1 & 3y \\ 0 & y^2-5y \end{bmatrix} = \begin{bmatrix} x+3 & y^2+2 \\ 0 & -6 \end{bmatrix}$

When two matrices are equal, their corresponding elements must be equal.

From position (1,1):

$2x + 1 = x + 3$

$x = 2$

From position (1,2):

$3y = y^2 + 2$

$y^2 - 3y + 2 = 0$

$(y-1)(y-2) = 0$

$y = 1 \text{ or } y = 2$

From position (2,2):

$y^2 - 5y = -6$

$y^2 - 5y + 6 = 0$

$(y-2)(y-3) = 0$

$y = 2 \text{ or } y = 3$

The common value from both equations is $y = 2$.

(A) $\rightarrow$ $x = 2, \, y = 2$ $\rightarrow$ (II)

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$\begin{bmatrix} 1 & 2 & -1 \\ x & 0 & 3 \\ y & 3 & 4 \end{bmatrix}$

A symmetric matrix satisfies $A = A^T$, meaning element at position $(i,j)$ equals element at position $(j,i)$.

From (1,2) = (2,1): $\quad 2 = x \implies x = 2$

From (1,3) = (3,1): $\quad -1 = y \implies y = -1$

From (2,3) = (3,2): $\quad 3 = 3$ ✓

(B) $\rightarrow$ $x = 2, \, y = -1$ $\rightarrow$ (I)

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$[x \quad 1]\begin{bmatrix} 1 & 0 \\ -2 & -3 \end{bmatrix}\begin{bmatrix} 5 & 2 \\ 0 & y \end{bmatrix} = O$

$[x \quad 1]\begin{bmatrix} 1 & 0 \\ -2 & -3 \end{bmatrix}$

$= [(x \cdot 1 + 1 \cdot (-2)), \quad (x \cdot 0 + 1 \cdot (-3))]$

$= [x - 2, \quad -3]$

$[x-2 \quad -3]\begin{bmatrix} 5 & 2 \\ 0 & y \end{bmatrix}$

$= [5(x-2) + 0, \quad 2(x-2) + (-3y)]$

$= [5(x-2), \quad 2(x-2) - 3y]$

Setting each entry to $0$:

$5(x - 2) = 0 \implies x = 2$

$2(2 - 2) - 3y = 0 \implies -3y = 0 \implies y = 0$

(C) $\rightarrow$ $x = 2, \, y = 0$ $\rightarrow$ (IV)

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$\begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ -1 & y/2 \end{bmatrix}$

Computing the LHS:

$\begin{bmatrix} x \cdot x + 0 \cdot 1 & x \cdot 0 + 0 \cdot 1 \\ 1 \cdot x + 1 \cdot 1 & 1 \cdot 0 + 1 \cdot 1 \end{bmatrix} = \begin{bmatrix} x^2 & 0 \\ x + 1 & 1 \end{bmatrix}$

Comparing corresponding elements:

From (1,1): $\quad x^2 = 4 \implies x = 2 \text{ or } x = -2$

From (2,1): $\quad x + 1 = -1 \implies x = -2$

From (2,2): $\quad 1 = y/2 \implies y = 2$

(D) $\rightarrow$ $x = -2, \, y = 2$ $\rightarrow$ (III)

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List-I	List-II
(A) $x = 2, \, y = 2$	(II)
(B) $x = 2, \, y = -1$	(I)
(C) $x = 2, \, y = 0$	(IV)
(D) $x = -2, \, y = 2$	(III)

Therefore, the correct answer is (A)-(II), (B)-(I), (C)-(IV), (D)-(III).