Match List-I with List-II List-I Matrix/equations List-II Values (A)…

The correct answer is Option (1) → (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

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

Equating: $2x+1 = x+3 \Rightarrow x=2$, and $y^2-5y = -6 \Rightarrow y^2-5y+6=0 \Rightarrow (y-2)(y-3)=0 \Rightarrow y=2,3$.

Also $3y=y^2+2 \Rightarrow y^2-3y+2=0 \Rightarrow (y-1)(y-2)=0 \Rightarrow y=1,2$.

Common solution: $x=2,y=2$. Matches (II).

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(B) $\begin{bmatrix} 1 & 2 & -1 \\ x & 0 & 3 \\ y & 3 & 4 \end{bmatrix}$ is symmetric.

For symmetry: $x=2$, $y=-1$. Matches (I).

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(C) $[x\ 1]\begin{bmatrix} 1&0 \\ -2&-3 \end{bmatrix}\begin{bmatrix} 5&2 \\ 0&y \end{bmatrix}=0$

First multiply: $[x\ 1]\begin{bmatrix} 1&0 \\ -2&-3 \end{bmatrix} = [x-2,\ -3-x]$.

Now $[x-2,\ -3-x]\begin{bmatrix} 5&2 \\ 0&y \end{bmatrix} = [5(x-2),\ 2(x-2)+(-3-x)y]$

= $[5x-10,\ 2x-4-xy-3y]$

Equating to [0,0]: $5x-10=0 \Rightarrow x=2$. Second: $2(2)-4-(2)y-3y=0 \Rightarrow 4-4-5y=0 \Rightarrow -5y=0 \Rightarrow y=0$.

So solution: $x=2,y=0$. Matches (IV).

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(D) $\begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ -1 & \frac{y}{2} \end{bmatrix}$

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

Equating: $x^2=4 \Rightarrow x=±2$; $x+1=-1 \Rightarrow x=-2$; $1=\frac{y}{2} \Rightarrow y=2$.

So solution: $x=-2,y=2$. Matches (III).