If $3\begin{bmatrix}x&y\\z&w\end{bmatrix} = \begin{bmatrix}x&6\\-1&2w\end{bmatrix} +\begin{bmatrix}4&x+y\\z+w&3\end{bmatrix}$, then the values of $x, y, z$ and $w$ are

$x=2, y = 4,z= 1, w = 3$

The correct answer is Option (3) → $x=2, y = 4,z= 1, w = 3$

Given:

$3\begin{bmatrix}x & y \\ z & w\end{bmatrix} = \begin{bmatrix}x & 6 \\ -1 & 2w\end{bmatrix} + \begin{bmatrix}4 & x+y \\ z+w & 3\end{bmatrix}$

Left side:

$3\begin{bmatrix}x & y \\ z & w\end{bmatrix} = \begin{bmatrix}3x & 3y \\ 3z & 3w\end{bmatrix}$

Right side:

$\begin{bmatrix}x + 4 & 6 + x + y \\ -1 + z + w & 2w + 3\end{bmatrix}$

Equating both sides:

$\begin{bmatrix}3x & 3y \\ 3z & 3w\end{bmatrix} = \begin{bmatrix}x + 4 & 6 + x + y \\ -1 + z + w & 2w + 3\end{bmatrix}$

Compare corresponding elements:

$3x = x + 4 \Rightarrow 2x = 4 \Rightarrow x = 2$

$3y = 6 + x + y \Rightarrow 3y = 6 + 2 + y \Rightarrow 3y = 8 + y \Rightarrow 2y = 8 \Rightarrow y = 4$

$3z = -1 + z + w \Rightarrow 3z - z = -1 + w \Rightarrow 2z = -1 + w \quad\text{(1)}$

$3w = 2w + 3 \Rightarrow w = 3$

Substitute $w = 3$ into (1):

$2z = -1 + 3 = 2 \Rightarrow z = 1$

Final values:

$x = 2,\quad y = 4,\quad z = 1,\quad w = 3$