If $\begin{bmatrix}1&0&0\\0&y+1&0\\0&0&1\end{bmatrix}\begin{bmatrix}2x\\-2\\z-3\end{bmatrix}=\begin{bmatrix}6\\4\\1\end{bmatrix}$ then $x+y+z$ is

4

The correct answer is Option (2) → 4

$\begin{bmatrix}1&0&0\\0&y+1&0\\0&0&1\end{bmatrix} \begin{bmatrix}2x\\-2\\z-3\end{bmatrix} = \begin{bmatrix}2x\\-2(y+1)\\z-3\end{bmatrix} = \begin{bmatrix}6\\4\\1\end{bmatrix}$

$2x = 6 \;\;\Rightarrow\;\; x = 3$

$-2(y+1) = 4 \;\;\Rightarrow\;\; y = -3$

$z - 3 = 1 \;\;\Rightarrow\;\; z = 4$

$x + y + z = 3 - 3 + 4 = 4$