If the matrix $\left[\begin{array}{ccc}-3 & a-b & 5 \\ 1 & 0 & 4 \\ a+b & c & 11\end{array}\right]$ is symmetric then the value of $a-b+c=$

A matrix $A$ is symmetric if $A = A^{T}$, which implies $A_{ij} = A_{ji}$ for all $i, j$.

Given the matrix $A = \begin{bmatrix} -3 & a-b & 5 \\ 1 & 0 & 4 \\ a+b & c & 11 \end{bmatrix}$, we equate the corresponding elements:

For $A_{12} = A_{21}$, we have $a-b = 1$.

For $A_{13} = A_{31}$, we have $5 = a+b$.

For $A_{23} = A_{32}$, we have $4 = c$.

We need to find the value of $a-b+c$.

Substituting the known values, we have $a-b = 1$ and $c = 4$.

Therefore, $a-b+c = 1 + 4 = 5$.

Answer: 5