$\text{If}\mathrm{ A}\text{ is a square matrix such that}{\mathrm{ A}}^2=2I,\text{then the value of} (A-\mathrm{I)}{}^2+(A+\mathrm{I\; )}{}^2-7A\text{ is}$

The correct option is -7A + 6I.

Let's break down the derivation step-by-step.

We are given that $A$ is a square matrix and $A^2=2I$, where $I$ is the identity matrix of the same order.

We need to find the value of the expression:
$(A-I{)}^2+(A+I{)}^2-7A$

First, let's expand the term $(A-I{)}^2$:
$(A-I{)}^2=(A-I\left)\right(A-I)=A^2-AI-IA+I^2$

Since multiplying any matrix by the identity matrix $I$ keeps the matrix unchanged (i.e., $AI=IA=A$) and $I^2=I$, we can simplify this to:
$(A-I{)}^2=A^2-2A+I$

Next, let's expand the term $(A+I{)}^2$ in the same manner:
$(A+I{)}^2=(A+I\left)\right(A+I)=A^2+AI+IA+I^2=A^2+2A+I$

Now, let's substitute these expanded forms back into our original expression:
$\left[\right(A-I{)}^2]+[(A+I{)}^2]-7A=(A^2-2A+I)+(A^2+2A+I)-7A$

Grouping the common terms together, we get:
$=A^2+A^2-2A+2A+I+I-7A$

Simplifying the terms:
$=2A^2+2I-7A$

Given the initial condition that $A^2=2I$, we can substitute $2I$ in place of $A^2$:
$=2\left(2I\right)+2I-7A$

Perform the scalar multiplication:
$=4I+2I-7A$

Combine the identity matrix terms:
$=6I-7A$

This can be rewritten as:
$=-7A+6I$

Therefore, the value of the given expression is indeed -7A + 6I.