The number of matrices $$A=\begin{bmatrix}a & b \\c & d \end{bmatrix}$$, where $$𝑎, 𝑏, 𝑐, d ∈ -1, 0, 1, 2, 3, … … , 10,$$ such that $$A=A^{-1}$$, is______.

If the condition is

$$A=A^{-1},$$

then

$$A^2=I$$

Let

$$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

where

$$a,b,c,d\in\{-1,0,1,2,\ldots,10\}$$

Now,

$$A^2= \begin{bmatrix} a^2+bc & ab+bd\\ac+cd & d^2+bc\end{bmatrix}$$

Since

$$A^2=I,$$

we get

$$a^2+bc=1\qquad\cdots(1)$$

$$d^2+bc=1\qquad\cdots(2)$$

$$b(a+d)=0\qquad\cdots(3)$$

$$c(a+d)=0\qquad\cdots(4)$$

From (1) and (2),

$$a^2=d^2$$

Hence,

$$a=d\quad \text{or}\quad a=-d$$

Case 1: $$a=-d$$

Then equations (3) and (4) are automatically satisfied.

Using

$$a^2+bc=1$$

Subcase (i): $$a=1,\ d=-1$$

$$1+bc=1$$

$$bc=0$$

Number of ordered pairs satisfying

$$bc=0$$

is

$$12+12-1=23$$

(Subtracting one for double counting $$b=c=0$$)

Subcase (ii): $$a=-1,\ d=1$$

Again,

$$bc=0$$

Hence, number of matrices is

$$23$$

Subcase (iii): $$a=0,\ d=0$$

Then,

$$bc=1$$

Possible ordered pairs are

$$(1,1),\ (-1,-1)$$

Hence, number of matrices is

$$2$$

Total from Case 1:

$$23+23+2=48$$

Case 2: $$a=d$$

From (3) and (4),

$$b(2a)=0,\qquad c(2a)=0$$

Subcase (i): $$a=d=1$$

Then,

$$b=0,\ c=0$$

giving

$$1$$

matrix.

Subcase (ii): $$a=d=-1$$

Again,

$$b=0,\ c=0$$

giving

$$1$$

matrix.

Total from Case 2:

$$1+1=2$$

Therefore, total number of matrices is

$$48+2=50$$

Hence, the required number of matrices is

$$\boxed{50}$$.