Given, $$D = \left| {\matrix{ 1 & 1 & 1 \cr 1 & {1 + x} & 1 \cr 1 & 1 & {1 + y} \cr } } \right|$$

Apply $$\,\,\,{R^2} \to {R_2} - {R_1}$$ $$\,\,\,\,$$

and $$\,\,\,\,$$ $$R \to {R_3} - {R_1}$$

$$\therefore$$ $$\,\,\,\,\,D = \left| {\matrix{ 1 & 1 & 1 \cr 0 & x & 0 \cr 0 & 0 & y \cr } } \right| = xy$$

Hence, $$D$$ is divisible by both $$x$$ and $$y$$

$$\left| {{A^2}} \right| = 25 \Rightarrow {\left| A \right|^2} = 25$$

$$ \Rightarrow {\left( {25\alpha } \right)^2} = 25 \Rightarrow \left| \alpha \right| = {1 \over 5}$$

$$A = \left[ {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right]\,\,\,\,B = \left[ {\matrix{ a & 0 \cr 0 & b \cr } } \right]$$

$$AB = \left[ {\matrix{ a & {2b} \cr {3a} & {4b} \cr } } \right]$$

$$BA = \left[ {\matrix{ a & 0 \cr 0 & b \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right] = \left[ {\matrix{ a & {2a} \cr {3b} & {4b} \cr } } \right]$$

Hence, $$AB=BA$$ only when $$a=b$$

$$\therefore$$ There can be infinitely many $$B's$$

for which $$AB=BA$$

$${A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)$$

$${A^2} - {B^2} = {A^2} + AB - BA - {B^2}$$

$$ \Rightarrow AB = BA$$