$$\Delta = 0 \Rightarrow \left| {\matrix{ 4 & k & 2 \cr k & 4 & 1 \cr 2 & 2 & 1 \cr } } \right| = 0$$

$$ \Rightarrow 4\left( {4 - 2} \right) - k\left( {k - 2} \right) + $$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\left( {2k - 8} \right) = 0$$

$$ \Rightarrow 8 - {k^2} + 2k + 4k - 16 = 0$$

$$ \Rightarrow {k^2} - 6k + 8 = 0$$

$$ \Rightarrow \left( {k - 4} \right)\left( {k - 2} \right) = 0,k = 4,2$$

$$\therefore$$ $$\,\,\,\,\,A' = A,B' = B$$

Now $$\,\,\,\left( {A\left( {BA} \right)} \right)' = \left( {BA} \right)'A'$$

$$ = \left( {A'B'} \right)A' = \left( {AB} \right)A = A\left( {BA} \right)$$

Similarly $$\left( {\left( {AB} \right)A} \right)' = \left( {AB} \right)A$$

So, $$A\left( {BA} \right)\,\,\,\,$$ and $$A\left( {BA} \right)\,\,\,\,$$ are symmetric matrices.

Again $$\left( {AB} \right)' = B'A' = BA$$

Now if $$BA=AB$$, then $$AB$$ is symmetric matrix.

$$\left[ {\matrix{ 1 & {...} & {...} \cr {...} & 1 & {...} \cr {...} & {...} & 1 \cr } } \right]\,\,$$ are $$6$$ non-singular matrices because $$6$$

blanks will be filled by $$5$$ zeros and $$1$$ one.

Similarly, $$\left[ {\matrix{ {...} & {...} & 1 \cr {...} & 1 & {...} \cr 1 & {...} & {...} \cr } } \right]\,\,$$ are $$6$$ non-singular matrices.

So, required cases are more than $$7,$$ non-singular $$3 \times 3$$ matrices.

$$D = \left| {\matrix{ 1 & 2 & 1 \cr 2 & 3 & 1 \cr 3 & 5 & 2 \cr } } \right| = 0$$

$${D_1}\left| {\matrix{ 3 & 2 & 1 \cr 3 & 3 & 1 \cr 1 & 5 & 2 \cr } } \right| \ne 0$$

$$ \Rightarrow $$ Given system, does not have any solution.

$$ \Rightarrow $$ No solution