Find the real value of x if \(\begin{vmatrix}9 & 3x^2 \\x & 1 \end{vmatrix}\) = \(\begin{vmatrix}x & 3x \\x & 9 \end{vmatrix}\)

1

By solving the determinants we get

9-3\( { x }^{ 3 } \) = 9x - 3\( { x }^{ 2 } \)

9- 9x = 3\( { x }^{ 3 } \) - 3\( { x }^{ 2 } \)

9(1-x) = 3\( { x }^{ 2 } \) (x-1)

0= 3\( { x }^{ 2 } \) (x-1) - 9(1-x)

0= (x-1) (3\( { x }^{ 2 } \)+9)

so when x-1 = 0 we have x=1

and when (3\( { x }^{ 2 } \)+9) = 0 

               3\( { x }^{ 2 } \) = -9

                \( { x }^{ 2 } \) = -3 which cannot yield any real solution for x.