Let A = $$\left( {\matrix{ {3 - t} \cr { - 1} \cr 0 \cr } \matrix{ {} \cr {} \cr {} \cr } \,\matrix{ 1 \cr {3 - t} \cr { - 1} \cr } \matrix{ {} \cr {} \cr {} \cr } \matrix{ 0 \cr 1 \cr 0 \cr } } \right)$$ and det A = 5, then

Let $$A = \left[ {\matrix{ {12} & {24} & 5 \cr x & 6 & 2 \cr { - 1} & { - 2} & 3 \cr } } \right]$$. The value of x for which the matrix A is not invertible is

Let $$A = \left( {\matrix{ a & b \cr c & d \cr } } \right)$$ be a 2 $$ \times $$ 2 real matrix with det A = 1. If the equation det (A $$ - $$ $$\lambda $$I₂) = 0 has imaginary roots (I₂ be the identity matrix of order 2), then

If $$\left| {\matrix{ {{a^2}} & {bc} & {{c^2} + ac} \cr {{a^2} + ab} & {{b^2}} & {ca} \cr {ab} & {{b^2} + bc} & {{c^2}} \cr } } \right| = k{a^2}{b^2}{c^2}$$,

then K =