Matrices and Determinants · Mathematics · VITEEE

If matrix $$A=\left[\begin{array}{ccc}0 & 2 b & -2 \\ 3 & 1 & 3 \\ 3 a & 3 & -1\end{array}\right]$$ is given to be symmetric, then the value of $$a b$$ is

The determinant of the matrix $$\left[\begin{array}{ccc}1 & 4 & 8 \\ 1 & 9 & 27 \\ 1 & 16 & 64\end{array}\right]$$ is

Suppose, $$A=\left[\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right]$$ is an adjoint of the matrix $$\left[\begin{array}{rrr}1 & 3 & -3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]$$. The value of $$\frac{a_1+b_2+c_3}{b_1 a_2}$$ is

For all values of $$\lambda$$, rank of matrix

$$A=\left[\begin{array}{ccc} { }^h C_0 & { }^4 C_3 & { }^5 C_4 \\ \lambda & 8 & 8 \lambda-6 \\ 1+\lambda^2 & 8 \lambda+4 & 2 \lambda+21 \end{array}\right]$$

If $$A=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$$, then $$\left(A-A^{\prime}\right)$$ is equal to (where, $$A^{\prime}$$ is transpose of matrix $$A$$ )

If $$A^{-1}=\left[\begin{array}{rr}5 & -2 \\ -7 & 3\end{array}\right]$$ and $$B^{-1}=\frac{1}{2}\left[\begin{array}{rr}9 & -7 \\ -8 & 6\end{array}\right]$$, then $$(A B)^{-1}$$ is equal to