Consider the system of equations: $x + ay = 0$,$y + az = 0$,and $z + ax = 0$. The set of all real values of $a$ for which the system has a unique solution is:

Let $A$ be a matrix such that $AB$ is a scalar matrix where $B = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$ and $\det(3A) = 27$. Then $3A^{-1} + A^2 =$

If $S$ is the set of distinct values of $b$ for which the following system of linear equations $x + y + z = 1$,$x + ay + z = 1$,and $ax + by + z = 0$ has no solution,then $S$ is:

$A$ trust fund has Rs. $30,000$ that must be invested in two different types of bonds. The first bond pays $5 \%$ interest per year,and the second bond pays $7 \%$ interest per year. Using matrix multiplication,determine how to divide Rs. $30,000$ among the two types of bonds if the trust fund must obtain an annual total interest of Rs. $1800$.

If the system of equations $kx + (k+1)y + (k-1)z = 0$,$(k-1)x + (k+2)y + kz = 0$,and $(k+1)x + ky + (k+2)z = 0$ has a non-trivial solution,then the sum of all possible values of $k$ is:

If the system of equations $x+y+z=5$, $x+2y+2z=6$, and $x+3y+\lambda z=\mu$ (where $\lambda, \mu \in R$) is solvable by the Matrix Inversion Method, then: