The cost of $4 \, kg$ onion,$3 \, kg$ wheat and $2 \, kg$ rice is $Rs \, 60$. The cost of $2 \, kg$ onion,$4 \, kg$ wheat and $6 \, kg$ rice is $Rs \, 90$. The cost of $6 \, kg$ onion,$2 \, kg$ wheat and $3 \, kg$ rice is $Rs \, 70$. Find the cost of each item per $kg$ using the matrix method.

For a system of simultaneous linear equations, if $A X=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$, $\operatorname{Adj} A=\left[\begin{array}{ccc}1 & -1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & 1\end{array}\right]$ and $\operatorname{det} A>0$, then $X=$

Let $S$ be the set of all $\lambda \in \mathbb{R}$ for which the system of linear equations
$2x - y + 2z = 2$
$x - 2y + \lambda z = -4$
$x + \lambda y + z = 4$
has no solution. Then the set $S$

Let $S$ be the set of all column matrices $\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right]$ such that $b_1, b_2, b_3 \in \mathbb{R}$ and the system of equations (in real variables)
$-x+2y+5z=b_1$
$2x-4y+3z=b_2$
$x-2y+2z=b_3$
has at least one solution. Then,which of the following system$(s)$ (in real variables) has (have) at least one solution for each $\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right] \in S$?
$(A)$ $x+2y+3z=b_1, 4y+5z=b_2$ and $x+2y+6z=b_3$
$(B)$ $x+y+3z=b_1, 5x+2y+6z=b_2$ and $-2x-y-3z=b_3$
$(C)$ $-x+2y-5z=b_1, 2x-4y+10z=b_2$ and $x-2y+5z=b_3$
$(D)$ $x+2y+5z=b_1, 2x+3z=b_2$ and $x+4y-5z=b_3$

If ${a_1}x + {b_1}y + {c_1}z = 0, {a_2}x + {b_2}y + {c_2}z = 0, {a_3}x + {b_3}y + {c_3}z = 0$ and $\left| \begin{matrix} {a_1} & {b_1} & {c_1} \\ {a_2} & {b_2} & {c_2} \\ {a_3} & {b_3} & {c_3} \end{matrix} \right| = 0$,then the given system has

$A$ 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. $1800$.