
Concept:
Let \(x\) be the number of Rs. 1,000 vouchers and \(y\) be the number of Rs. 500 vouchers.
The equations can be written in matrix form as: \[ AX=B \] The solution is: \[ X=A^{-1}B \] where \[ A^{-1}=\frac{1}{|A|}\operatorname{adj}(A) \]
Step 1: Form the equations
The total number of vouchers is: \[ x+y=60 \] The second equation is: \[ x+3y=100 \]
Step 2: Write in matrix form
The system can be represented as: \[ \begin{bmatrix} 1 & 1 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 60 \\ 100 \end{bmatrix} \]
Step 3: Find \(A^{-1}\)
Let: \[ A= \begin{bmatrix} 1 & 1 \\ 1 & 3 \end{bmatrix} \] The determinant is: \[ |A|=(1)(3)-(1)(1) \] \[ |A|=2 \] The adjoint of \(A\) is: \[ \operatorname{adj}(A)= \begin{bmatrix} 3 & -1 \\ -1 & 1 \end{bmatrix} \] Therefore: \[ A^{-1} = \frac{1}{2} \begin{bmatrix} 3 & -1 \\ -1 & 1 \end{bmatrix} \]
Step 4: Solve for \(x\) and \(y\)
Using: \[ X=A^{-1}B \] \[ \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 3 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 60 \\ 100 \end{bmatrix} \] \[ = \frac{1}{2} \begin{bmatrix} 180-100 \\ -60+100 \end{bmatrix} \] \[ = \frac{1}{2} \begin{bmatrix} 80 \\ 40 \end{bmatrix} \] Therefore: \[ x=40,\qquad y=20 \]
Final Answer:
There are: \[ \boxed{40\text{ vouchers of Rs. }1,000} \] and \[ \boxed{20\text{ vouchers of Rs. }500} \]
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
