The cost of 4 , kg onion,3 , kg wheat and 2 , | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the cost of onions,wheat,and rice per $kg$ be $Rs \, x$,$Rs \, y$,and $Rs \, z$ respectively.The given situation can be represented by a syste

Vedclass · Accepted Answer

$x=5, y=8, z=8$

Vedclass · Answer

(A) Let the cost of onions,wheat,and rice per $kg$ be $Rs \, x$,$Rs \, y$,and $Rs \, z$ respectively.The given situation can be represented by a system of equations as:$4x + 3y + 2z = 60$$2x + 4y + 6z = 90$$6x + 2y + 3z = 70$This system of equations can be written in the form of $AX = B$,where$A = \begin{bmatrix} 4 & 3 & 2 \ 2 & 4 & 6 \ 6 & 2 & 3 \end{bmatrix}, X = \begin{bmatrix} x \ y \ z \end{bmatrix}, B = \begin{bmatrix} 60 \ 90 \ 70 \end{bmatrix}$$|A| = 4(12 - 12) - 3(6 - 36) + 2(4 - 24) = 0 + 90 - 40 = 50 
eq 0$Calculating the adjoint of $A$:$A_{11} = 0, A_{12} = 30, A_{13} = -20$$A_{21} = -5, A_{22} = 0, A_{23} = 10$$A_{31} = 10, A_{32} = -20, A_{33} = 10$$adj(A) = \begin{bmatrix} 0 & -5 & 10 \ 30 & 0 & -20 \ -20 & 10 & 10 \end{bmatrix}$$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{50} \begin{bmatrix} 0 & -5 & 10 \ 30 & 0 & -20 \ -20 & 10 & 10 \end{bmatrix}$$X = A^{-1}B = \frac{1}{50} \begin{bmatrix} 0 & -5 & 10 \ 30 & 0 & -20 \ -20 & 10 & 10 \end{bmatrix} \begin{bmatrix} 60 \ 90 \ 70 \end{bmatrix} = \frac{1}{50} \begin{bmatrix} 250 \ 400 \ 400 \end{bmatrix} = \begin{bmatrix} 5 \ 8 \ 8 \end{bmatrix}$Thus,$x=5, y=8, z=8$. The cost of onions is $Rs \, 5/kg$,wheat is $Rs \, 8/kg$,and rice is $Rs \, 8/kg$.

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.

Similar Questions

The set of values of $k$ for which the system of simultaneous equations $x+y+kz=1$,$2x+2y=3$,and $x+2y+2kz=k$ has no real solution is

If the system of equations
$2x + y - z = 5$
$2x - 5y + \lambda z = \mu$
$x + 2y - 5z = 7$
has infinitely many solutions,then $(\lambda + \mu)^2 + (\lambda - \mu)^2$ is equal to

Let $A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$ such that $AX = B$,then $X =$

The number of solutions of the system of equations $2x + y - z = 7$,$x - 3y + 2z = 1$,and $x + 4y - 3z = 5$ is:

Let $x = \alpha, y = \beta, z = \gamma$ be the unique solution of the system of simultaneous linear equations $2x + 3y - 2z + 4 = 0$,$3x - 4y + 3z + 5 = 0$,and $kx - 2y + z + 3 = 0$. If $\alpha = -2$,then $k =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module