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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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The cost of 4 kg onion , 3 kg wheat and 2 kg rice of 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 cost of each item per kg by matrix method.

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Toolbox:
  • $A^{-1}=\large\frac{1}{|A|}$$adj \;A$
  • $adj \;A=\begin{bmatrix}A_{11} & A_{21}& A_{31}\\A_{12} & A_{22}& A_{32}\\A_{13} & A_{23}& A_{33}\end{bmatrix}$
Step 1:
Let the cost of onion is $Rs.x$
Let the cost of wheat is $Rs.y$
Let the cost of rice is $Rs.z$
Thus the above given situation can be represented as
$4x+3y+2z=60$
$2x+4y+6z=90$
$6x+2y+3z=70$
$A=\begin{bmatrix}4 & 3 & 2\\2 & 4 & 6\\6 & 2 & 3\end{bmatrix}$
$X=\begin{bmatrix}x\\y\\z\end{bmatrix}$
$B=\begin{bmatrix}60\\90\\70\end{bmatrix}$
$|A|=4(12-12)-3(6-36)+2(4-24)$
$\quad=0+90-40$
$\quad=50\neq 0$
Step 2:
$A_{11}=0$
$A_{12}=30$
$A_{13}=-20$
$A_{21}=-5$
$A_{22}=0$
$A_{23}=10$
$A_{31}=10$
$A_{32}=-20$
$A_{33}=10$
$adj \;A=\begin{bmatrix}0 & -5 & 10\\30 & 0 & -20\\-20 & 10 & 10\end{bmatrix}$
$\Rightarrow X=\large\frac{1}{50}$$\begin{bmatrix}0 & -5 & 10\\30 & 0 & -20\\-20 & 10 & 10\end{bmatrix}\begin{bmatrix}60\\90\\70\end{bmatrix}$
Step 3:
$A^{-1}=\large\frac{1}{|A|}$$adj \;A=\large\frac{1}{50}$$\begin{bmatrix}0 & -5 & 10\\30 & 0 & -20\\-20 & 10 & 10\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\large\frac{1}{50}$$\begin{bmatrix}0-450+700\\1800+0-1400\\-1200+900+700\end{bmatrix}$
$\quad\;\;=\large\frac{1}{50}$$\begin{bmatrix}250\\400\\400\end{bmatrix}$
$\quad\;\;=\begin{bmatrix}5\\8\\8\end{bmatrix}$
$x=5,y=8,z=8$
Step 4:
Hence the cost of onions is $Rs.5/kg.$
The cost of wheat is $Rs.8/kg.$
The cost of rice is $Rs.8/kg.$
answered May 15, 2013 by sreemathi.v
 

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