Question

A shop sells three commodities. X purchases 2 units of q and sells 3 units of p and 3 units r. Y purchases 1 units of r and sells 2 units of p and 1 unit of q. Z purchases 3 unit of q and sells 4 units of p and 2 units of r. In the process X, Y, and Z earns Rs. 8,000, Rs.1000, and Rs. 4,000 respectively when price of item is p, q, and r per unit respectively (Consider selling units is positive earning and buying the units negative earnings)
i)Write system of simultaneous equations
ii)Find det.|A|
iii)Write transpose of A
Using matrices, find prices per unit of the three commodities

Answer 1

Solution.

Given: Selling units are positive earnings and buying the units negative earnings.

$X$ purchases $2$ units of $q$ and sells $3$ units of $p$ and $3$ units $r.$ $X$ earns Rs. $8000$

$\Rightarrow 3p-2q+3r=8000\quad.....(1)$

$Y$ purchases $1$ units of $r$ and sells $2$ units of $p$ and $1$ unit of $q.$ $Y$ earns Rs. $1000$

$\Rightarrow 2p+q-r=1000\quad.....(2)$

$Z$ purchases $3$ unit of $q$ and sells $4$ units of $p$ and $2$ units of $r.$ $Z$ earns Rs. $4000$

$\Rightarrow 4p-3q+2r=4000\quad.....(3)$

i) System of simultaneous equations.

Here the system of simulations equations is given by

$\begin{array}{cccccccc}3p&-&2q&+&3r&=&8000&\quad...(1)\\2p&+&q&-&r&=&1000&\quad...(2)\\4p&-&3q&+&2r&=&4000&\quad...(3)\end{array}$

In matrix form the system of linear equation can be written as

$\quad\quad AX=B$

where:

$A=\begin{pmatrix} 3&-2&3 \\ 2&1&-1 \\ 4&-3&2\end{pmatrix}$

$X=\begin{pmatrix}p\\q\\r\end{pmatrix},\quad B=\begin{pmatrix}8000\\1000\\4000\end{pmatrix}$

ii) Finding $det(A)$.

We have found: $A=\begin{pmatrix} 3&-2&3 \\ 2&1&-1 \\ 4&-3&2\end{pmatrix}$

$\Rightarrow det(A)=\left|\begin{array}{ccc} 3&-2&3 \\ 2&1&-1 \\ 4&-3&2\end{array}\right|$

$\Rightarrow det(A)=3(2-3)+2(4+4)+3(-6-4)$

$\quad$(Expanding along $R_{1})$.

$\Rightarrow det(A)=3(-1)+2(8)+3(-10)$

$\Rightarrow det(A)=-3+16-30$

$\Rightarrow \boxed{det(A)=-17}$

iii) Finding $A^{T}$.

We have found: $A=\begin{pmatrix} 3&-2&3 \\ 2&1&-1 \\ 4&-3&2\end{pmatrix}$

$\Rightarrow A^{T}={\begin{pmatrix} 3&-2&3 \\ 2&1&-1 \\ 4&-3&2\end{pmatrix}}^{T}$

$\Rightarrow \boxed{A^{T}=\begin{pmatrix}3&2&4\\-2&1&-3\\3&-1&2\end{pmatrix}}$

iv) Finding solution.

Here, the system of simulations equations is given by

$\quad 3p-2q+3r=8000\quad.....(1)$

$\quad 2p+q-r=1000\quad.....(2)$

$\quad 4p-3q+2r=4000\quad.....(3)$

$\therefore D=det{A}=-17$

$D_{1}=\left|\begin{array}{ccc}8000&-2&3\\1000&1&-1\\4000&-3&2\end{array}\right|$

$\quad=8000(2-3)+2(2000+4000)+3(-3000-4000)$

$\quad$(Expanding along $R_{1})$.

$\quad=8000(-1)+2(6000)+3(-7000)$

$\quad=-8000+12000-21000$

$\quad=-17000$

$D_{2}=\left|\begin{array}{ccc}3&8000&3\\2&1000&-1\\4&4000&2\end{array}\right|$

$\quad=3(2000+4000)-8000(4+4)+3(8000-4000)$

$\quad$(Expanding along $R_{1})$.

$\quad=3(6000)-8000(8)+3(4000)$

$\quad=18000-64000+12000$

$\quad=-34000$

$D_{3}=\left|\begin{array}{ccc}3&-2&8000\\2&1&1000\\4&-3&4000\end{array}\right|$

$\quad=3(4000+3000)+2(8000-4000)+8000(-6-4)$

$\quad$(Expanding along $R_{1})$.

$\quad=3(7000)+2(4000)+8000(-10)$

$\quad=21000+8000-80000$

$\quad=-51000$

The solution of the given system of equations is given by:

$\quad p=\frac{D_{1}}{D},\:q=\frac{D_{2}}{D},\:r=\frac{D_{3}}{D}$

$\Rightarrow p=\frac{-17000}{-17},\:q=\frac{-34000}{-17},\:r=\frac{-51000}{-17}$

$\Rightarrow \boxed{p=1000,\:q=2000,\:r=3000}$

Therefore price of per unit $p$ commodities is Rs. $1000$, price of per unit $q$ commodities is Rs. $2000$ and price of per unit $r$ commodities is Rs. $3000$.

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