Question

A diet is to contain $30$ units of vitamin $A$, $40$ units of vitamin $B$ and $20$ units of vitamin $C$. Three types of foods $F_{1}$, $F_{2}$ and $F_{3}$ are available. 1 unit of food $F_{1}$ contains $3$ units of vitamin $A$, $2$ units of vitamin $B$, $1$ unit of vitamin $C$. $1$ unit of food $F_{2}$ contains $1$ unit of vitamin $A$, $2$ units of vitamin $B$ and $1$ unit of vitamin $C$. $1$ unit of food $F_{3}$ contains $5$ units of vitamin $A$, $3$ units of vitamin $B$ and $2$ units of vitamin $C$. Represent the above situation algebraically and find the diet contains each types of food by using matrix method.

• A. $5, 15, 0$
• B. $5, 15, 10$
• C. $5, -15, 0$
• D. $-5, 15 , 0$

Answer 1

Answer: (A)

$5, 15, 0$

Answer 2

Let the diet contains $x$ units of food $F_{1}, y$ units of food $F_{2}$ and $z$ units of food $F_{3}$.
According to the given condition, we make the following equations
$3x + y + 5z = 30, 2x + 2y + 3z = 40$ and $x + y + 2z = 20$
These algebraic equations can be written in matrix form as $\left[\begin{matrix}3&1&5\\\ 2&2&3\\\ 1&1&2\end{matrix}\right]\left[\begin{matrix}x\\\ y\\\ z\end{matrix}\right]=\left[\begin{matrix}30\\\ 40\\\ 20\end{matrix}\right]$ i.e., $AX = B\quad...\left(i\right)$
Now, $|A|=\left|\begin{matrix}3&1&5\\\ 2&2&3\\\ 1&1&2\end{matrix}\right|=3\left(4-3\right)-1\left(4-3\right)+5\left(2-2\right)=3-1+0=2\ne0$
$\therefore\quad$ There exists a unique solution $X=A^{-1}B$.
$\therefore\quad adj\left(A\right)=\left[\begin{matrix}1&3&-7\\\ -1&1&1\\\ 0&-2&4\end{matrix}\right]$
$\therefore\quad A^{-1}=\frac{adj\,\left(A\right)}{\left|A\right|}=\frac{1}{2}\left[\begin{matrix}1&3&-7\\\ -1&1&1\\\ 0&-2&4\end{matrix}\right]$
$\Rightarrow\quad\left[\begin{matrix}x\\\ y\\\ z\end{matrix}\right]=\frac{1}{2}\left[\begin{matrix}1&3&-7\\\ -1&1&1\\\ 0&-2&4\end{matrix}\right]\left[\begin{matrix}30\\\ 40\\\ 20\end{matrix}\right]$
On solving, we get $x = 5$, $y = 15$ and $z = 0$.