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)

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 givencondition, 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
$\left[\begin{array}{ccc}3& 1& 5\\ 2& 2& 3\\ 1& 1& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}30\\ 40\\ 20\end{array}\right]$
i.e., $AX=B...\left(i\right)$
Now, $|A|=\left|\begin{array}{ccc}3& 1& 5\\ 2& 2& 3\\ 1& 1& 2\end{array}\right|=3\left(4-3\right)-1\left(4-3\right)+5\left(2-2\right)=3-1+0=2\ne 0$
$\therefore$ There exists a unique solution $X=A^{-1}B$.
$\therefore adj\left(A\right)=\left[\begin{array}{ccc}1& 3& -7\\ -1& 1& 1\\ 0& -2& 4\end{array}\right]$
$\therefore A^{-1}=\frac{adj\left(A\right)}{|A|}=\frac{1}{2}\left[\begin{array}{ccc}1& 3& -7\\ -1& 1& 1\\ 0& -2& 4\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{2}\left[\begin{array}{ccc}1& 3& -7\\ -1& 1& 1\\ 0& -2& 4\end{array}\right]\left[\begin{array}{c}30\\ 40\\ 20\end{array}\right]$
On solving, we get $x=5$, $y=15$ and $z=0$.